Minimum Size Subarray Sum

Identifying Problem Isomorphism

“209. Minimum Size Subarray Sum” can be approximately mapped to “713. Subarray Product Less Than K”.

“209. Minimum Size Subarray Sum” asks for the minimum size of a contiguous subarray for which the sum of its elements is greater or equal to a target ’s’, while “713. Subarray Product Less Than K” requires finding the number of contiguous subarrays where the product of all the elements in this subarray is less than ‘k’.

The isomorphism is in the concept of working with contiguous subarrays and evaluating a cumulative property of them (sum in one case, product in the other). Both problems involve finding a subset of the array that meets a certain condition, using a sliding window approach. However, the problems differ in the specifics of the condition (sum vs product, greater than/equal vs less than) and what they ask to return (the minimum length of such a subarray vs the count of such subarrays), making this an approximate rather than exact mapping.

10 Prerequisite LeetCode Problems

Here are 10 problems that cover concepts which is useful for tackling problem 209:

1. 1. Two Sum
2. 26. Remove Duplicates from Sorted Array
3. 27. Remove Element
4. 35. Search Insert Position
5. 53. Maximum Subarray
6. 88. Merge Sorted Array
7. 121. Best Time to Buy and Sell Stock
8. 167. Two Sum II - Input array is sorted
9. 189. Rotate Array
10. 283. Move Zeroes

These cover array manipulation, understanding the properties of sorted arrays, and various problem-solving strategies like two pointers, binary search, etc. which are useful in solving more complex problems.

Problem Boundary

Establishing the boundaries of a problem is a crucial step in problem-solving for several reasons:

1. Understanding Problem Scope: Establishing boundaries helps us understand the extent and limitations of a problem. It sets a clear picture of what inputs to consider and what outputs to expect, and it helps us eliminate unnecessary considerations that are outside the boundaries.

2. Defining Input/Output Space: The boundaries of a problem define the possible range of input and output. This can help us select appropriate data structures and algorithms for our solution. For instance, knowing that an array input will never exceed a certain size might allow us to use an algorithm that is less efficient for large inputs but simpler or more convenient for smaller ones.

3. Error Handling: By knowing the boundaries, we can anticipate and handle edge cases and potential errors in the input data. This can help improve the robustness and reliability of our solution.

4. Performance Optimization: In some cases, understanding the boundaries can also lead to performance optimizations. For instance, if we know that an array is of a fixed size, we might be able to parallelize or otherwise optimize operations on that array.

5. Testing: When boundaries are established, we can create better test cases that include edge cases at the boundaries. These test cases can help ensure that our solution works correctly in all scenarios within the established boundaries.

In the context of the “Minimum Size Subarray Sum” problem, knowing that the input array size is between 1 and 105, and each element is a positive integer between 1 and 104, helps us select an approach that can handle this range of input effectively. It also helps us to construct appropriate test cases to validate our solution.

Establishing boundaries in a problem is crucial to correctly understanding the problem’s scope and requirements. Boundaries in a problem can be defined by the constraints and requirements of the problem. In the case of the “Minimum Size Subarray Sum” problem, the boundaries can be established as follows:

1. Input Boundary: The input to this problem is an array nums of positive integers and a positive integer target. The constraints on these inputs establish our boundaries:

   • The length of nums must be between 1 and 105, inclusive. This means that an array with less than 1 element or more than 105 elements is out of bounds.
   • Each element of nums is a positive integer that must be between 1 and 104, inclusive. This means that a negative number, zero, or any number greater than 104 is out of bounds for individual array elements.
   • target is a positive integer that should be less than or equal to 109. This means any negative number, zero, or any number greater than 109 is out of bounds for target.

2. Functional Boundary: This problem requires us to find the minimal length of a subarray from nums whose sum is greater than or equal to target. This establishes the functional boundaries:

   • The problem requires to return the length of the smallest such subarray. Returning the actual subarray or its sum is out of bounds.
   • If there is no such subarray that satisfies the condition, the function should return 0. Returning anything other than 0 in this case, like -1 or any error message, is out of bounds.

3. Output Boundary: The output of this problem is an integer representing the minimal length of a subarray from nums whose sum is greater than or equal to target.

   • Since the length of a subarray can’t be negative, the output will always be a non-negative integer.
   • The maximum possible length of the subarray is the length of nums itself, which is at most 105 according to the input constraints. Thus, any output greater than 105 would be out of bounds.

These are the boundaries of this problem, and the solution needs to respect them to solve the problem correctly and efficiently.

Also, understanding the problem’s boundaries allows us to design a solution that is just right - not over-engineered, not under-engineered. It ensures that we provide a solution that fits the problem’s needs exactly, without wasting time or resources on unnecessary functionality.

Over-engineering a solution might result in a more complex, less understandable, and harder to maintain codebase. On the other hand, under-engineering might result in a solution that doesn’t fully solve the problem or doesn’t handle all the edge cases correctly.

So, comprehending the boundaries keeps us on track, and we end up devising a solution that perfectly aligns with the problem requirements.

In the context of our “Minimum Size Subarray Sum” problem, understanding the problem boundaries makes sure we consider the possible input space and construct a solution that works for all cases within those boundaries, handling edge cases and achieving the desired output efficiently. It helps us decide on the right algorithmic approach (sliding window in this case) which is efficient and effective to solve the problem.

Problem Classification

Problem Domain: This problem belongs to the domain of “Arrays” and “Sliding Window”.

What Components:

1. An array of positive integers (nums).
2. A positive integer target.
3. The task is to find the minimal length of a contiguous subarray whose sum is greater than or equal to target.
4. If no such subarray exists, we return 0.
5. Constraints are given for target, the length of the array, and the elements of the array.

Categorization:

1. Data Structures: The problem involves working with arrays, which is a fundamental data structure. We’re dealing with an array of integers.

2. Algorithms/Techniques: The problem hints at two potential solutions: one with a time complexity of O(n) and another with a time complexity of O(n log(n)). This suggests that we might use strategies like “Two Pointers” or “Sliding Window” (for the O(n) solution) and potentially “Binary Search” in some way (for the O(n log(n)) solution).

3. Search: The problem involves finding a specific subset of array elements that satisfy a certain condition, which is essentially a search operation.

4. Optimization Problem: This problem asks for the minimum length of a subarray that satisfies a given condition, which is a type of optimization.

In conclusion, this is an optimization problem involving array manipulation, which can potentially be solved using the two pointers or sliding window technique for the O(n) solution, and perhaps binary search for the O(n log(n)) solution.

Distilling the Problem to Its Core Elements

1. Fundamental Concept: The fundamental concept this problem is based upon is the “Sliding Window” concept in the realm of Array or String manipulation problems. In a “Sliding Window” problem, you have a dataset (here an array), and you are trying to find or calculate something within a contiguous subset of the data (a “window”). The window “slides” through the data until you’ve covered all of it.

2. Simplest Description: Imagine you have a list of positive numbers and you want to find the shortest sequence in that list that adds up to a certain target value or more. If no such sequence exists, then you simply return 0.

3. Core Problem and Simplified Statement: The core problem is finding the shortest sequence of numbers in the list that adds up to at least a certain target number. The simplified problem statement could be: “Given a list of positive numbers, find the smallest number of consecutive numbers that, when added together, equal or exceed a specified target number.”

4. Key Components:

   • An input array of positive integers
   • A target positive integer
   • The goal of finding the minimal length of a subarray (continuous part of the array) such that the sum of its elements is greater than or equal to the target
   • Return the length of that subarray. If no such subarray exists, return 0

5. Minimal Set of Operations:

   • Initialize two pointers at the start of the array, along with a minimum length variable (initialized to infinity or some large value) and a sum variable (initialized to 0).
   • Start a loop to “slide” the window through the array:
     • Add the value at the end pointer to the sum.
     • While the sum is greater than or equal to the target, update the minimum length variable if the current window size (end pointer - start pointer + 1) is less than the minimum length found so far. Then, subtract the value at the start pointer from the sum and move the start pointer forward.
   • Continue the process until the end pointer has gone through the entire array.
   • If the minimum length variable is still at its initialized value (meaning no valid subarray was found), return 0. Otherwise, return the minimum length.

Visual Model of the Problem

Visualization for problems like this one is often best achieved with a diagram, often using arrows to indicate the “sliding window” in the array. Let’s visualize the problem using the first example from the problem statement:

target = 7, nums = [2,3,1,2,4,3]

You can think of the array as a sequence of boxes. We’ll denote the start and end of the window with arrows:

   2  3  1  2  4  3
 ↑
start/end

Initially, start and end both point at the first element. The sum of elements within the window is 2, which is less than the target (7). So, we expand the window to the right:

   2  3  1  2  4  3
 ↑     ↑
start end

The sum of the elements within the window is now 5 (2 + 3), which is still less than the target. We keep expanding:

   2  3  1  2  4  3
 ↑        ↑
start    end

The sum now is 6 (2 + 3 + 1), which is still less than 7. We expand once more:

   2  3  1  2  4  3
 ↑           ↑
start       end

The sum is 8 (2 + 3 + 1 + 2), which is greater than the target (7). Now, we try to minimize the window while maintaining a sum greater or equal to the target. So, we move the start of the window to the right:

   2  3  1  2  4  3
      ↑     ↑
    start  end

The sum is 6 (3 + 1 + 2), which is less than the target, so we need to expand the end of the window to the right:

   2  3  1  2  4  3
      ↑        ↑
    start     end

Now the sum is 10 (3 + 1 + 2 + 4), which is greater than the target. So, we move the start of the window to the right:

   2  3  1  2  4  3
         ↑  ↑
       start end

The sum is 7 (1 + 2 + 4), which is equal to the target. We move the start of the window to the right:

   2  3  1  2  4  3
            ↑  ↑
          start end

The sum is 6 (2 + 4), which is less than the target, so we expand the end of the window to the right:

   2  3  1  2  4  3
            ↑     ↑
          start  end

The sum is 9 (2 + 4 + 3), which is greater than the target. We try to minimize the window by moving the start of the window to the right:

   2  3  1  2  4  3
               ↑  ↑
             start end

The sum is 7 (4 + 3), which is equal to the target. Now, the end of the window can’t be expanded any further

The sum is 7 (4 + 3), which is equal to the target. Now, the end of the window can’t be expanded any further because we’ve reached the end of the array, so we stop.

Now, among all the windows whose sum was greater or equal to the target, we find the shortest one. The shortest window that met the conditions was [4, 3], which had a length of 2, so that’s our answer. Here’s how that window looked:

   2  3  1  2  4  3
               ↑  ↑
             start end

The sum of the numbers within this window is equal to our target, and the length of the window is the minimum possible for windows meeting these conditions.

This visualization helps us understand the “sliding window” concept and how it applies to this problem. We essentially expand the window until we’ve met the conditions, then try to minimize the window while still meeting the conditions. If we can’t meet the conditions with a minimized window, we go back to expanding the window.

Problem Restatement

The problem is asking us to find the shortest sequence of continuous numbers within a larger list of numbers, where the sum of the numbers in this sequence is at least as large as a given target number.

More specifically, we are given a list of positive numbers and a positive target number. We need to identify the smallest possible subsequence within the list where the sum of its numbers is equal to or larger than the target number. If there’s no such subsequence, we should return 0.

This subsequence should be a “contiguous” sequence, meaning the numbers appear next to each other in the original list.

In terms of constraints, the target number will be between 1 and 10^9. The list will have between 1 and 10^5 elements, each between 1 and 10^4.

We’re also asked to consider a follow-up: if we can solve the problem in linear time (O(n)), can we also find a solution that would have a time complexity of O(n log(n))?

Abstract Representation of the Problem

Let’s create an abstract representation of the problem.

We’re given a sequence of positive elements (which we’ll call S), and a target number (T). A subsequence of S is defined as a contiguous set of elements in S.

The problem requires us to find the minimal length of any subsequence in S such that the sum of elements in this subsequence is greater than or equal to T. If no such subsequence exists, we should return 0.

Here, the length of a subsequence is defined as the number of elements in it, and the sum of a subsequence is the sum of all its elements.

In the context of this problem, the sequence S is a list of positive integers, and the target T is a positive integer. However, the problem could theoretically apply to any ordered set of positive elements, and any positive target.

Our objective is to design an algorithm that can efficiently find the minimal length of such a subsequence.

In addition, a secondary goal is to explore alternative approaches to the problem that trade off different time complexities - specifically, to find an algorithm with time complexity O(n log(n)) if we have already found one with O(n).

Terminology

Here are some key terms related to this problem:

1. Array: An array is a data structure that contains a group of elements. The elements can be identified by their array index - an integer which represents the position of an element in the array. In this problem, the array is referred to as “nums”.

2. Subarray: A subarray is a contiguous part of an array. The subarray must be located in consecutive positions within the original array. For example, in the array [1, 2, 3, 4, 5], [2, 3, 4] is a subarray. In this problem, we’re interested in finding a subarray that meets certain criteria.

3. Target: In the context of this problem, the target is a specific number that the sum of elements in the subarray should be greater than or equal to.

4. Time Complexity: This refers to the computational complexity that describes the amount of computational time taken by an algorithm to run, as a function of the size of the input to the program. It’s commonly expressed using Big O notation. In this problem, there’s a goal to develop a solution with O(n) time complexity, and another with O(n log(n)) time complexity.

5. Binary Search: Binary search is a search algorithm that finds the position of a target value within a sorted array by repeatedly dividing the search interval in half. It’s relevant here because one of the goals of the problem is to find a solution with a time complexity of O(n log(n)), and binary search is one common way to achieve that time complexity.

6. Sliding Window: This is a technique for efficiently processing arrays or lists using two pointers. At any point, the difference between the two pointers is the “size of the window”, and we can slide the window through the list by moving the pointers. This technique may be useful in this problem to find the subarray with the minimal length.

Problem Simplification and Explanation

Imagine you are standing in a hallway filled with lockers, and each locker contains a certain amount of apples. You want to gather at least a specific number of apples (say, 7 apples for instance), and to do this, you can start at any locker and keep moving to the next ones, collecting all the apples in each locker you pass by. However, you want to reach your apple goal in the smallest number of lockers. So you might skip some lockers at the beginning if you know that you’ll reach your goal quicker by starting somewhere else.

The lockers are like the array of numbers, each number representing the quantity of apples in each locker. The target number of apples is the ’target’ in this problem. A ‘subarray’ is like a stretch of lockers you’ve chosen to gather apples from, starting from one locker and ending at another locker.

Your task is to find the shortest stretch of lockers that will give you at least the target number of apples. In more technical terms, you’re trying to find the minimum length subarray whose sum is greater than or equal to the target.

In this scenario, there might be many ways to gather your target number of apples. The challenge is finding the most efficient one - the one that involves opening the fewest lockers. This requires you to take a strategic approach to choosing your starting locker, which is where the ‘sliding window’ concept might come in handy.

This problem involves elements of array manipulation, understanding subarrays, binary search, and the sliding window technique. It’s a complex problem because it requires not just understanding the individual concepts, but also understanding how to use them in combination to arrive at the most efficient solution.

Constraints

Let’s break down the problem and its constraints to find some characteristics that can be leveraged for a solution.

1. Positive Numbers: The problem statement specifies that the array consists of positive integers. This can be exploited as the sum of any subarray will not decrease if we move from left to right.

2. Subarray Constraint: The problem is asking for a contiguous subarray, which makes it ideal for using the sliding window technique, a powerful method for array traversal problems. In this case, the window size varies as we need to find a minimum size subarray. Thus, the window can expand and shrink based on the sum of elements.

3. Target Constraint: The target is a positive integer and we are looking for a subarray sum that is greater than or equal to this target. This means that once we reach a sum equal to or larger than the target, we can start reducing the window size from the left to find a possible smaller subarray with the same condition.

4. Element Magnitude: The elements of the array are at most 10^4. This might not immediately present an obvious advantage, but it sets an upper bound to the values we are dealing with. If the target is considerably large compared to this upper bound, we know that we will need at least target/10^4 elements in our subarray.

5. Array Size: The array size can go up to 10^5. This indicates that a solution with a time complexity of O(n^2) or worse would probably not be efficient enough. This again points towards an O(n) or O(n log n) solution, which are typical time complexities when using a sliding window or binary search approach.

Using these characteristics, we can come up with an efficient solution using a sliding window or binary search approach that operates within the given constraints.

From analyzing the constraints, here are the key insights we get:

1. Use of Sliding Window: The problem involves finding a contiguous subarray, and all the numbers are positive. This is a strong indicator that a sliding window approach can be useful. Once we have a subarray sum that is equal to or greater than the target, we can try to reduce the size of this subarray by incrementing the left pointer of our window, potentially discovering a smaller satisfying subarray.

2. Efficiency Required: The maximum size of the input array can be quite large (up to 10^5). This means that the solution needs to be efficient. A brute force approach with a time complexity of O(n^2) or worse will likely not be feasible. The problem itself hints at an O(n) or O(n log(n)) solution, confirming the need for an efficient algorithm.

3. Positive Numbers: Since the array consists of only positive numbers, the sum of elements within a window will not decrease unless the window shrinks. This characteristic ensures that once the sum of a window becomes less than the target (after shrinking), we don’t need to consider this window (and any smaller window starting at the same position) anymore.

4. Bound on Individual Elements: The elements of the array are at most 10^4. This can help set expectations about the minimum number of elements we might need to reach the target sum. If the target is much larger than this value, we know that we will need at least target/10^4 elements.

These insights should guide the approach towards solving the problem, and help in deciding which data structures and algorithms to use.

Case Analysis

The problem statement specifies that the given array is an array of positive integers, and the target is also a positive integer. This implies that the array cannot be empty or of length 0.

An empty array or an array of length 0 has no elements in it, hence it would be impossible to form a subarray with a sum equal to or greater than the target. The only valid output for an empty array would be 0, indicating that no such subarray exists.

The purpose of the problem is to find the minimum size subarray that sums to a target from a given array of positive integers, so an empty array would not contribute to understanding or solving the problem.

Furthermore, test cases are used to validate the logic of the solution and an empty array does not provide useful information in this case. In general, it is beneficial to use test cases that cover a broad range of possibilities within the problem constraints to ensure the solution is comprehensive and handles all edge cases.

So, in this problem context, arrays of length 0 were not chosen for the test cases as they don’t provide any meaningful information and they don’t fall within the problem’s constraints.

Let’s go through some test cases.

1. Case: Minimal Array Size

   • Input: nums = [1], target = 1
   • Output: 1
   • Explanation: Here the array consists of a single element which is equal to the target. Hence, the minimal length of the subarray would be 1.

2. Case: Multiple Possible Subarrays

   • Input: nums = [2,3,1,2,4,3], target = 7
   • Output: 2
   • Explanation: Multiple subarrays in this case have a sum that is greater than or equal to the target, including [2,3,2], [3,4], [2,4,3], and [4,3]. The shortest of these subarrays is [4,3] with a length of 2.

3. Case: No Possible Subarray

   • Input: nums = [1,1,1,1,1,1], target = 7
   • Output: 0
   • Explanation: No subarray can sum to the target, hence the answer is 0.

4. Case: All Elements Are The Same

   • Input: nums = [3,3,3,3,3], target = 6
   • Output: 2
   • Explanation: The minimal length subarray in this case is [3,3].

5. Case: Single Element Larger Than Target

   • Input: nums = [1,2,3,4,5,10], target = 6
   • Output: 1
   • Explanation: In this case, we have a single element (10) that is larger than the target. Hence, the smallest subarray is just that single element.

6. Case: Array Is in Descending Order

   • Input: nums = [10, 9, 8, 7, 6], target = 15
   • Output: 2
   • Explanation: Even though the array is in descending order, we still find the smallest subarray [10, 9] that sums to a value greater than or equal to target.

Through these test cases, you can see that we have covered a range of scenarios. These scenarios account for different array lengths, different array values, and different target values. By creating and understanding these test cases, you get a better grasp of the problem at hand.

Identification of Applicable Theoretical Concepts

This problem falls under the domain of array manipulation and the use of sliding window technique.

1. Sliding Window Technique: The Sliding Window technique is an algorithmic paradigm that manages a subarray, or “window”, of an array or list, and cleverly updates the window by removing an element from the front and adding an element to the end. This can be particularly useful in this problem, where we’re looking for a subarray that meets a certain criteria.

2. Prefix Sums: Another mathematical concept that could be useful here is the idea of prefix sums, which is a method used to pre-calculate the sum of elements of an array up to a certain index. However, it’s not very applicable in this case as the elements can be in any order and the array is not sorted.

3. Binary Search: In some cases, binary search can be used to solve this problem in O(n log n). This is mentioned as a follow-up in the problem statement, but the implementation is more complex than the sliding window approach.

4. Cumulative Sum: In a broader sense, this problem could be seen as a special case of a more general concept of cumulative sums, where we are interested in the sum of all values in a range within an array.

By leveraging these concepts, we can break the problem down into more manageable chunks and simplify the process of finding the shortest subarray with a sum at least as large as the target value. These algorithmic concepts allow us to traverse and manipulate the array in a way that is more efficient than a simple brute force approach, which could involve checking every possible subarray and would be far less efficient.

Identify Invariant

In computer science, an invariant is a condition that can be relied upon to be true during execution of a program or during some portion of it. It is a logical assertion that is always held to be true during a certain phase of execution.

In the context of the minimum size subarray problem, the invariant is the condition that the sum of elements within the current sliding window is always less than the target sum when the window is to be expanded (right pointer is to be moved right), and it’s always greater than or equal to the target sum when the window is to be shrunk (left pointer is to be moved right).

So, during the process of solving this problem:

1. If the sum of the current window elements is less than the target, we expand the window to the right by moving the right pointer, hoping to capture more elements in the window so that the sum might become equal to or larger than the target.

2. If the sum of the current window elements is greater than or equal to the target, we shrink the window from the left by moving the left pointer to the right, trying to potentially minimize the size of the subarray meeting the target sum condition.

You can identify this invariant by understanding the problem and figuring out what condition must hold true throughout the process to achieve the desired goal. In this case, it was the relationship between the sum of elements within the sliding window and the target sum.

Identify Loop Invariant

A loop invariant is a condition or set of conditions that remain unchanged throughout the execution of a loop. In other words, the condition is initially true before the loop begins and remains true after each iteration of the loop.

For the minimum size subarray problem, the loop invariant can be thought of as:

1. “The sum of elements from the left pointer (inclusive) to the right pointer (inclusive) in the array is always greater than or equal to the target sum at the end of each iteration when we’re trying to shrink the window.”

2. “The sum of elements from the left pointer (inclusive) to the right pointer (inclusive) in the array is always less than the target sum at the end of each iteration when we’re trying to expand the window.”

These conditions ensure that at each iteration, we are keeping track of the smallest size of the subarray whose sum is greater than or equal to the target, or we’re trying to expand our window to find such a subarray. In essence, the loop invariant is a formal statement about the relationship between the variables involved in the loop that is maintained with each iteration, thereby helping to eventually solve the problem.

Problem Breakdown and Solution Methodology

The approach I would suggest for solving this problem is known as the “Sliding Window” technique. Here’s how we can go about it:

1. Initialization: We first initialize two pointers left and right both at the starting index of the array. This pair of pointers represents our current “window”. We also initialize a variable min_length to store the minimum length of the subarray found and current_sum to store the sum of the elements within the current window.

2. Expanding the Window: We start expanding our window towards right i.e., increment the right pointer and add the nums[right] to the current_sum.

3. Shrinking the Window: Whenever the current_sum becomes greater than or equal to target, we start shrinking our window from the left i.e., increment the left pointer and subtract nums[left] from current_sum. During the shrinking process, we keep updating our min_length variable.

4. Iterating: We repeat steps 2 and 3, expanding and shrinking our window until the right pointer has traversed the entire array.

5. Checking and Returning: If we never find a subarray that sums to at least target, that means no such subarray exists, so we return 0. Otherwise, we return min_length.

Let’s take an example to visualize this approach: Assume nums = [2, 3, 1, 2, 4, 3] and target = 7.

• Initially, left = right = 0, min_length = infinity, current_sum = 0.

• We start expanding our window towards right until current_sum >= target.

• After expanding, left = 0, right = 4, current_sum = 12, min_length = infinity.

• Now we start shrinking our window from left until current_sum < target.

• After shrinking, left = 3, right = 4, current_sum = 6, min_length = 2.

• Now again start expanding the window until right pointer has traversed the entire array.

• Finally, min_length = 2 which is the answer.

This process, also known as the “Sliding Window” technique, helps us solve this problem in a time-efficient manner. For any change in the problem’s parameters, this approach can easily be modified by adjusting how the window expands or shrinks. For instance, if the problem were to find a subarray sum that is exactly equal to target, we would only start shrinking the window when current_sum == target. This makes the approach flexible and versatile.

Inference of Problem-Solving Approach from the Problem Statement

The Sliding Window approach is typically a good fit for problems that ask you to find some kind of optimal subarray within an array, particularly when the array is ordered or the problem involves sum or product of the array elements. Here are the signs in the problem statement that led me to consider the Sliding Window approach:

1. Subarray: The problem asks for the minimal length of a contiguous subarray with a sum that is greater than or equal to a target number. This is a key indicator that we might be able to use a sliding window to keep track of a subarray.

2. Optimization: The problem isn’t just asking for any subarray with a sum greater than or equal to the target; it’s asking for the subarray with the smallest length. This suggests we need a way to compare multiple potential solutions. With the sliding window approach, we can effectively compare different subarrays by sliding the window and updating a variable that keeps track of our best solution so far.

3. Sum of elements: The condition for the subarray is based on the sum of its elements. This makes the sliding window approach a natural fit, since we can efficiently compute the sum of the elements in the current window by adding the new element when we expand the window to the right and subtracting the first element when we shrink the window from the left. This avoids having to recompute the sum of the whole subarray from scratch each time we change the window, which would be much less efficient.

4. Non-negative numbers: The problem specifies that all the numbers in the array are positive. This is important because if the numbers could be negative, we could not guarantee that adding a new number or removing an existing number would increase or decrease the sum, respectively. But with all positive numbers, we know that adding a new number will always increase the sum and removing a number will always decrease the sum. This property is essential for the sliding window approach to work.

Remember, these are just indicators, and they suggest that sliding window could be a good approach to try. But just because a problem has these characteristics doesn’t mean sliding window is necessarily the best or only solution. Conversely, even if a problem doesn’t obviously have these characteristics, that doesn’t mean you couldn’t use a sliding window approach; you might just need to be more creative in how you define your window or what you track within it.

Stepwise Refinement

Here’s a step-by-step breakdown of how we can solve this problem using a Sliding Window approach:

1. Initialize Variables:

   • Start with two pointers, left and right, both set to the beginning of the array.
   • Initialize a variable, windowSum, to keep track of the current sum within the window (subarray from left to right). Initially, this will be 0.
   • Initialize a variable, minLength, to keep track of the minimum length of subarray we’ve found that meets the problem’s condition. Initially, set it to INT_MAX or some large value as we haven’t found any valid subarray yet.

2. Expand the Window:

   • Iterate over the array by incrementing the right pointer and adding the current number pointed by right to windowSum.
   • Continue expanding the window by moving the right pointer to the right and adding the new number to windowSum until windowSum is greater than or equal to the target.

3. Shrink the Window:

   • When windowSum is greater than or equal to target, try to shrink the window from the left by moving the left pointer to the right and subtracting the number pointed by left from windowSum.
   • Keep shrinking the window until windowSum is less than the target again.

4. Update the Minimum Length:

   • After each step of shrinking the window, check if the length of the current window (subarray) is less than minLength.
   • If it is, update minLength to be the length of the current window.

5. Repeat the Process:

   • Repeat steps 2-4 until the right pointer has moved past the end of the array. At this point, we’ve checked every possible subarray.

6. Return the Result:

   • After the loop, minLength will contain the length of the smallest subarray whose sum is greater than or equal to the target.
   • If minLength is still INT_MAX (or the initial large value), this means we didn’t find any subarray that meets the condition, so return 0. Otherwise, return minLength.

These steps represent a high-level view of the Sliding Window approach. This approach is efficient because we only go over the array once with the right pointer, so the time complexity is O(n).

Let’s break down the high-level solution into more granular, actionable steps:

1. Initialization:

   • Define left and right pointers, both initially set to 0.
   • Define a windowSum to 0 which will store the sum of the numbers within the window (subarray from left to right).
   • Define a minLength to a large number (e.g., INT_MAX). This variable will hold the length of the smallest subarray that meets the condition.

2. Iterate over the array:

   • Run a loop which continues until the right pointer has moved past the end of the array.
   • In each iteration, add the right-th element in the array to windowSum and then increment right.

3. Check and shrink the window:

   • Within the main loop, start an inner loop that continues as long as windowSum is greater than or equal to the target.
   • In each iteration of the inner loop, calculate the length of the current window (which is right - left) and compare it with minLength. If it’s smaller, update minLength.
   • Then, subtract the left-th element in the array from windowSum and increment left. This shrinks the window from the left.

4. Edge Case Handling:

   • After the main loop, check if minLength is still the large number (e.g., INT_MAX). If it is, this means there was no valid subarray, so return 0.

5. Return Result:

   • If minLength is not the initial large number, return minLength as it represents the length of the smallest subarray whose sum is greater than or equal to the target.

This approach efficiently explores all possible subarrays and keeps track of the smallest valid one, resulting in a time complexity of O(n).

In this problem, it’s a bit difficult to isolate independent parts because the solution is based on the “sliding window” approach which inherently involves continuous interaction between different parts of the input array. The calculations for the window sum, the window size, and the adjustments of the window boundaries are all intertwined and dependent on each other.

However, if we want to break down the problem, we can consider the following aspects:

1. Finding the Sum of the Window: This part involves managing the ‘windowSum’ variable, which keeps track of the sum of the elements currently in the window.

2. Adjusting the Window Size: This part involves manipulating the ’left’ and ‘right’ pointers which define the boundaries of the window.

3. Comparing and Updating Minimum Length: This part involves comparing the current window size to the ‘minLength’, and updating ‘minLength’ if the current window size is smaller and its sum is at least the target.

4. Checking for Valid Subarray: After the main loop, checking whether a valid subarray exists (i.e., if ‘minLength’ has been updated from its initial value) can be considered an independent part.

Remember that although we can conceptually isolate these parts, in the actual solution implementation, they are interconnected and must be managed together within the main loop.

There are repeatable patterns in the sliding window approach for this problem:

1. Sliding the Window: The primary repeatable pattern is the process of moving the ‘right’ pointer forward until the sum of the elements within the window is at least the target. This is repeated for each new window position until we reach the end of the array.

2. Shrinking the Window: Once the sum within the window is at least the target, we try to shrink the window from the left to minimize its size while maintaining the condition that the sum is at least the target. This pattern is repeated whenever the sum within the window is greater than or equal to the target.

3. Updating the Minimum Length: After every window adjustment (either sliding or shrinking), we check if the current window size is smaller than the current minimum length, and update the minimum length if it is. This pattern is repeated for each window adjustment.

These patterns are repeatedly applied throughout the entire array, constituting the main logic of the sliding window approach to this problem.

Solution Approach and Analysis

Let’s look at this problem as if we are trying to identify a segment of a road (the array) where the sum of the segment’s length (the sum of the elements in the subarray) is greater than or equal to a specific value (the target).

The approach I would take is known as the ‘sliding window’ technique. This approach is like using a sliding scale or moving spotlight over our road, expanding and contracting our spotlight to find the smallest possible segment that still meets our criteria.

Here’s a step-by-step breakdown:

1. Initialization: Start with the left and right pointers both at the starting point of the array. This is like shining our spotlight at the very beginning of the road. The sum is also initialized to zero, as we haven’t yet added any segment lengths.

2. Find the Initial Valid Window: Move the right pointer to the right, adding the element at the right pointer to the sum at each step, until the sum becomes greater than or equal to the target. This is equivalent to expanding our spotlight to the right until the total length of the road covered is at least as long as our target length.

3. Contract the Window: Once we’ve found a valid window, we’ll try to reduce its size from the left to find the minimum length. We move the left pointer to the right, subtracting the element at the left pointer from the sum at each step, until the sum becomes less than the target. This is like contracting our spotlight from the left.

4. Keep Expanding and Contracting: Repeat steps 2 and 3, moving the right pointer when the sum is less than the target, and moving the left pointer when the sum is greater than or equal to the target, until the right pointer reaches the end of the array.

5. Find the Minimum Length: Keep track of the minimum length of the window (i.e., the distance between the right and left pointers) at each step. This is the minimum segment of the road that meets our length criteria.

6. Handle No Solution: If no valid window is found, return 0.

Let’s consider an example. Let nums = [2, 3, 1, 2, 4, 3], target = 7.

• We start with a spotlight on the first element (2). Since it’s not >= 7, we expand our spotlight to the right until the sum is >= 7, giving us [2,3,1,2] (sum = 8). The length of this segment is 4.
• Next, we try to contract from the left. Moving the left pointer one step to the right gives us [3,1,2] (sum = 6), which is less than 7, so we stop contracting.
• We then expand again to the right, moving the right pointer one step to get [3,1,2,4] (sum = 10).
• We contract from the left to get [1,2,4] (sum = 7) and then [2,4] (sum = 6), at which point we stop contracting since the sum is less than 7.
• We expand one more time to the right to get [2,4,3] (sum = 9), and then contract from the left to get [4,3] (sum = 7).
• Since we can’t contract any further without going below the target, we stop here. The minimum length of a subarray with a sum >= 7 is 2.

Note that if we increase the target, the minimum length would increase. If we decrease the target, the minimum length could potentially decrease. And if the

array elements were to change in such a way that they are all smaller, then the length would increase (since we’d need more elements to reach the target), whereas if they are larger, then the length would decrease (since we’d need fewer elements to reach the same target). Hence, the specifics of the array and the target value will directly influence the solution.

The beauty of this approach is its efficiency. Instead of checking every possible subarray, which would take quadratic time, we are using a linear process - each element is visited at most twice, once by the right pointer and once by the left. So, the time complexity is O(n), where n is the length of the array.

Thought Process

This problem presents us with an array of positive integers and a target number. We’re asked to find the shortest subarray whose sum is greater than or equal to the target number.

Let’s break down the problem:

1. Identify the Key Terms: The problem is asking for a “minimal length” subarray, which suggests we’re looking for the shortest possible subarray that meets the requirements. We’re also given that the subarray’s sum needs to be “greater than or equal to target.” This means we need a method to calculate subarray sums and compare them with the target.

2. Make Observations: The sum of the subarray is a continuously increasing quantity as we add more elements from the array. The fact that we’re looking for a “greater than or equal” sum suggests the possibility of overshooting the target with a large sum, which could lead to longer than necessary subarrays. Hence, the need for minimizing the length of the subarray.

3. Generate Insight: Since we’re dealing with an array (a linear data structure), one of the most common methods for traversing and solving such problems is a sliding window approach. This approach is also suitable because we need to maintain a contiguous subarray, which is exactly what a “window” over an array represents. As we slide this window, we can increase or decrease its size, effectively tracking different subarrays and their sums.

4. Devise a Plan: Implement a sliding window technique. Start with a window at the start of the array and keep expanding it until the sum of elements within the window is greater than or equal to the target. At that point, start shrinking the window from the left until the sum is less than the target. Repeat these steps, tracking the minimum length of the window that still achieves the sum greater than or equal to the target.

Here is the Python code implementing the above insights and approach:

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class Solution:
    def minSubArrayLen(self, target: int, nums: List[int]) -> int:
        min_len = float('inf')  # Initialize min length to infinity
        left = 0
        total = 0
        
        for right in range(len(nums)):  # Expand the window to the right
            total += nums[right]
            
            # Contract the window from the left
            while total >= target:
                min_len = min(min_len, right - left + 1)
                total -= nums[left]
                left += 1
                
        # If min_len is still infinity, no valid subarray was found
        return min_len if min_len != float('inf') else 0

This function returns the minimum length of a subarray with sum greater than or equal to the target. If no such subarray exists, it returns 0. The time complexity is O(n), and the space complexity is O(1), making it an efficient solution.

Establishing Preconditions and Postconditions

1. Parameters:

   • The method takes two parameters: target and nums.
   • target is an integer and nums is a list of integers.
   • target represents the target sum that we want our subarray’s sum to meet or exceed. nums is the array of positive integers in which we’re trying to find the subarray.

2. Preconditions:

   • Before this method is called, nums must be a list of positive integers and target must be a positive integer.
   • nums cannot be an empty list. The length of nums must be between 1 and 105 (inclusive), and the integers in nums should be between 1 and 104 (inclusive). The target should be a positive integer less than or equal to 109.
   • There’s no specific state that the program must be in.

3. Method Functionality:

   • This method is expected to find the minimal length of a contiguous subarray of nums whose sum is greater than or equal to target. If no such subarray exists, it returns 0.
   • It processes the input array nums and interacts with the target to find and return the length of the minimal qualifying subarray.

4. Postconditions:

   • After the method has been called and returned, we have determined the length of the shortest subarray in nums that has a sum of at least target.
   • The return value represents the minimum length of the subarray. If it is 0, it indicates that no such subarray exists.
   • The method doesn’t have any side effects; it does not change the input parameters or any other state.

5. Error Handling:

   • The method assumes that preconditions are met, i.e., the inputs are in the correct format and range. If they are not, the behavior is undefined.
   • It does not explicitly throw an exception or return a special value if the preconditions are not met. If nums is an empty list, for instance, the method would still return 0, which could be misinterpreted as “no qualifying subarray exists” when it’s actually an input error. So, proper input validation should be done before calling this function.

Problem Decomposition

1. Problem Understanding:

   • We have an array of positive integers and a target value. We want to find the smallest subarray such that the sum of its elements is greater than or equal to the target. If no such subarray exists, we return 0.

2. Initial Breakdown:

   • We need to iterate through the array, maintaining a running sum and tracking the length of the subarray.

3. Subproblem Refinement:

   • Within the iteration, we need to handle two cases: when the running sum is less than the target and when the running sum is greater than or equal to the target.
   • If the sum is less than the target, we add the next element to the sum.
   • If the sum is greater than or equal to the target, we subtract the first element of the current subarray from the sum and move the start of the subarray forward.

4. Task Identification:

   • The key tasks that are repeated are: calculating the running sum, comparing it with the target, and adjusting the start and end points of the current subarray.

5. Task Abstraction:

   • These tasks are already quite abstract and make sense in the context of the problem. The running sum is a common concept, and adjusting the start and end points of the subarray is specific to this problem but also quite straightforward.

6. Method Naming:

   • We could define a method minimum_size_subarray(nums, target) to solve the entire problem. Inside this method, the iteration could be in a loop titled find_minimum_subarray(), making its purpose clear.

7. Subproblem Interactions:

   • These tasks interact closely with each other and need to be performed in the order outlined above. The comparison with the target depends on the calculation of the running sum, and the adjustment of the subarray depends on the result of the comparison.

Based on this analysis, we can now create a Python function that implements our solution:

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def min_subarray_len(target, nums):
    min_len = float('inf') 
    left = 0 
    value_sum = 0 

    for right in range(len(nums)): 
        value_sum += nums[right] 

        while value_sum >= target: 
            min_len = min(min_len, right-left+1) 
            value_sum -= nums[left] 
            left += 1 

    return min_len if min_len != float('inf') else 0

This function implements the sliding window approach we described earlier. It iterates through the nums list, maintaining a running sum (value_sum) and continuously adjusting the start (left) and end (right) of the current subarray to find the minimum length subarray that meets the problem requirements.

From Brute Force to Optimal Solution

Let’s start with the brute force solution.

1. Brute Force Solution:

The brute force solution to this problem is to generate all possible subarrays of the input array, calculate the sum of each one, and check if it is greater than or equal to the target. We keep track of the length of the shortest subarray whose sum meets the requirement.

In Python, it would look something like this:

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def min_subarray_len(target, nums):
    min_len = float('inf') 
    
    for i in range(len(nums)):
        for j in range(i, len(nums)):
            subarray_sum = sum(nums[i:j+1])
            if subarray_sum >= target:
                min_len = min(min_len, j-j+1)
    
    return min_len if min_len != float('inf') else 0

The problem with this approach is its inefficiency. In the worst-case scenario, we are generating and summing O(n^2) subarrays (where n is the length of the input array), which results in a time complexity of O(n^2). This can be extremely slow for large input arrays.

2. Optimized Solution:

A more efficient way to solve this problem is to use the “sliding window” technique. The idea is to maintain a running sum of a “window” or subarray of elements. We start with a window containing the first element of the array. Then, we expand the window to the right (by adding the next element to the sum) until the sum is greater than or equal to the target. Once the sum is large enough, we try to shrink the window from the left (by subtracting the first element of the window from the sum), while keeping the sum greater than or equal to the target. The goal is to find the smallest window that meets this condition.

Here’s how we can implement this in Python:

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def min_subarray_len(target, nums):
    min_len = float('inf') 
    left = 0 
    value_sum = 0 

    for right in range(len(nums)): 
        value_sum += nums[right] 

        while value_sum >= target: 
            min_len = min(min_len, right-left+1) 
            value_sum -= nums[left] 
            left += 1 

    return min_len if min_len != float('inf') else 0

With this approach, we are only going through the array once, so the time complexity is O(n). The space complexity is O(1), since we are not using any additional data structures that scale with the size of the input.

In conclusion, by analyzing the problem and finding an underlying pattern, we were able to greatly optimize the brute force solution. It’s a great example of how understanding the problem and looking for patterns can lead to significant performance improvements.

Coding Constructs

1. High-level problem-solving strategies or techniques: The code uses a sliding window technique, which is a subset of the Two Pointers technique. This approach helps to solve problems related to subarrays in linear time by maintaining a ‘window’ of elements in the array.

2. Explaining the purpose of this code to a non-programmer: Imagine you have a list of positive numbers, and you’re trying to find the shortest continuous sequence of these numbers that, when added together, are equal to or larger than a given target number. This code is like a measuring tape that starts from the beginning of the list and stretches towards the end until it finds a sequence long enough. Then, it tries to shorten the sequence from the start while still meeting the criteria.

3. Logical elements or constructs used in this code: The code uses a pair of iterators (left and right), a continuous sum, an update to this sum based on the window expansion or contraction, a minimum length variable to keep track of the smallest valid sequence found, and a comparison operation to update this minimum length.

4. Algorithmic approach used by this code in plain English: The code starts with a ‘window’ at the beginning of the array. It then expands this window to the right until the sum of the elements inside the window is at least the target. When that happens, it tries to shrink the window from the left, aiming to maintain the sum but decrease the length. The process continues, always expanding the window if the sum is too small, and contracting it when possible while the sum remains large enough. The shortest window length that still meets the sum requirement is kept track of and returned.

5. Key steps or operations this code is performing on the input data:

   • Initialize variables for the window boundaries (left and right), the current sum, and the minimum length.
   • Iterate over the array with the right boundary, adding each element to the sum.
   • If the sum is large enough, enter a loop that shrinks the window from the left until the sum is smaller than the target, each time updating the minimum length if the current window is shorter.
   • After traversing the entire array, check if we found a valid window. If we did, return the length of the shortest one. If not, return 0.

6. Algorithmic patterns or strategies used by this code: The code uses the Sliding Window pattern, which is an approach for efficiently processing arrays or lists by maintaining a subset of elements (a ‘window’) and updating it as the window ‘slides’ through the data structure.

Language Agnostic Coding Drills

1. Dissection into distinct coding concepts:

   • Variable Initialization: Understanding how to create and initialize variables. This is necessary for setting up our pointers, sum, and minimum length.

   • Loops: Knowing how to set up and control loops, both “for” and “while” types. This is crucial for iterating over the array and moving the window.

   • Conditional Statements: Being able to write “if” statements to compare values. We need this to check if the sum is large enough or if the current window is the smallest found.

   • Arithmetic Operations: Being able to add and subtract values. This is needed for expanding and contracting the sum as the window moves.

   • Indexing Arrays: Understanding how to access elements in an array using their index. This is required for adding and subtracting elements from the sum.

2. List of concepts in order of increasing difficulty, with descriptions:

   • Variable Initialization: This is a beginner-level concept. It involves understanding the syntax for creating variables and assigning values to them.

   • Arithmetic Operations: This is a beginner-level concept. It requires knowledge of basic arithmetic operations like addition and subtraction.

   • Indexing Arrays: This is an intermediate-level concept. It involves accessing elements of an array, which requires understanding of arrays and how indexing works.

   • Conditional Statements: This is an intermediate-level concept. It involves comparing values and making decisions based on those comparisons, which requires a logical mindset and understanding of comparison operators.

   • Loops: This is an advanced-level concept in this context. It involves not only understanding the syntax of loops but also the logic to control the flow of the program and manipulate variables accordingly.

3. Problem-solving approach from problem statement to final solution:

   • Start by understanding the problem. We’re given an array of positive integers and a target value, and we’re looking for the minimum length of a continuous subarray that sums to at least the target.

   • Initialize variables. We’ll need two pointers to represent the left and right boundaries of our window, a variable to store the current sum of the window’s elements, and a variable to store the minimum length of a valid subarray.

   • Start a loop over the array. On each iteration, expand the window to the right by incrementing the right pointer and adding the corresponding array element to the sum.

   • Inside this loop, if the sum becomes large enough, start a nested loop to contract the window from the left. On each iteration, decrement the sum by the value at the left pointer, then increment the left pointer. Update the minimum length if the current window length is shorter.

   • After the loop, check if we found a valid subarray. If we did, return the length of the shortest one. If not, return 0.

   • Each of the coding drills corresponds to a piece of this approach. Variables are used throughout, arithmetic operations are used to manipulate the sum, indexing is used to access array elements, conditionals are used to check the sum and update the minimum length, and loops are used to move the window.

Targeted Drills in Python

• Variable Initialization: This is how we initialize variables in Python.

  1 2	var1 = 0 # an integer variable var2 = 0.0 # a float variable

• Arithmetic Operations: Here’s a simple arithmetic operation in Python.

  1 2	sum = 3 + 5 # addition operation difference = 10 - 2 # subtraction operation

• Indexing Arrays: Accessing elements in an array in Python.

  1 2	arr = [1, 2, 3, 4, 5] print(arr[0]) # prints the first element

• Conditional Statements: Writing conditionals in Python.

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  if 5 > 3: print("5 is greater than 3") else: print("5 is not greater than 3")

• Loops: Looping over an array in Python.

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  arr = [1, 2, 3, 4, 5] for i in arr: print(i)

• Maintaining a Running Sum: This drill is vital for our problem because we need to keep track of the sum of the window’s elements as we expand and contract it.

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  arr = [2, 3, 1, 2, 4, 3] sum = 0 for i in arr: sum += i print(sum)

• Updating Minimum Value: It’s essential for our problem as we need to update the minimum length of the valid subarray as we traverse the array.

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  min_len = float('inf') # initialized to infinity lengths = [3, 1, 4, 2] for l in lengths: if l < min_len: min_len = l print(min_len) # prints 1

The individual concepts and problem-specific drills can be assembled to create the final solution as follows:

• Start with the initialization of the variables (left, right, sum, and min_len).

• Then, start a for loop to traverse the array. This is where the loop concept comes into play.

• Inside this loop, expand the window by incrementing the right pointer and adding the corresponding array element to the sum (uses indexing, arithmetic operations, and the running sum drill).

• If the sum becomes large enough (conditional statement), start another loop to contract the window from the left. On each iteration, decrement the sum by the value at the left pointer, then increment the left pointer. Update the minimum length if the current window length is shorter (uses all concepts and the updating minimum value drill).

• After the loop, check if we found a valid subarray. If we did, return the length of the shortest one. If not, return 0.

Note that this description is language-agnostic and focuses on the logic of assembling the drills into the final solution. You’d need to implement this logic in your chosen programming language.

Q&A

This problem can be efficiently solved using the sliding window technique. We maintain a window that expands and contracts while keeping track of the sum of its elements. The algorithm starts by initializing two pointers, ’left’ and ‘right’, to the start of the array. Additionally, we initialize variables ‘minLength’ and ‘currentSum’ to infinity and 0, respectively. We iterate over the array using the ‘right’ pointer until it reaches the end. At each step, we add the current element ’nums[right]’ to ‘currentSum’. While ‘currentSum’ is greater than or equal to the target, we update the ‘minLength’ if necessary and contract the window by subtracting ’nums[left]’ from ‘currentSum’ and moving the ’left’ pointer to the right. Finally, if ‘minLength’ is still infinity, it means there is no subarray whose sum is greater than or equal to the target, so we return 0. Otherwise, we return the ‘minLength’. The time complexity of this algorithm is O(n) since we iterate over each element of the array once. The space complexity is O(1) since we only use a constant amount of extra space to store variables.

In simple terms, this problem is like looking at a long row of numbers and using a moving window to find the smallest group of numbers that add up to a certain value or more. Imagine you have a telescope and you’re moving it along the row of numbers.

The telescope (our “window”) starts at the beginning. You keep adding the next number to your total, moving the telescope to the right to see more numbers. If the total is big enough (reaches or surpasses the target), you start moving the left side of the telescope to the right too, leaving out numbers from your total.

The goal is to find the smallest possible length of the telescope that still gives you a total that is enough. If you’ve looked at all the numbers and can’t find any group that is enough, then your answer is zero.

This method is pretty efficient because you only have to look at each number once (this is what we mean by time complexity is O(n)). Plus, you don’t need to keep much extra information beyond a few numbers (that’s why space complexity is O(1)).

To solve this problem, we can use the sliding window technique. The idea is to maintain a window that expands and contracts while keeping track of the minimum length of a subarray with a sum greater than or equal to the target.

Here’s the step-by-step algorithm:

Initialize two pointers, left and right, both pointing to the start of the array nums. Initialize the variables minLen and currentSum to track the minimum length and the sum of the current window, respectively. Iterate the right pointer from 0 to n-1, where n is the length of nums. Add nums[right] to the currentSum. While the currentSum is greater than or equal to the target: Update minLen by taking the minimum value between minLen and the length of the current window (right - left + 1). Subtract nums[left] from the currentSum. Increment the left pointer to contract the window. After the iteration, if minLen is still the initial value (indicating no subarray with sum >= target was found), return 0; otherwise, return minLen.

In really simple terms, this problem is like having a stretchy, magic ruler that can only measure a continuous row of numbers. We’re looking to find the shortest possible length of this ruler where the numbers it covers add up to at least a certain value.

Imagine we put our stretchy ruler on a line of numbers. First, we only cover the first number. Then, we keep stretching the ruler to the right to include more numbers and add them all up. If the total is enough (reaches or goes above our target number), we start shrinking the ruler from the left, but keep the total above the target. The goal is to find the shortest possible length of the ruler that still gives a total that’s enough.

We keep a note of the shortest length we’ve found as we move along the line. If we get to the end and haven’t found any group of numbers that adds up to enough, then we say the shortest length is zero.

This method works quite quickly because we only need to look at each number once. Plus, it doesn’t need much extra space to do the calculations, just enough to keep track of a few numbers.

Python prefix sum and 2 pointers approach. Minimise the distance between the two pointers

The prefix sum approach in combination with two pointers can be an effective method to solve this problem.

The prefix sum approach is where you maintain an array where each element is the sum of all the numbers before it, including the current number. This makes it easier to calculate the sum of any subarray, because you can just subtract the prefix sum at the start of the subarray from the prefix sum at the end of the subarray.

The two pointer approach involves having two pointers, left and right, that represent the start and end of a subarray.

Here’s a high-level idea of how we can solve this problem using the prefix sum and two pointers approach:

1. First, calculate the prefix sum of the array.

2. Initialize two pointers, left and right, both set to the start of the array.

3. Initialize a variable minLength to infinity, this will keep track of the minimum length of subarray we find.

4. Loop while the right pointer has not reached the end of the array.

   • Expand the right pointer until the sum between left and right pointers (calculated using the prefix sum array) is greater than or equal to s.

   • Once the sum is greater than or equal to s, try to minimize the window by moving the left pointer to the right, until the sum is less than s.

   • Update minLength to the minimum of minLength and the current length of the window (right - left + 1).

5. After the loop, if minLength is still infinity, return 0, otherwise return minLength.

This approach will ensure we find the minimal length of a contiguous subarray whose sum is greater than or equal to s, while keeping the time complexity to O(n) and space complexity to O(1), assuming the space for input array is not counted.

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class Solution:
    def minSubArrayLen(self, target: int, nums: List[int]) -> int:
        prefix = [0] + list(accumulate(nums))
        i, j = 0, 0
        res = float('inf')

        while j < len(prefix):
            curr = prefix[j] - prefix[i]
            if curr >= target:
                res = min(res, j - i)
                i += 1
            else:
                j += 1
        
        return res if res != float('inf') else 0

The provided Java code uses the sliding window technique to find the minimal length of a subarray whose sum is greater than or equal to the target. It initializes two pointers, start and end, along with variables for tracking the minimum length (minLength) and the current subarray sum (subarraySum).

The code iterates through the array, incrementing the end pointer and adding the corresponding element to the subarraySum. It then enters a nested while loop that slides the window by incrementing the start pointer and subtracting the element at that position from subarraySum until the sum is less than the target.

Within the nested loop, the code calculates the length of the current subarray (end - start + 1) and updates minLength if the current length is smaller. Finally, the code returns the minimum length if it has been updated during the process or 0 if no subarray satisfies the given condition.

The sliding window technique allows for an efficient solution to this problem with a time complexity of O(n), where n is the length of the input array.

Here’s a simplified explanation:

This code is using a trick called the ‘sliding window’ to find the smallest group of numbers in a list that adds up to a certain target number or more. It starts with a window of just one number, the first number in the list, and then slides the window to the right, making it bigger and bigger until the numbers in the window add up to the target.

Once the numbers inside the window add up to the target or more, it tries to make the window smaller from the left side, as long as the numbers still add up to the target or more. During this process, it keeps track of the smallest window it has found that meets the criteria.

It repeats this sliding and shrinking process all the way through the list. If it finds at least one group of numbers that adds up to the target, it tells you the size of the smallest such group it found. If it doesn’t find any, it just tells you ‘0’.

Thanks to the sliding window trick, it can do this quickly even for a big list, checking each number only once.

Use two pointers: one adding to the sum and another subtracting. As all numbers are positive, then sum will always be increasing with adding a number and deceasing when subtracting.

In simple language:

This code uses two markers, like on a number line, to pick out groups of numbers in a list. One marker moves along adding numbers to a total, making the group bigger. Because all the numbers are positive, the total only gets bigger when we do this.

The other marker subtracts numbers from the total, making the group smaller. Whenever the total is big enough (at least as big as the target number), we try moving this second marker to make the group as small as possible while keeping the total big enough.

This way, by moving the markers along the list, the code can find the smallest possible group of numbers that adds up to the target or more. If it doesn’t find any such group, it tells you ‘0’.

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class Solution {
    public int minSubArrayLen(int target, int[] nums) {
        int lo = 1;
        int hi = nums.length;

        int ans = 0;

        while(lo <= hi){
            int mid = lo + (hi - lo) / 2;

            if(isPossible(nums, mid, target)){
                ans = mid;
                hi = mid - 1;
            }
            else{
                lo = mid + 1;
            }
        }

        return ans;
    }

    private boolean isPossible(int[] nums, int sz, int target){
        int sum = 0;

        for(int i = 0; i < sz; i++){
            sum += nums[i];
        }

        if(sum >= target) return true;

        for(int i = 1; i < nums.length - sz + 1; i++ ){
            sum = sum + nums[i + sz - 1] - nums[i - 1];
            if(sum >= target) return true;
        }

        return false;
    }
}

This is an interesting implementation that takes a binary search approach to solving the problem. The binary search is being performed on the size of the subarray, which ranges from a minimum of 1 to a maximum of the length of the given array. In each iteration, it checks whether it’s possible to find a subarray of the current size (mid) whose sum is equal to or greater than the target. If it’s possible, the solution reduces the size by moving towards the lower half of the search space; if not, it expands the size by moving towards the upper half.

1. Efficiency: This approach involves multiple passes over the array to calculate sums of subarrays, which can make it slower than a sliding window approach. The time complexity of this approach is O(n log n) because of the binary search on the size (log n) and summing subarrays for each size (n). A sliding window approach can solve the problem in linear time, O(n), which makes it a more efficient solution.

2. Clarity: The code is clean, and the method names clearly communicate what each method does. The variable names could be more descriptive. For example, “sz” could be renamed to “subarraySize” or “windowSize” for clarity.

3. Robustness: This solution correctly handles edge cases such as an empty array or target that is larger than the sum of all elements in the array. If no subarray meets the condition, the binary search will eventually lead to the condition lo > hi, and the function will return ans as 0.

4. Space Complexity: The space complexity of this solution is O(1) which is excellent, as it only uses a few variables and does not use any additional data structures.

In conclusion, while this is a valid approach, the problem could be solved more efficiently using the sliding window technique, which would reduce the time complexity to O(n).

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public int minSubArrayLen(int target, int[] nums) {
    
    TreeMap<Integer, Integer> sumToIndex = new TreeMap<>();
    
    int sum = 0, result = Integer.MAX_VALUE;
    
    sumToIndex.put(0,-1);
    
    for( int i = 0; i < nums.length; i++ ){
        sum+=nums[i];
        int required = sum-target;
        
        Map.Entry<Integer, Integer> requiredIndex = sumToIndex.floorEntry(required);
        
        if( requiredIndex != null ){
            result = Math.min(result, i-requiredIndex.getValue());
            if ( result == 1 ) return 1;
        }
        
        sumToIndex.put(sum, i);
    }
    
    return result == Integer.MAX_VALUE ? 0 : result;
}

This code solves the problem using a slightly different approach - using a TreeMap data structure in Java. A TreeMap in Java is a Red-Black tree-based NavigableMap implementation. It is sorted according to the natural ordering of its keys, or by a Comparator provided at the TreeMap’s creation time, depending on which constructor is used.

1. Efficiency: This code uses a TreeMap to store the cumulative sum up to a certain index. This allows for the usage of floorEntry() method, which returns the greatest key less than or equal to the given key, or null if there is no such key. So, we are looking for the greatest cumulative sum that is less than or equal to required (which is current sum - target). However, the time complexity of TreeMap operations (like put and floorEntry) can be O(log(n)). So, in the worst-case scenario, this code may operate at O(nlog(n)) complexity. This is not the most efficient solution for this problem, as it can be solved in O(n) time using a sliding window technique.

2. Clarity: The code is relatively clear with good variable naming. However, it might be a little confusing for someone unfamiliar with TreeMap and its floorEntry() method. Some comments explaining the logic would improve readability.

3. Robustness: The solution appears to handle edge cases correctly. It properly returns 0 if there is no subarray with a sum that is equal to or exceeds the target (when result remains as Integer.MAX_VALUE). The early exit when result equals 1 is also a nice touch for optimization.

4. Space Complexity: This solution has a space complexity of O(n), as in the worst-case scenario, each prefix sum (up to index i) is stored in the TreeMap.

In summary, this is a valid but slightly less efficient solution to the problem. It demonstrates a clever use of TreeMap to keep track of the prefix sums and their corresponding indices, but the use of a TreeMap leads to a higher time complexity than necessary. The problem could be solved in a more time-efficient manner using a two-pointer or sliding window approach.

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var minSubArrayLen = function(target, nums) {
    let minLength = nums.length+1;
    let runningSum = 0;
    let i = 0;
    let j = 0;

    while(i < nums.length){
        runningSum += nums[i];

        while(runningSum >= target){
            minLength = Math.min(minLength, i-j+1);
            runningSum -= nums[j];
            j++;
        }

        i++;
    }

    return minLength == nums.length+1? 0: minLength;
};

This code in JavaScript uses the sliding window technique to solve the problem. Here is a critique:

1. Efficiency: This is an efficient solution to the problem. The sliding window technique allows us to traverse the array only once, so the time complexity is O(n), where n is the number of elements in the array.

2. Clarity: The code is clear and easy to understand. The variable names runningSum, minLength, i, and j are appropriately chosen and self-explanatory. The while loops and if conditions follow a logical sequence that aligns with the sliding window technique.

3. Robustness: This solution seems to handle edge cases properly. If no subarray meets the target sum, it correctly returns 0. The while loop condition ensures that we don’t run into an Index Out of Bound Error.

4. Space Complexity: This solution uses a fixed number of variables, so the space complexity is O(1).

Overall, this is an efficient and effective solution. It implements the sliding window technique correctly and has considered edge cases as well. The code is clean, and the logic is straightforward, making it easy to understand.

Sliding Window approach. Let ini represent the index of the first element of the sliding window. Let fin represent the index of the last element of the sliding window. Note ini can be at most one step ahead of fin, in which fin catches up to ini indicating that for the next outer iteration the sliding window will contain solely of nums[fin] == nums[ini].

In very simple terms, this approach is like playing with a telescope on the array of numbers. Here, ini and fin are like the two ends of the telescope. ini represents the start of the telescope, and fin represents the end of the telescope.

During each step, fin moves one step forward, just like stretching the telescope to see further. If the sum of numbers within the telescope is greater than or equal to the target, we then try to shrink the telescope from the ini side, just like shortening the telescope to see closer.

There might be a situation where ini and fin meet, which means the telescope is at its shortest length. In the next step, both ini and fin will move forward, like shifting the whole telescope one step further. In this way, we slide our “telescope” - the window - over the array to find the smallest possible length that sums to the target.

A subarray is a contiguous non-empty sequence of elements within an array

I used greedy algorithm, but I got wrong answer!

The minimum size subarray problem is a problem where a greedy approach would typically not fail, because the problem lends itself to a sliding window technique, which is a form of a greedy algorithm. The goal of the problem is to find the smallest contiguous subarray that sums to a target or greater, and since the problem only involves adding positive integers, expanding and contracting the window in a greedy manner will always lead to the smallest possible subarray.

However, if we modify the problem slightly such that we have negative numbers or we are asked to find a subarray that sums exactly to the target, a greedy approach can fail.

For example, consider the array [-1, 2, 3, -4, 5] and the target is 4.

A greedy approach might start at the first element and start expanding the window. It sees the subarray [-1, 2, 3] which sums to 4 and it stops, giving the length as 3.

However, the correct answer should be the subarray [4] which has a length of 1.

Therefore, in the modified problem where we need an exact sum and the array can contain negative numbers, a greedy approach can fail.

Having an understanding helps me personally in remembering and applying it in some other problem.

From my understanding the O(N) sliding window approach where neither pointer is ever decremented works because:

Increasing the size of the window keeps or breaks the “constraint” in one direction (the new sum can only be >= previous window) Decreasing the size of the window keeps or breaks the constraint in the opposite direction of increasing it (the new sum can only be <= previous window) If any of those don’t apply the solution wont work, which is the case for negative numbers (anything could happen with the sum depending on specific values). So for negative windows we must reset the end of the window to start of the window (so try window size 1,2,…n at each index) each time we increment start for an O(N^2) solution.

Do you have a better way to articulate this?

You’ve articulated the idea for the sliding window approach quite well! The sliding window approach indeed works by exploiting the property that changing the size of the window (i.e., moving the pointers) only modifies the sum in one direction. It does so by either:

1. Increasing the window size: When we move the ’end’ pointer to the right, we’re increasing the size of the window, hence increasing the sum of elements within the window or keeping it the same (in case of zero values). This is because we’re only adding non-negative values.

2. Decreasing the window size: When we move the ‘start’ pointer to the right, we’re decreasing the size of the window, hence decreasing the sum of elements within the window or keeping it the same. This is because we’re only removing non-negative values.

It’s crucial to note that this approach’s effectiveness stems from the non-negativity of elements. If we introduce negative numbers, these properties don’t hold anymore because adding a negative number decreases the sum, and removing a negative number increases the sum.

The complexity of this approach is O(N) because each pointer only moves from left to right once. If we have negative numbers in the array, we would need a more complex approach (like resetting the window for each starting point), leading to a higher time complexity, often O(N^2), depending on the exact method used.

Isn’t O(n) linear better then O(nlogn) ?

Yes, in terms of time complexity, O(n) is better than O(n log n). Here’s why:

• O(n) complexity, also known as linear complexity, means that the time it takes to run the program increases directly proportionally with the size of the input. So, if the size of your input doubles, the time it takes to run the program also doubles.

• O(n log n) complexity means that the time it takes to run the program increases linearithmically with the size of the input. This is slower than linear time because as the input size increases, the program doesn’t just take n times longer to run (like it would with linear time complexity), but n times longer multiplied by the logarithm of n.

So, if you have the choice between an algorithm that is O(n) and one that is O(n log n), the O(n) algorithm would generally be more efficient.

Using sliding window is more intuitive than using binary search to solve this problem. Is it faster than binary search?

In terms of time complexity, sliding window and binary search can both operate at O(n log n) under certain circumstances, but for the minimum size subarray sum problem, the sliding window technique can solve the problem in O(n) time, which is faster than binary search.

The reason why the sliding window solution can be more efficient in this case is because it takes advantage of the property of the problem that the array is non-negative. The window can expand until the sum is over the target and then contract to find the minimal length. Each element in the array is visited at most twice (once for expanding the window and once for contracting the window), so the total operation count is proportional to 2n, leading to an O(n) time complexity.

On the other hand, binary search for this problem involves checking whether a subarray of a certain size can meet the condition, and if so, reducing the size. This would require checking all subarrays of a certain size, which would make the process less efficient.

Therefore, in the context of this problem, sliding window can be both more intuitive and faster than binary search.

What does shortest subarray that its sum >= s mean? for example this testcase 64 [2,3,1,2,4,3,1,2,3,4,0,-2,-34,3,65] why the anwser is 0 rather than 1 cuz {65} is a subarray and 65>= 64 ?

Could anyone elaborate this for me ?

The “shortest subarray that its sum >= s” means that you are trying to find the smallest length of the contiguous subarray (i.e., a sequence of numbers next to each other in the array) whose sum is greater than or equal to a given number ’s’.

In your test case, the sum is given as 64, and the array is [2,3,1,2,4,3,1,2,3,4,0,-2,-34,3,65].

A subarray is a contiguous sequence of elements within an array. So, for instance, [3, 1, 2] is a subarray of your input, but [2, 1, 3] is not because even though these numbers exist in your array, they are not contiguous in the order presented.

Now, the subarray [65] indeed sums to 65 which is greater than 64. Therefore, the length of the shortest subarray whose sum is greater than or equal to 64 is indeed 1 in this case.

If you got an answer of 0, there might have been an error in the code that was used to solve the problem. The length of a subarray can’t be 0 because a subarray must contain at least one element. In this case, the subarray [65] is the correct answer, and its length is 1.

In such questions for sliding window, it is better to have a dry run on a page or code editor, which gives you a clear idea of when to increment the pointers, and when to expand or shorten the window.

Doing a dry run (manually tracing through a program’s execution step by step) is an excellent way to understand how the sliding window technique works, and can indeed help clarify when and how to move the pointers and adjust the window size.

Let’s do a dry run of a simple example using the sliding window approach:

Given an array [2,3,1,4,5] and target sum is 7, find the minimal length of a subarray sum >= 7.

1. Initialize two pointers, start and end, both at 0. Initialize the variables minLength (set to infinity or array length + 1 at first) and currentSum, starting at 0.

2. The main loop starts. Increase the end pointer, moving to the right, and add the corresponding element to the currentSum.

   currentSum: 2 window: [2] start: 0, end: 0

3. currentSum is less than target, increase end pointer:

   currentSum: 5 window: [2,3] start: 0, end: 1

4. currentSum is still less than target, increase end pointer:

   currentSum: 6 window: [2,3,1] start: 0, end: 2

5. currentSum is still less than target, increase end pointer:

   currentSum: 10 window: [2,3,1,4] start: 0, end: 3

6. Now, currentSum is greater than or equal to the target. Calculate the length of the window (end - start + 1 = 4), and compare it to the minLength. If it’s less, update minLength.

7. Try to make the window smaller by moving the start pointer to the right (and subtracting the corresponding element from currentSum), while keeping the currentSum greater than or equal to the target:

   currentSum: 8 window: [3,1,4] start: 1, end: 3

8. Repeat step 7:

   currentSum: 7 window: [1,4] start: 2, end: 3

9. Now, the window size is 2 which is smaller than the previous minLength 4. So, update minLength to 2. Try to move the start pointer again, but now currentSum will be less than the target. So, increase the end pointer in the next round of the main loop.

Following this dry run, the correct answer of minimal length subarray with sum >= 7 is 2.

Dry runs like this can be instrumental in understanding and applying the sliding window technique, and they’re a highly recommended step when developing and debugging your solutions.

Here is a table summarizing the 9 steps from the dry run of the sliding window technique:

In the table, “window” represents the elements in the current sliding window, “start” and “end” are the pointers defining the window’s limits, “currentSum” is the sum of elements within the window, and “minLength” is the smallest length of subarray whose sum is greater than or equal to the target found so far.

How do we infer that we should not sort the array from the problem statement?

The problem statement asks for the “minimum size subarray” with a sum that is greater than or equal to a given target. In this context, a subarray is a contiguous portion of an array, and sorting the array would disrupt the original order of elements. Therefore, sorting would change the positions of the elements and make it impossible to correctly identify the original subarray that meets the problem’s requirements.

Furthermore, the problem is asking for a subarray, not a subset. A subset of an array could include any elements from the array in any order, while a subarray is always a sequence of elements in their original order. Therefore, we infer from the use of the word “subarray” that maintaining the original order of elements is important.

Lastly, the nature of this problem – finding a contiguous subarray with a certain property (in this case, a sum greater than or equal to a target) – is a common type of problem that often involves techniques such as the sliding window approach, which operates on the array in its original order.

So, from the use of terms like “subarray”, the requirement for a contiguous sequence of elements, and the nature of this type of problem, we can infer that we should not sort the array.

Are the 209 and 862 problems the same They are not. #209 contains only positive integers. #862 can also include negative integers so, it requires a whole different approach. It’s a good idea to solve both of them to understand the reason why a technique can or cannot be applied in each particular case.

Yes, you’re correct. The problems 209 (“Minimum Size Subarray Sum”) and 862 (“Shortest Subarray with Sum at Least K”) on LeetCode are similar in their overall goal – both problems ask for the shortest (or minimum size) subarray that meets a certain sum condition. However, they have different constraints, which affect the solution strategies.

Problem 209 deals with an array of positive integers and a target sum, and the task is to find the smallest contiguous subarray with a sum that is equal to or greater than the target sum. Because all the integers in the array are positive, we can use a two-pointer or sliding window strategy to solve this problem efficiently.

On the other hand, problem 862 involves an array that can contain both positive and negative integers, and the task is to find the shortest subarray with a sum that is at least K. The presence of negative integers in this problem means that a simple two-pointer or sliding window strategy will not work, as the sum of a “window” of numbers can decrease even when we add numbers to it. This makes problem 862 more complex than problem 209.

So, while the two problems share a similar overall goal, the constraints and therefore the solution strategies are quite different. It’s a great observation that tackling both can provide a deeper understanding of how similar problem statements can require different approaches due to their specific conditions and constraints.

Those who were asking about how to binary search on this as this is not sorted or binary search will be even slower First of all leetcode knows that thinking of Sliding Window is simpler and it want us to think of all possible solutions. Secondly, Binary Search is possible on this nums(even it is unsorted and we aren’t allowed to solve it ) as we have to find subarray of nums which is >= tar, and SINCE ALL THE NUMBER ARE POSITIVE SO PREFIX SUM OF NUMbs will be always in increasing order

Steps :

Start s = 0, e = n-1; get mid if(isValid(mid, tar, nums)) ans = mid +1, e = mid -1; else s = mid +1;

Here Think it like this: int mid = s + (e - s)/2; // we are selecting a window of size m if(isValid(mid, target, pref, nums)){ // checking if any subarray is possible of wiindow size mid res = mid + 1; e = mid -1; // if yes then trydecreasing window size } else s = mid +1; // if no then try increasing window size

here isValid is a function which check all subarray sum of size mid.

PS : IT won.t work if array contains negative elements give it a try : https://leetcode.com/problems/shortest-subarray-with-sum-at-least-k/

You’re correct that the binary search approach can be used in this problem, particularly because the array has all positive integers, which guarantees the prefix sum array will always be in increasing order.

To simplify the explanation:

The binary search here is not used to find an element in an array like in classic use-cases. Instead, it’s used to find the minimal length of a subarray whose sum is equal or greater than a given target.

The binary search range is the possible length of the subarray, from 1 to the length of the input array.

• Start with two pointers s = 0 (start) and e = n-1 (end), where n is the length of the array.
• For each iteration, calculate the middle point ‘mid’ and check if there exists a subarray of length ‘mid’ whose sum is equal or greater than the target (using the ‘isValid’ function).
• If such a subarray exists, update the answer to ‘mid’ and then move the ’end’ pointer to ‘mid - 1’ to continue searching for a smaller subarray length.
• If no such subarray exists, move the ‘start’ pointer to ‘mid + 1’ to search for a longer subarray length.

This approach continues until the binary search process finishes. This algorithm is indeed more complicated than the sliding window approach.

It’s important to note that the binary search approach wouldn’t work if the array contains negative elements since the prefix sum array will no longer be strictly increasing. In those cases, other techniques, such as maintaining a deque as in problem 862 on LeetCode (“Shortest Subarray with Sum at Least K”), are needed.

Applying binary search in this way demonstrates the flexibility of this algorithmic strategy beyond simple sorted array search tasks.

In the first glance, it seems to Two Pointers approach

Yes, you’re correct. At first glance, this problem lends itself to the two pointers approach, which is often referred to as the “sliding window” technique in these types of problems.

The idea here is to have two pointers, typically called ‘start’ and ’end’, that represent the bounds of a subarray in your input array. As you progress through the array, you adjust these pointers to ‘slide’ your window along the array, expanding and contracting it as needed.

For this specific problem, the goal is to find the shortest subarray whose sum is equal to or exceeds a target value. You start with a window at the beginning of the array and keep adding elements to your current sum until it equals or exceeds the target. At this point, you’ve found a valid subarray, but it might not be the shortest one.

To find the potentially shorter valid subarray, you start contracting the window from the ‘start’ side, subtracting elements from your sum, until your sum is less than the target. You then continue the process until you’ve gone through the whole array.

The beauty of this approach is that you only need to pass through the array once, making it an efficient solution with a time complexity of O(n), where n is the size of the input array.

Do you have this case? And I think below case is wrong!!

Input: 11, [1,2,3,4,5] Output: 0 Expected: 3

It appears that there might be a misunderstanding or a mistake with the output you’ve posted. The problem is asking for the length of the smallest contiguous subarray with a sum that is equal to or greater than a specified target.

Given the input you’ve provided (target = 11, array = [1,2,3,4,5]), the smallest contiguou```s subarray that sums to 11 or more is [3,4,5], which indeed has a length of 3. Thus, the expected output should indeed be 3.

If the output was 0, it would indicate that no subarray sums to the target or more, which is not the case here. Therefore, if this is the output you’re seeing, it suggests there is an error in the code being used to solve this problem.

Remember, the problem is not asking for the subarray itself or the sum of the subarray. It is asking for the length (the number of elements) of the smallest subarray that meets the given criteria.

It’s mentioned that sum is greater then equal to [3,4,5]>=11 which is true

Yes, you’re absolutely correct. The subarray [3,4,5] indeed sums up to 12, which is greater than the target value of 11. The length of this subarray is 3, so the correct answer should be 3.

The problem asks for the length of the smallest (shortest) contiguous subarray where the sum is equal to or greater than a target value. In this example, [3,4,5] is the smallest subarray that satisfies this condition, so the length of this subarray, which is 3, is the correct answer.

It’s important to note that the subarray has to be contiguous, meaning the elements have to be next to each other in the array.

In summary, if the problem asks for the minimum size of the subarray whose sum is greater than or equal to the target, and you find a subarray whose sum meets this condition, the correct answer would be the size (or length) of that subarray.

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class Solution {
public:
    int minSubArrayLen(int target, vector<int>& nums) {
        int len=0,ans=INT_MAX,i=0,j=0,sum=0,n=nums.size(),b=0;
        while(j < n){
            cout << j;
            while(j < n){
                sum += nums[j];
                b += nums[j];
                len++;
                j++;
                if(sum >= target)break;
            }
            while(sum - nums[i] >= target){
                i++;
                sum -= nums[i];
                len--;
            }
            ans=min(ans,len);
        }
        if(b < target)return 0;
        return ans;
    }
};

The issue in your code lies in the second while loop where you’re decrementing the window and sum without ensuring that the index ‘i’ is within the valid range. Also, you are subtracting nums[i] from sum after incrementing i, which could potentially lead to incorrect results.

Try adjusting the code as follows:

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class Solution {
public:
    int minSubArrayLen(int target, vector<int>& nums) {
        int len = 0, ans = INT_MAX, i = 0, j = 0, sum = 0, n = nums.size();
        while(j < n){
            sum += nums[j];
            len++;
            j++;

            while(sum >= target){
                ans = min(ans, len);
                sum -= nums[i];
                i++;
                len--;
            }
        }
        return (ans == INT_MAX) ? 0 : ans;
    }
};

Here’s what’s changed:

1. The second while loop is now checking sum >= target instead of sum-nums[i] >= target. We’re still sliding the window to the right when the total is greater than or equal to the target.

2. Inside the second while loop, sum -= nums[i]; is now before i++; ensuring that we’re decrementing the sum by the correct amount.

3. The final return statement is adjusted to handle the case where no subarray sums to the target. If no such subarray is found, ans will remain at its initial value of INT_MAX, and the function should return 0.

Can anyone tell me what’s wrong in my code

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class Solution {
public:
	int helper(vector& nums, int target, int i) {
		if (target <= 0) {
			return 0;
		}

		if (i >= nums.size()) {
			return INT_MAX/2;
		}

		int l1 = 1 + helper(nums, target - nums[i], i + 1);
		int l2 = helper(nums, target, i + 1);
		return min(l1, l2);
	}

	int minSubArrayLen(int target, vector<int>& nums) {
		int x = helper(nums, target, 0);
		if (x >= INT_MAX/2) {
			return 0;
		}

		return x;
	}
};

The code you’ve posted appears to be attempting a recursive approach to solving the problem. The helper function is trying to explore all possible subarrays by recursively reducing the target by the value of the current element (l1), and moving to the next element, or skipping the current element and moving to the next one (l2). Then it chooses the smaller length subarray between the two.

However, this approach doesn’t fit the problem for two reasons:

1. It does not preserve the order of the array elements: The problem statement specifies that you need to find the smallest contiguous subarray that sums to at least the target value. Contiguous means the elements are in their original order in the array. However, the recursive approach you’re using can mix up the order of the elements, as it’s effectively trying all combinations of elements, not just contiguous subarrays.

2. It’s inefficient: The time complexity of this approach is O(2^n), where n is the number of elements in the array. This is because each recursive call branches into two new recursive calls, leading to an exponential number of total function calls. This could cause your program to run very slowly, or even exceed the time limit, for larger input arrays.

You might want to reconsider your approach and consider using a technique like a sliding window or prefix sum which can solve this problem with a time complexity of O(n).

How is this a sliding window problem? The window length can be anything and is not fixed.

In a sliding window problem, the window size does not always have to be fixed. What’s more important is that there is a contiguous subarray, or a “window,” that you’re considering at any given time.

In this particular problem, the “window” refers to a contiguous subarray of the original array. The goal is to find the shortest such subarray that has a sum greater than or equal to a target value.

You start by initializing a window at the beginning of the array and continue to expand it (by moving the end of the window to the right) as long as the sum of the elements in the window is less than the target. When the sum is greater than or equal to the target, you attempt to contract the window (by moving the start of the window to the right) while ensuring the sum still remains greater than or equal to the target.

In this way, the window “slides” through the array, and its size dynamically changes depending on the specific values in the array and their sum relative to the target. Therefore, this problem is categorized as a sliding window problem.

For nlogn, apply BS on fixed sized window!

The Binary Search (BS) approach mentioned here for the minimum size subarray problem would work like this:

1. Instead of directly searching for the answer in the array, you create a search space of possible answers, which is the size of the subarrays. The smallest subarray size is 1 and the largest size is the length of the array.

2. Then, you use Binary Search to progressively refine your search space. You take the midpoint of the search space as a potential subarray size and check if it’s possible to have a subarray of that size with sum greater or equal to the target.

3. If it is possible, then the midpoint becomes your new high bound (since you’re looking for the minimum subarray length). If it’s not possible, the midpoint + 1 becomes your new low bound.

4. You continue this process until the search space is empty. The smallest valid subarray size you found is your answer.

The binary search approach is O(n log n) because for each mid value you have to check whether there’s a subarray of length mid with sum >= target, which takes O(n) time, and you have to do this O(log n) times.

This approach can work for this problem, but it’s less intuitive than the sliding window approach, which is simpler and more efficient (O(n)). Furthermore, applying Binary Search to this problem may require understanding that the prefix sum array of the original array (since all numbers are positive) will be in non-decreasing order, creating a situation where Binary Search can be applied.

for target = 11 and nums =[1,2,3,4,5] answer should be 0 but in test cases its 3 can someone explain how?

subarray with sum value greater than or equal to 11

[1,2,3,4,5] 15 >= 11 [2,3,4,5] 14 >= 11 [3,4,5] 12 >= 11 the min subarray is [3,4,5] so the answer is 3

why they don’t describe that the sub array should be continuous

In the problem statement for minimum size subarray, the term “subarray” implicitly refers to a contiguous segment of the array. By definition, a subarray is a contiguous subset of an array, which means that the elements in the subarray must be in the same order they appear in the original array, with no breaks or gaps.

So, if you see a problem that asks for a “subarray,” you can assume that it is asking for a contiguous segment unless it explicitly specifies otherwise. Therefore, it is not necessary for every problem to explicitly state that the subarray should be contiguous. If non-contiguous subsets of an array are allowed, the problem will typically use the term “subsequence” instead of “subarray.”

very good problem on arrays , it took O(2N) complexity to solve , can anyone tell me a better way to solve it.My Approach is based on two pointers ,and i am shrinking the window once i get more or equal to the target

Your approach of using a sliding window with two pointers is a good one and it is actually one of the most optimal ways to solve this problem! This method allows you to scan through the array once, making it a O(n) time complexity solution, where n is the number of elements in the array.

Here’s a little clarification on why it’s O(n) and not O(2n):

In Big O notation, we drop constants. This is because Big O notation describes the rate of growth of an algorithm’s runtime, rather than the exact runtime. So even if each element of the array might be visited twice (once when expanding the window, and once when shrinking), we still consider this to be linear time complexity, or O(n), as the number of operations grows linearly with the size of the input array.

If you want to further optimize this problem, you could consider optimizing the space complexity if you’re storing extra data. But in terms of time complexity, the two-pointer sliding window approach is considered optimal for this problem.

explain this one to me it is the wrong test case

target =213 nums =[12,28,83,4,25,26,25,2,25,25,25,12] Subarray : [83, 26, 25, 25, 25, 25, 12] Output : 7 Expected : 8

I believe there might be a misunderstanding. If we take a look at the subarray [83, 26, 25, 25, 25, 25, 12] from the ’nums’ array, we can see that the sum of these numbers equals 221, which is indeed greater than or equal to the target value 213.

The subarray length is 7, which seems to be the expected output.

The confusion might stem from the expectation of 8 as the output, but from the given ’nums’ array, there is no need to have 8 elements to reach or exceed the target of 213.

Therefore, the output of 7 is correct as it represents the smallest length of subarray with sum greater than or equal to the target value.

It is a subarray problem. You can’t delete elements in the middle.

Yes, you’re right. In the context of this problem, a subarray means a contiguous segment from the original array, so you cannot remove elements in the middle of it. The solution must involve finding a subarray (a consecutive sequence in the array) such that the sum of its elements is equal to or larger than the target. The order of elements in the subarray will be exactly the same as their order in the original array.

If you have already solved this problem, do not submit as it is. Just without having a look at the code reset to default and check how much you still remember. (Best way of revising questions is leetcode daily challenges) ✌

That’s excellent advice! By doing so, you reinforce the concepts and problem-solving techniques you’ve learned. Regular practice is key to mastering any skill, including competitive programming. This not only helps you remember the techniques, but also enables you to apply them to a variety of similar problems. If you keep practicing and reinforcing your skills in this manner, you’ll find it easier to solve new problems over time. Additionally, working on problems under the daily challenges can provide you with a consistent practice schedule and expose you to a variety of different problems, further improving your problem-solving abilities.

minimal length of a subarray whose sum is greater than or equal to target

The statement “minimal length of a subarray whose sum is greater than or equal to target” refers to the problem of finding the smallest possible contiguous subarray (a segment of the array) in which the sum of the elements is equal to or exceeds a given value (the target).

For example, if we have an array [2, 3, 1, 2, 4, 3] and a target of 7, the smallest subarray with a sum that meets or exceeds the target is [4, 3], which has a length of 2. There are other subarrays that meet or exceed the target sum, such as [2, 3, 1, 2] or [3, 1, 2, 4], but these are longer and therefore do not meet the condition of being the smallest possible subarray.

This is a common type of problem in coding interviews, and there are several different strategies and algorithms that can be used to solve it, including sliding window and prefix sum approaches.

I’m trying my hand in backtracking (it’s not a topic I’m very confident in), and I’m unsure if this question can be solved through the approach?

The test case I’m getting stuck in is nums=[12,28,83,4,25,26,25,2,25,25,25,12] and target = 213.

Here’s the code:

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int result=INT_MAX;
void loop(int index, int target, vector&nums, int count){
	if(target <= 0){
		result = min(result,count);
		return;
	}
	for(int i = index; i < nums.size(); i++){
		target -= nums[i];
		count += 1;
		loop(i + 1, target, nums, count);
		target += nums[i];
		count -= 1;
	}
	return;
}
int minSubArrayLen(int target, vector& nums) {
	loop(0, target, nums, 0);
	return result == INT_MAX ? 0 : result;
}

I’d be grateful if anyone can point out the mistake!

While backtracking can theoretically solve this problem, it is not the most efficient approach due to the high time complexity (O(2^n) in the worst case, where n is the number of elements in the array). The backtracking approach tries all possible combinations of subarrays, which is not necessary for this problem and results in a lot of redundant computation.

In your code, the issue arises from the fact that you’re recursively exploring all possible subarrays starting from each index in the array. But the problem specifically asks for contiguous subarrays. Your approach doesn’t ensure contiguity because it explores all subarrays starting from each index, without considering whether these subarrays are contiguous or not.

The more optimal way to solve this problem is by using a “sliding window” or “two pointers” approach, which solves the problem in O(n) time.

Here’s the general idea of the sliding window approach:

1. Start with a window that includes the first element of the array.
2. Then expand the window from the right until the sum of the window elements is equal to or greater than the target.
3. When the sum becomes equal to or greater than the target, try to shrink the window from the left side as much as possible while maintaining a sum that is equal to or greater than the target.
4. Keep track of the minimum length of such a window during this process.
5. Move the right end of the window one position at a time, repeating steps 3 and 4, until the right end reaches the last element of the array.

This approach ensures that you’re only looking at contiguous subarrays and that you’re efficiently minimizing the window size whenever possible.

why pick and non pick concept is not apply? can anyone resolve ?

The “pick and non-pick” concept, typically used in backtracking and recursive solutions, involves making a decision at each step whether to “pick” or “not pick” an element for a particular solution.

In this problem, however, we are not selecting individual, non-contiguous elements from the array. Instead, we’re trying to find the smallest contiguous subarray with a sum greater than or equal to the target. The elements in a contiguous subarray are inherently linked – you can’t pick one and not pick the one next to it, because doing so would break the contiguity of the subarray.

Therefore, the concept of “pick and non-pick” doesn’t apply here. Instead, the problem is more suited to a “sliding window” or “two pointers” approach, where you adjust the start and end points of the subarray to find the smallest subarray that meets the criteria.

The sliding window approach involves initializing two pointers at the beginning of the array and then moving them through the array such that the sum of the subarray between the two pointers is as close to the target as possible. The window (subarray) “slides” through the array, growing and shrinking as necessary, to find the smallest subarray with a sum >= target.

The follow up of this question is confusing. If you can come up with a O(N) solution, why would you look for a slower O(N logN) solution? If there’s a typo and they mean following up with a O(logN) solution, this practically implies applying binary search. I don’t know if there is a clever way of doing such a binary search on the nums array. Some preprocessing is necessary for binary search. Sorting the array would make it O(N logN) and accumulating the nums values on another array would make it O(N). So, I am not sure O(logN) is achievable at all if that’s what they meant.

The follow-up question in problem #209 is asking if it is possible to solve the problem in O(n log n) time, if the initial input array is unsorted. The sliding window approach, which is O(n), works when the input array is sorted. The follow-up is suggesting a scenario where you have an unsorted array, and you still need to solve the problem.

You might think that sorting the array first (which takes O(n log n) time) and then applying the sliding window approach would be a solution, but remember that sorting the array would change the order of the original elements, and in this problem, the subarray must be contiguous based on the original ordering of the array.

Instead, an O(n log n) solution would involve a different approach, such as using a data structure like a TreeMap (in Java), which maintains its elements in sorted order and allows for operations like floorEntry(). These operations take O(log n) time, and you’d have to do them for each of the n elements in the array, hence the O(n log n) time complexity.

However, it’s worth noting that a faster solution (like the O(n) sliding window approach) is generally preferred over a slower one (like the O(n log n) TreeMap approach). The follow-up is really just a thought experiment to see if you understand the problem well enough to come up with a different solution under different conditions.

Sliding window approach looks good. Make sure to recompute the min as you remove the elements from back/left side of the array, too.

Absolutely! That’s the key idea behind the sliding window approach. When you’re looking for a contiguous subarray that meets a certain condition, the sliding window approach can be very efficient. Here’s how it might look:

1. Start with a window that only contains the first element in the array.
2. Keep adding elements to the window until the condition is met (in this case, the sum of the elements in the window is greater than or equal to the target).
3. When the condition is met, try removing elements from the start of the window one by one, each time checking if the condition is still met with the smaller window.
4. Whenever the condition is still met with the smaller window, update the minimum window size.
5. Repeat steps 2-4 until you’ve gone through all the elements in the array.

The important thing here is that you’re constantly checking if you can make the window smaller, not just when you’re adding elements, but also when you’re removing them. This ensures that you find the smallest possible window that meets the condition.

I could not figure out why my answer wasn’t matching the expected answer. After a lot of confusion, I learned that a subarray has to be contiguous. It would have been helpful to have that in the explanation. In math a subset does not have to be contiguous, so I didn’t know it would be different for an array.

You’re right. It’s an important detail that can lead to misunderstanding if not explicitly stated. When we’re talking about subarrays in computer science, we are usually referring to contiguous sections of an array. This is different from the concept of a subset in mathematics, where elements do not have to be contiguous.

In the context of these problems, a subarray is a contiguous block of elements within the array. This is important because the solutions often involve techniques like sliding windows or prefix sums, which rely on the elements being contiguous.

I understand the confusion and it’s a good point to remember for future programming problems. Always assume that subarray refers to contiguous elements unless explicitly stated otherwise.

My code is not working can someone please see to it

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class Solution {
public:
	int minSubArrayLen(int target, vector& nums) {
		int left=0;
		int right=1;
		int sum=nums[0];
		while(sum < target){
			sum += nums[right];
			right++;
		}
		int window = right-left;
		cout << left << " "<< right << " " << window << " " << sum << " ";
		while(left!=nums.size()){
			cout<<"a";
			if(right==nums.size()&&sum<=target){
				cout<<"p";
				cout<<"qq";
				break;
				break;
			}

			if(sum<=target){
				cout<<"b";
				if(sum==target&&right-left<window)window=right-left;
				if(left<nums.size()){
					sum-=nums[left];
					left++;
				}
				if(right<nums.size()){
					sum+=nums[right];
					right++;
				}
				cout<<" "<<left<<" "<<right<<" "<<window<<" "<<sum<<" "; 
            
			}
			else{
				cout<<"c";
				if(right-left<window)window=right-left;
				if(left<nums.size()){
					sum-=nums[left];
					left++;
				}
				cout<<" "<<left<<" "<<right<<" "<<window<<" "<<sum<<" "; 
			}
		}
		return window;
	}
};

After reviewing your code, it seems like you’re trying to implement a sliding window approach to solve the problem. However, there are a few issues:

1. Your outer while loop (while(left!=nums.size())) may lead to an infinite loop in some cases, because you’re not incrementing left in every iteration of the loop.

2. Your code is not considering the possibility that the sum may not reach the target at all. This happens when you’re trying to calculate the initial sum before the main loop.

3. The condition in your second if statement in the main loop (if(sum<=target)) is misleading. If the sum is equal to the target, you would want to update your minimum window size, decrement the sum by nums[left] and increment left. But if the sum is less than the target, you would want to increment right (if it’s less than nums.size()) and add nums[right] to the sum.

4. The ’else’ part should handle when the sum is greater than the target, and not the other way around as it is in your code.

Here is a revised version of your code:

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class Solution {
public:
    int minSubArrayLen(int target, vector<int>& nums) {
        int left = 0;
        int right = 0;
        int sum = 0;
        int window = nums.size() + 1;

        while (right < nums.size()) {
            while (sum < target && right < nums.size()) {
                sum += nums[right++];
            }

            while (sum >= target) {
                window = min(window, right - left);
                sum -= nums[left++];
            }
        }

        return window == nums.size() + 1 ? 0 : window;
    }
};

This code keeps expanding the window until the sum is greater than or equal to the target, and then it shrinks the window from the left until the sum is less than the target, while keeping track of the smallest window size. If no valid window is found, it returns 0.

Can anyone have a look at my solution?

Since the desired subarray must be a subarray of the original array, we can find it by keep popping the smaller end value (the left most value or right most value) out while the sum is greater.

For example, target = 7, nums = [2, 3, 1, 2, 4, 3] subarray = [3, 1, 2, 4, 3] [1, 2, 4, 3] [2, 4, 3] [4, 3]

Your approach of starting with the entire array and shrinking it from one end when the sum is greater than or equal to the target is an interesting idea. However, it doesn’t always work as expected.

The primary issue with your approach is that it doesn’t account for situations where the sum of a smaller subarray could still meet the target value. Your algorithm is biased towards the beginning of the array, and may result in subarrays that are unnecessarily long.

Take, for example, the array [1, 2, 3, 4, 100, 1, 1, 1, 1, 1] with a target of 7. Your algorithm would start by shrinking the array from the left until it ends up with the subarray [100, 1, 1, 1, 1, 1], which has a length of 6. However, the actual minimum length subarray with a sum greater than or equal to 7 is [100], which only has a length of 1.

Instead, consider using a sliding window approach. This method involves maintaining a “window” of elements that shifts rightward through the array. The window expands to the right as long as the sum is less than the target, and it shrinks from the left when the sum is greater than or equal to the target. Throughout this process, you keep track of the minimum window size where the sum is greater than or equal to the target. This ensures that you consider all possible subarrays.

The solution says that the time complexity of O(n^2) is accepte: however, it gives time limit exceeded for my solution: as far as I understand my code should have quadratic time complexity in n: am I missing something?

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class Solution:
	def minSubArrayLen(self, target: int, nums: List[int]) -> int:

    if  sum( nums ) < target:
        return 0
  
    min_sofar = int( 1e5 ) 

    for st in range( len( nums ) ):
        current_sum = nums[st]
        current_len = 1
        end  = st

        while current_sum < target and current_len < min_sofar:
            end += 1

            if end  >= len( nums ):
                break

            current_sum += nums[end]
            current_len += 1

        if current_sum >= target:
            if current_len < min_sofar:
                min_sofar = current_len

    return min_sofar
    return min_len_sofar

Yes, your solution indeed has a time complexity of O(n^2). However, LeetCode has a certain time limit for each test case of a problem. If your solution exceeds this limit, it results in a Time Limit Exceeded error. Quadratic time complexity can often be too slow for larger inputs, and it seems that’s the case for your solution here.

In this problem, a quadratic time complexity solution might not be efficient enough for larger test cases because it will check every subarray in the worst-case scenario.

The problem can be solved in O(n) time complexity using a sliding window approach. In the sliding window approach, you start with a window at the leftmost part of the array and move it to the right, while keeping track of the sum of the elements in the window. If the sum is less than the target, you expand the window from the right. If the sum is more than or equal to the target, you shrink the window from the left and update the minimum length of the window that has been found so far. This approach will traverse the array only once, hence the time complexity is linear.

It’s a good practice to think about the time and space complexity of your solution before implementation. It can help you understand whether your approach will be efficient enough for all test cases.

Follow up: If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log(n)).

Is this approach related to prefix sum + binary search ? did this with another complexity approach

The follow-up part is indeed asking for a prefix sum + binary search approach.

Here is a high-level description of this approach:

1. First, construct a prefix sum array where each element is the sum of all the elements before it, inclusive. This is O(n).
2. Then, for each prefix sum, use binary search to find the smallest subarray length that can be achieved for each index in the array. This is O(n log(n)).

So the time complexity of this approach is O(n) for the prefix sum construction + O(n log(n)) for the binary search, hence it’s O(n log(n)) in total.

This approach is different from the sliding window approach, which has a time complexity of O(n). This follow-up question encourages you to think about the problem from a different angle and use a different algorithmic paradigm to solve it. It’s a good practice to solve problems in multiple ways, as it broadens your understanding and ability to tackle similar problems in the future.

It’s important to note that this O(n log n) approach may not work if the array contains negative numbers, as the prefix sums would not be sorted in that case, and we cannot apply binary search on an unsorted array. So, this approach is only valid for arrays with non-negative numbers.

why in the case of target = 213, nums = [12,28,83,4,25,26,25,2,25,25,25,12] answer is 8? Why not 7?

In the case of target = 213 and nums = [12,28,83,4,25,26,25,2,25,25,25,12], the minimum size subarray with a sum of at least 213 is indeed of size 7: [83,4,25,26,25,2,25], which sums to 214.

However, I suspect that if you’re getting an answer of 8 in some solution, it could be due to an off-by-one error or a misunderstanding in how the sliding window works.

In this problem, the subarray is a contiguous segment of the array. The goal is to find the smallest contiguous segment that sums to at least the target. This problem is usually solved with a sliding window or two pointers approach.

Ensure to check the window sum at each step and adjust the window accordingly. If the sum is smaller than the target, expand the window. If the sum is equal to or larger than the target, contract the window from the left and update the minimum length if it’s smaller.

Always make sure to verify the implementation against test cases, and adjust as necessary based on the results.

target = 213 nums = [12,28,83,4,25,26,25,2,25,25,25,12]

Output = 7 Expected = 8

83 + 25 X 4 + 28 + 2 = 213.

That is 7 numbers that are equal to the target.

The smallest subarray in this case is indeed of length 7, which includes the numbers [83,4,25,26,25,2,25] summing to 214, which is greater than the target 213.

If the expected output is 8, it could be due to a specific implementation of the solution or a mistake in the problem setup. However, based on the problem statement, the smallest subarray with a sum of at least the target is the correct answer, and in this case, that would be of length 7.

I would suggest revisiting the source of this problem or the solution that gave an expected output of 8, as there might be some additional constraints or requirements not included in the problem statement as we know it.

If I have already figured out the O(N) solution, why will I try to think of a O(NlgN) solution?

It’s generally true that a solution with lower time complexity is preferable. However, the purpose of asking for a solution with higher time complexity could be pedagogical. That is, the exercise may be designed to make you explore and understand other possible approaches to solving the problem.

Moreover, even if a solution has a higher time complexity, it could have other desirable characteristics. For instance, it could require less memory, be easier to implement, or be more robust in certain scenarios. So, the goal of the exercise might be to have you think about these trade-offs.

However, in the case of this problem, considering that the O(n) solution is more efficient and doesn’t have particular downsides, there might be a misunderstanding or a typo in the question’s follow-up instruction. Generally, you’d want to find a more efficient solution, not a less efficient one.

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if(nums[0]==target)
	return 1;

    int left=0;
    int right=1;
    int size=Integer.MAX_VALUE;
    int sum=nums[left]+nums[right];
    while(right < nums.length-1)
    {
        if(sum<target)
        {
            right++;
            sum=sum+nums[right];
        }
        else if(sum>=target)
        {
            if(((right-left)+1)<size)
            {
                size=right-left+1;
            }
            sum=sum-nums[left];
            left++;
        }
    }

    if(size==Integer.MAX_VALUE)
    return 0;


    return size;

}
can any one explain how to remove the array out of bound exccption in this problem;

The array out of bounds exception happens when an array is accessed with an index that is out of its bounds, which are 0 to (array length - 1). In your code, an array out of bounds exception might occur because you are incrementing the `right` pointer and immediately using it to access the `nums` array without checking if it's within the array bounds.

You can prevent the array out of bounds exception by rearranging your code slightly. You can increment the `right` pointer only after checking it won't exceed the array bounds. Also, to make sure all sums are calculated correctly, check if the `sum` is greater than or equal to `target` in a separate while loop after the first one. 

Here is a version of your code that shouldn't throw an array out of bounds exception:

```java
if(nums[0] == target)
    return 1;

int left = 0;
int right = 0;
int size = Integer.MAX_VALUE;
int sum = nums[left];

while(right < nums.length) {
    if(sum < target) {
        right++;
        if (right < nums.length) {
            sum = sum + nums[right];
        }
    } else {
        if((right - left + 1) < size) {
            size = right - left + 1;
        }
        sum = sum - nums[left];
        left++;
    }
}

while(sum >= target) {
    if((right - left + 1) < size) {
        size = right - left + 1;
    }
    sum -= nums[left];
    left++;
}

if(size == Integer.MAX_VALUE)
    return 0;

return size;

This code now checks that right does not exceed the length of the nums array before incrementing it and accessing nums[right]. It also has an additional while loop to check if decreasing the sum can find a shorter subarray that still satisfies the problem constraints.

For finding the length of current sub-array (j-1): If you have 2 pointer i and j with j >= i and you want to figure out the length of the subarray:

len = j-i+1 but here we have only

len = j-i This is because, we have already incremented j by 1 position but not i, so no need to add 1.

Yes, you’re absolutely correct.

The sliding window technique is based on two pointers i and j where j is usually the fast pointer and i is the slow one. The subarray’s length is indeed j - i + 1 in general, because array indexing starts at 0, so you need to add 1 to include both ends of the subarray.

However, in your specific situation, j has already been incremented in the loop before calculating the length. Hence, there’s no need to add 1 to find the length, it will indeed be j - i.

Always remember to consider the specifics of your algorithm’s logic and the actual position of your pointers when calculating such lengths or other dependent values.

We need to understand that inner loop is not dependent to outer loop. That means both i and j pointer can go till n and they are independent so it is n+n operation hence O(n).

If inner loop is dependent to outer loop value then we should think:

If for each outer loop value, how many times inner loop operation will be executed. Sum them up and get the time complexity n*(Total independent operations)

Yes, you’re exactly correct. In many problems that use the sliding window technique or two pointers technique, the two pointers are independent, and both can traverse the entire array, making the time complexity O(2n) or, more simply, O(n).

Contrast this with a nested for-loop structure where the inner loop is dependent on the outer loop. For example:

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for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
        // some constant-time operation
    }
}

In this case, for each value of i, the inner loop executes n - i times. Summing over all i gives a time complexity of O(n^2).

But for the sliding window/two pointers technique, the two loops are not nested and the inner loop doesn’t reset each time the outer loop increments, so the time complexity is linear, not quadratic.

The pointer i moves no more than n steps. This is because i does not start from 0 for each iteration, instead it starts from its position from the last iteration.

Exactly, in a sliding window approach, the pointer i does not reset to 0 for each iteration but rather continues from its last position. This is why the time complexity is linear O(n), not quadratic.

In every iteration of the main loop, we are moving the i pointer to the right until the sum of elements in the current window is less than the target. Then we try to extend the window by moving the j pointer to the right.

Hence, even though there are two pointers, both i and j traverse the array only once, making the total number of operations proportional to the size of the array. This leads to a time complexity of O(n). This method is more efficient than a brute force method that would have a time complexity of O(n^2), where n is the size of the array.

The from pointer only moves n times, so it should be O(n)

Yes, you’re correct.

In the sliding window approach, both the start and end pointers of the window only traverse the array once. The start pointer moves from left to right and the end pointer also moves from left to right. So, even though you are using two pointers, the overall time complexity is O(n) as each pointer moves a total of n times, and n + n operations can be simplified to O(n) in Big O notation.

This makes the sliding window approach quite efficient for problems like this one, where you’re looking for a specific sum within a contiguous subarray.

It seems like that two while loops give O(N^2). But actually, for every j, i is not starting from 0. Pointer i will not go back, so in the worst case, pointer j moves forward n times and pointer i also moves forward n times.

Is this one a DP solution? Or is there any general category it fits?

This problem does not fall under the Dynamic Programming (DP) category. Dynamic programming is typically used when the problem can be broken down into overlapping sub-problems that can be solved independently.

The solution to this problem is best characterized as a Sliding Window or Two Pointers approach. The idea is to maintain a window in the array and adjust its size and position based on the sum of the elements in the window compared to the target.

The two pointers represent the left and right boundaries of the window. We begin with the window at the leftmost side of the array, and then we adjust the boundaries as follows:

• If the sum of the window’s elements is less than the target, we expand the window to the right (i.e., increase the right pointer).
• If the sum is equal to or greater than the target, we try to shrink the window from the left (i.e., increase the left pointer), and update the minimum length if the current window’s length is less than the previously found minimum.

This process is repeated until the entire array has been traversed by the right pointer.

While Dynamic Programming problems usually involve making decisions based on previously computed states or values, Sliding Window problems like this one involve continuously adjusting a range in an array or list based on certain conditions.

I also used 2 pointer but in my case only 13/15 cases are passing. Can some one please tell me what i am doing wrong ?

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public class Solution {
    public int minSubArrayLen(int s, int[] nums) {
        int sum =0;
        int len = nums.length;
        int minLen = len;
        for(int el:nums) {
            sum += el;
        }
        if(sum < s) {
            return 0;
        }
        int ptr1 = 0;
        int ptr2 = len - 1;
        while(ptr1 < ptr2) {
            if(nums[ptr1] <= nums[ptr2]) {
                if(sum - nums[ptr1] >= s) {
                 sum = sum - nums[ptr1];
                 ptr1++;
                 minLen--;   
                 continue;
                } 
                return minLen;
            } else {
                if(sum - nums[ptr2] >= s) {
                 sum = sum - nums[ptr2];
                 ptr2--;
                 minLen--; 
                 continue;
                } 
                return minLen;
            }
        }
        return minLen;
    }
}

Your approach is to remove elements from either end of the array depending on which one is smaller, while making sure that the sum of the elements in the current window is still equal to or greater than the target. This is not the correct approach because the minimal length subarray isn’t always located at the two ends of the array.

For example, consider this case: s = 15, nums = [5, 1, 2, 3, 10, 7]

The minimal length subarray summing to at least 15 is [10, 7], which is in the middle of the array, but your algorithm will return [5, 1, 2, 3, 10], since it starts removing elements from the ends.

The sliding window approach to this problem is to keep a window of elements from the array, always try to reduce its size from the left end while its sum is still equal to or greater than s, and when the sum becomes less than s, expand the window from the right end. Here’s a correct version of the code:

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public class Solution {
    public int minSubArrayLen(int s, int[] nums) {
        int sum = 0;
        int len = nums.length;
        int minLen = Integer.MAX_VALUE;

        int left = 0;
        int right = 0;
        while (right < len) {
            sum += nums[right];
            while (sum >= s) {
                minLen = Math.min(minLen, right - left + 1);
                sum -= nums[left];
                left++;
            }
            right++;
        }
        return minLen == Integer.MAX_VALUE ? 0 : minLen;
    }
}

This code initializes two pointers, left and right, to the start of the array. It moves the right pointer to expand the window and increase the sum until it’s no longer less than s. Then, it tries to shrink the window from the left by moving the left pointer and decreasing the sum, as long as the sum is still equal to or greater than s. The minimum length of the window during these operations is kept in minLen. If no valid window is found, it returns 0.

Hi, I love the clean code.However there may be an improvement. In the case that array has one element which is larger than s, like ([4, 5, 6], s = 3 ) we can break out of the entire loop directly when i = 0, your code would still do the iteration ,which is unnecessary.

You’re correct that in cases where a single element in the array is greater than or equal to the target s, we could optimize by stopping the iteration as soon as we find such an element.

Here’s how you could modify the code to incorporate this improvement:

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public class Solution {
    public int minSubArrayLen(int s, int[] nums) {
        int sum = 0;
        int len = nums.length;
        int minLen = Integer.MAX_VALUE;

        int left = 0;
        int right = 0;
        while (right < len) {
            sum += nums[right];
            while (sum >= s) {
                minLen = Math.min(minLen, right - left + 1);
                // If a single element is larger than or equal to s, return 1.
                if(minLen == 1) {
                    return 1;
                }
                sum -= nums[left];
                left++;
            }
            right++;
        }
        return minLen == Integer.MAX_VALUE ? 0 : minLen;
    }
}

In this version of the code, after updating minLen, we check if minLen is equal to 1. If it is, we know we’ve found a single element in the array that’s greater than or equal to s, so we return 1 immediately. This improvement optimizes for the specific scenario you mentioned and will reduce the time complexity in such cases from O(n) to O(k), where k is the position of the first single element that’s larger than or equal to s.

Just consider an extreme case, what if j-i =Integer.MAX_VALUE? so the min should be Integer.MAX_VALUE and not be zero, is that right?

You’re correct that in an extreme case where the array’s length is equal to Integer.MAX_VALUE and no subarray sum is greater than or equal to the target, j-i would be Integer.MAX_VALUE. However, since this difference represents the length of the subarray, it’s not possible for the length of a subarray to be Integer.MAX_VALUE unless the whole array is that long, and still, no subarray sum matches the target.

In the context of this problem, we initialize minLen as Integer.MAX_VALUE as a way to easily update minLen the first time we find a sum that’s greater than or equal to s. If minLen remains Integer.MAX_VALUE at the end of the function, it means we’ve never found a sum greater than or equal to s, so we return 0.

The return value of 0 is meant to indicate that no such subarray exists, following the problem’s instruction: “If there isn’t one, return 0 instead.” If you wanted to indicate that no subarray met the condition by returning Integer.MAX_VALUE, you’d have to confirm that this is acceptable based on the context or requirements you’re working within.

I love this solution but i think we can add a small improvement here:

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public int minSubArrayLen(int target, int[] nums) {
	int minCount = Integer.MAX_VALUE;
	int sum = 0; 
	int start = 0;
	int end = 0;
	while(end < nums.length) {
		if(nums[end] >= target) return 1; // This will exit loop if a number greater than or equal to target is spotted. We need not iterate further in that case
		sum += nums[end++];
		while(sum >= target){
			minCount = Math.min(minCount, end-start);
			sum -= nums[start++];
		}
	}
	return minCount > nums.length ? 0: minCount;
}

Yes, adding the line if(nums[end] >= target) return 1; is indeed a good optimization for the cases where an individual number in the array is greater than or equal to the target. It’s a clever way to short-circuit the evaluation and quickly provide an answer in those scenarios.

Your solution still retains its O(N) time complexity because the early return only happens in specific cases, but in those cases, it can provide an answer much faster than the standard solution would. Overall, it’s a smart way to enhance the efficiency of the solution for those specific inputs. Well done!

can someone help me finding why i am getting wrong output

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class Solution {
public:
	int minSubArrayLen(int k, vector& A) {
		int i=0,j=0,sum=0;
		int maxi=INT_MIN;
		int N=A.size();
		while(j<A.size())
		{
			sum+=A[j];
			if(sum<k)
			{
				j++;
			}
			else if(sum==k)
			{
				maxi=max(maxi,j-i+1);
				j++;
			}
			else if(sum>k)
			{
				while (sum > k) {
					sum -= A[i];
					i++;
				}
				j++;
			}

		}
		return maxi;
	}
};

I believe the issue lies in how you’re handling the maximum length of the subarray (maxi). The question asks for the minimal length of a subarray with sum at least s, not the maximum length.

So, instead of finding the maximum length (maxi=max(maxi,j-i+1)), you should be trying to find the minimum length. Initialize maxi to a large value (e.g., INT_MAX) and then update it as maxi=min(maxi,j-i+1) when the sum is greater than or equal to k.

In addition, your code does not handle the case when no qualifying subarray can be found. If maxi remains INT_MAX (i.e., no subarray with a sum at least k was found), you should return 0 as per the problem’s instructions.

Here is the modified version of your code:

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class Solution {
public:
    int minSubArrayLen(int k, vector<int>& A) {
        int i=0,j=0,sum=0;
        int minLen=INT_MAX;
        int N=A.size();
        while(j<A.size()) {
            sum += A[j];
            while(sum >= k) {
                minLen = min(minLen, j-i+1);
                sum -= A[i];
                i++;
            }
            j++;
        }
        return minLen == INT_MAX ? 0 : minLen;
    }
};

In this version, we start by moving the j pointer and adding up the numbers. When the sum is at least k, we start moving the i pointer and subtracting the numbers until the sum is less than k again. This way, for each j, we’re finding the smallest i such that the sum of the subarray from i to j is at least k. This gives us the shortest subarray ending at j with a sum at least k. We keep track of the smallest length we find in minLen.

If no such subarray is found (i.e., minLen is still INT_MAX), we return 0.

why are we returning 0 at the end when its equal to max ?

In this problem, returning 0 signifies that no subarray with a sum greater than or equal to the target was found.

The reason we check if the minimum length is equal to Integer.MAX_VALUE or INT_MAX at the end is because we initially set the minimum length to be Integer.MAX_VALUE or INT_MAX (a placeholder value). If no valid subarray is found throughout the algorithm, this initial value remains unchanged. Hence, before returning the result, we check if our result is still Integer.MAX_VALUE or INT_MAX, and if it is, we return 0 as per the problem’s instructions (indicating that no valid subarray was found).

Therefore, we aren’t returning 0 because it’s equal to the maximum, but because no valid subarray was found during the execution of the algorithm.

break when min is 1

Yes, you’re correct!

In the context of this problem, if you ever find a subarray of length 1 where that single element is greater than or equal to the target, you can immediately break out of your loop because you won’t find a smaller subarray that meets the criteria.

This can help to optimize your solution by avoiding unnecessary computations after you’ve already found the smallest possible valid subarray.

Here’s an example of how you could integrate that into a code snippet:

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for end in range(len(nums)):
    sum += nums[end]
    while sum >= target:
        min_len = min(min_len, end - start + 1)
        if min_len == 1:  # We found the smallest possible subarray
            return 1
        sum -= nums[start]
        start += 1

In this example, nums is your input list of numbers and target is the target sum. This is a sliding window approach where start and end are the indices of the start and end of your current subarray, and sum is the sum of the current subarray. The variable min_len keeps track of the minimum length of a subarray found so far.

Indeed the 2 pointer solution is the most intuitive one. Here’s my take

We use a sliding window here to find the best solution. p1 and p2 are the window’s left and right edges. Expand the right edge (p2) and keep track of the sum of the current window. If the sum exceeds target, we have a possible solution. Store it in ‘min’ if its smaller than any previously found minimum window size. Now there’s no point in expanding the window further as its not going to be better than the window that we just now found. So start shrinking the left edge of the window (p1) and decrement the sum as we drop elements from the window. Check if sum is still >= target after shrinking the window. If yes, we found a better possibly solution. Store it in ‘min’ is its indeed better i.e. size of the current window is lesser than the minimum window found till now. Return min after we have covered the complete array.

Indeed the 2 pointer solution is the most intuitive one. Here’s my take

We use a sliding window here to find the best solution. p1 and p2 are the window’s left and right edges. Expand the right edge (p2) and keep track of the sum of the current window. If the sum exceeds target, we have a possible solution. Store it in ‘min’ if its smaller than any previously found minimum window size. Now there’s no point in expanding the window further as its not going to be better than the window that we just now found. So start shrinking the left edge of the window (p1) and decrement the sum as we drop elements from the window. Check if sum is still >= target after shrinking the window. If yes, we found a better possibly solution. Store it in ‘min’ is its indeed better i.e. size of the current window is lesser than the minimum window found till now. Return min after we have covered the complete array.

Can someone please help me understand this:

In most of the top solutions: Once we get a valid window, we still keep looking for subarrays of equal or greater lengths. Shouldn’t we just do searches for subarrays of smaller lengths after we have a valid subarray?

Here’s my solution, where I tried to take care of this. Please let me know what you think of it.

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/**
 * @param {number} s
 * @param {number[]} nums
 * @return {number}
 */
var minSubArrayLen = function (target, nums) {
	let s = 0; 		// start idx
	let e = 0;		// end idx
	let windowSum = 0;  // window sum

	// Get a window where the condition sum >= target is satisfied. For this I keep incrementing the end idx, untill we get one such window
	while (e < nums.length) {
		windowSum += nums[e];
		if (windowSum >= target)
			break;
		e++;
	}

	// Now if we reach end of array and the total sum is still not greater/ equal to target sum, then no subarray exists and we return 0
	if (windowSum < target) 
		return 0;

	/**
		Here, as we have got a valid window, we need to look at windows having smaller lengths. And this can be done by:
		i. Incrementing just the start idx, whenever we get a new valid window. (This is bound to be shorter that the previous window)
		ii. Incrementing both the start and end idx, untill we reach the end of array.
	 */
	
	let res = e - s + 1;
	while (e < nums.length) {
		if (windowSum >= target) {
			res = e - s + 1;
			windowSum -= nums[s];
			s++;
		} else {
			windowSum -= nums[s];
			s++; e++;
			if (e < nums.length)
				windowSum += nums[e];
			else
				break;
		}
	}
	return res;
};

Your solution is correct and you have a good understanding of the problem.

When we have a valid window, we indeed want to search for subarrays of smaller lengths. This is why we increment the start index s to shrink the current subarray once we have a valid subarray whose sum is greater than or equal to the target. This makes the current subarray smaller.

However, there is no guarantee that the next subarray with a sum greater than or equal to the target will also be smaller. This is why we also have to increment the end index e to explore larger subarrays when the current subarray’s sum falls below the target.

It’s a balance of searching smaller subarrays by incrementing s and ensuring we find all valid subarrays by incrementing e. Your implementation takes care of both these aspects well.

Your solution is an excellent implementation of the sliding window pattern and it has a time complexity of O(n), which is optimal for this problem. The logic in your code is sound and the comments make it easy to understand what each part of the code is doing. Good job!

Can anyone explain why this approach doesn’t work here :(

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class Solution {
    public int minSubArrayLen(int s, int[] nums) {
        int len = nums.length;
        if(len == 0) return 0;
        int[] sumDP = new int[len];
        sumDP[0] = nums[0];
        for(int i = 1; i < len; i++)    sumDP[i] = sumDP[i-1] + nums[i];
        if(sumDP[len-1] < s)    return 0;
        int leftPtr = 0, rightPtr = len-1;
        while( leftPtr <= rightPtr){
            int currSum = sumDP[rightPtr] - (leftPtr == 0 ? 0 : sumDP[leftPtr-1]);
            if(currSum < s)     return rightPtr - leftPtr + 2;
            if(nums[leftPtr] <= nums[rightPtr])     leftPtr++;
            else                                    rightPtr--;
        }
        return 1;
    }
}

The approach you’re using is a variant of the two-pointer method where you increment the left pointer if the left element is smaller, or decrement the right pointer if the right element is smaller. This approach does not work for this problem because we’re not interested in the size of the elements themselves, but the sum of the elements in the subarray.

The problem requires us to find the minimum size subarray with sum at least equal to s. By incrementing and decrementing pointers based on the value of the elements rather than their sum, you’re not necessarily minimizing the size of the subarray with sum at least s.

For example, consider the array [2, 1, 6, 3] and s = 9. Your approach would first compare 2 and 3, decide to move the left pointer, compare 1 and 3, decide to move the left pointer again, and finally compare 6 and 3, decide to move the right pointer. This leads to a subarray [6], which does not have a sum at least s.

What you should do instead is to initialize both pointers at the start of the array, then increment the right pointer until the sum is at least s, and then increment the left pointer while the sum is still at least s. This way, you always maintain the smallest valid subarray. If you reach the end of the array with the right pointer, you can stop the algorithm. This approach will ensure that you always find the smallest subarray with sum at least s if such a subarray exists.

Can someone explain why the second while loop condition is sometimes “if the subarray/substring is valid” and sometimes “if the subarray/substring is invalid”? For example, in this question, we start shrinking the window when sum >= s (meet requirement). In problem 1208, we start shrinking the window when “cost > maxCost” (this is when the substring first becomes invalid).

The difference between the two conditions in the second while loop is due to the nature of the problems. In the problem of minimum size subarray with a sum at least s (here), the goal is to find the smallest valid subarray, so we continue shrinking the window (i.e., move the left pointer forward) while the sum of the subarray is still valid (i.e., sum >= s). This way, we always maintain the smallest possible subarray with a sum at least s.

On the other hand, in problem 1208 (“Get Equal Substrings Within Budget”), the task is to find the largest length of a substring that consists of equal letters when both operations (i.e., move the left pointer forward and the right pointer forward) are allowed, but the total cost should not exceed maxCost. Therefore, we continue expanding the window (i.e., move the right pointer forward) until the total cost is over maxCost. At this point, the substring becomes invalid, so we start shrinking the window (i.e., move the left pointer forward) until the substring becomes valid again (i.e., cost <= maxCost). This way, we always maintain the largest possible substring with a total cost not exceeding maxCost.

In both cases, the second while loop allows us to adjust the size of the subarray or substring to meet the problem’s requirement. The specific condition depends on whether we want to maximize or minimize the size of the subarray/substring and what makes a subarray/substring valid or invalid.

Anyone has idea, how to solve if negative number is also included in o(n)?

When the array contains negative numbers, the sliding window approach may not work. This is because, when you have a negative number in your array, adding it to your sum could decrease the sum instead of increasing it. So, simply sliding the window whenever the current sum is greater than or equal to the target doesn’t guarantee you’ve found the smallest subarray that sums to the target.

Here’s a broad idea to tackle this using a prefix sum array and binary search:

1. Compute a prefix sum array: Iterate through the input array, adding each element onto a running total, and store these cumulative sums in an array. The prefix sum array will have the same size as the input array.

2. For each prefix sum prefixSum[i], use binary search to find the smallest j such that prefixSum[j] - prefixSum[i] >= target.

3. Keep track of the smallest such j - i over all i.

This approach works because the prefix sum array is monotonically non-decreasing (thanks to non-negative integers in the array). The time complexity of this approach is O(n log n) because we perform a binary search for each of the n elements in the array.

Remember that dealing with negative numbers usually adds a level of complexity to problems that involve finding sums, because you have to consider whether including a negative number would bring you closer to or further from your target sum. In some cases, it may be easier to simply filter out the negative numbers before processing the array, if the problem allows for this.

Just a thought: we can init min to array length + 1 since min can’t be larger than array length anyway.

Yes, initializing min to array length + 1 is a clever trick to simplify the code. This way, you don’t have to handle the special case where no valid subarray is found separately at the end of the function. Instead, you can just return min directly, because if min is still equal to array length + 1 at the end of the function, it means no valid subarray was found, and according to the problem specification, in this case you should return 0.

Here’s how this trick can be used:

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public int minSubArrayLen(int target, int[] nums) {
    int min = nums.length + 1;
    int sum = 0;
    int left = 0;

    for (int right = 0; right < nums.length; right++) {
        sum += nums[right];
        while (sum >= target) {
            min = Math.min(min, right - left + 1);
            sum -= nums[left++];
        }
    }

    return min == nums.length + 1 ? 0 : min;
}

Note that we don’t need to check if min has changed at the end of the function. If min is still equal to array length + 1, it means no valid subarray was found, and we return 0. If min has changed, it means a valid subarray was found, and we return its length.

I think you don’t need to obtain the min during the while loop. you can put that line after the loop and this can improve the efficiency a lot

min = Math.min(min, end - start + 2)

In this problem, the purpose of the min operation inside the while loop is to keep track of the smallest subarray length that satisfies the condition (sum >= target) as the array is being traversed. The min operation is necessary in each iteration of the while loop, as the length of the subarray that satisfies the condition can decrease as elements are removed from the sum.

Here’s how the process works:

• When the sum becomes greater than or equal to the target, the while loop starts executing.
• Inside the loop, the size of the current window is compared with the smallest window seen so far, and the smaller one is retained. This is done by the min operation.
• An element is then removed from the start of the current window,

I think the confusion might arise from the statement “min = Math.min(min, end - start + 2)”. The “+2” may seem like we’re expanding the window size after finding a valid window. However, remember that “end” and “start” are indices, and in zero-based index languages like Java or Python, the length of a subarray from “start” to “end” (inclusive) is “end - start + 1”. So “end - start + 2” is incorrect if you’re trying to calculate the current window size.

The “min = Math.min(min, end - start + 1)” operation should be done each time a valid window is found (i.e., the sum is greater than or equal to the target), because we want to keep track of the smallest valid window.

Here’s the correct version of that part:

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while (sum >= target) {
    min = Math.min(min, end - start + 1);
    sum -= nums[start++];
}

Now, regarding efficiency: Since the “min” operation and the “sum -= nums[start++]” operation are both O(1), it wouldn’t significantly affect the efficiency whether the “min” operation is inside the while loop or after it. The time complexity of the entire algorithm is still O(n).

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void helper(int i, vector & Nums, vector < vector > & h, vector & k) {
  if (i == Nums.size()) {
    h.push_back(k);
    return;
  }
  k.push_back(Nums[i]);
  helper(i + 1, Nums, h, k);
  k.pop_back();
  helper(i + 1, Nums, h, k);

}
int minSubArrayLen(int target, vector < int > & nums) {
  vector < vector < int >> ans;
  vector < vector < int >> G;
  vector < int > h;
  helper(0, nums, G, h);
  int size = INT_MAX;
  for (int i = 0; i < G.size(); i++) {
    int sum = 0;
    for (int j = 0; j < G[i].size(); j++) {
      sum = sum + G[i][j];
    }
    if (sum >= target) {
      int size1 = G[i].size();
      size = min(size, size1);
    }
  }
  if (size == INT_MAX) {
    return 0;
  }
  return size;

}

[12,28,83,4,25,26,25,2,25,25,25,12] output:7 expected:8

Why i am getting the wrong answer in this case? Please anybody tell me .

The code you’ve shared is an implementation of a brute force solution to find the smallest length subarray with a sum greater than or equal to a given target. This approach generates all possible subarrays and checks each one to see if its sum is greater than or equal to the target.

The problem is that the code doesn’t guarantee that the minimum length subarray found is contiguous (i.e., its elements are in consecutive positions in the original array). In your code, you are considering all subsets of the array, not just the contiguous subarrays.

Let’s consider the array [12,28,83,4,25,26,25,2,25,25,25,12] and the target value 213. The smallest contiguous subarray that sums up to 213 or more is [83,4,25,26,25,2,25,25], which is 8 elements long. However, your code might return a non-contiguous subarray [83,4,25,26,25,2,25], which is 7 elements long.

You need to change your approach to solve this problem. An efficient solution is to use a two-pointer (or sliding window) technique. This approach maintains a window of elements and adjusts its size and position based on the sum of its elements compared to the target. This guarantees that you are only considering contiguous subarrays. Here’s a simple implementation:

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int minSubArrayLen(int target, vector<int>& nums) {
    int n = nums.size();
    int left = 0, sum = 0, minLen = INT_MAX;
    for (int right = 0; right < n; right++) {
        sum += nums[right];
        while (sum >= target) {
            minLen = min(minLen, right - left + 1);
            sum -= nums[left++];
        }
    }
    return (minLen != INT_MAX) ? minLen : 0;
}

In this code, we continuously add elements to sum until it is greater than or equal to target. Once sum is large enough, we subtract elements from the left side of the window until sum is less than target again. We repeat this process, maintaining the smallest window size (minLen) seen so far. If no subarray meets the condition, the function returns 0. This approach has a time complexity of O(n), making it much more efficient for large inputs.

Tidbits

Keep a window of sum >= target and track its size

At first glance, it seems to be Two Pointers approach

I used greedy algorithm, but I got the wrong answer!

Python prefix sum and 2 pointers approach. Minimise the distance between the two pointers

A subarray is a contiguous non-empty sequence of elements within an array

The goal is to find subarray (continguous subsequence), and your answer skips some numbers

The sliding window is 2 pointer. One for the start and one for the end

Damn, missed the “greater than” in this question

I set the window length to array length.

The from pointer only moves n times, so it should be O(n)

break when min is 1

Similar Problems

1. Maximum Subarray (LeetCode 53): This problem also involves finding a subarray with certain properties (maximum sum), which requires a similar sliding window approach.

2. Subarray Product Less Than K (LeetCode 713): In this problem, the objective is to find the number of subarrays whose product is less than a given number. The sliding window technique is employed here as well.

3. Subarray Sum Equals K (LeetCode 560): This problem requires finding the number of subarrays whose sum equals a certain value. Similar to the minimum size subarray problem, the problem-solving approach involves tracking sums of subarrays.

4. Maximum Size Subarray Sum Equals k (LeetCode 325): The goal here is to find the maximum size subarray that sums up to k. This is the reverse of the minimum size subarray problem but requires a similar approach.

5. Minimum Window Substring (LeetCode 76): This problem requires finding the smallest window in a string that contains all characters of another string. It involves a variant of the sliding window technique.

6. Longest Substring Without Repeating Characters (LeetCode 3): Here, you need to find the longest subarray (substring, in this case) with no repeating characters. This problem utilizes the sliding window approach to maintain the property of having unique characters.

7. Longest Substring with At Most Two Distinct Characters (LeetCode 159): This problem requires identifying the longest substring with at most two distinct characters. It uses a variant of the sliding window technique similar to the given problem.

8. Fruit Into Baskets (LeetCode 904): This problem requires finding the longest sequence of two types of fruits that can be collected. It employs the sliding window technique to maintain the property of having at most two types of fruits.

9. Shortest Unsorted Continuous Subarray (LeetCode 581): This problem involves finding a continuous subarray that if sorted, results in the entire array being sorted. The problem-solving strategy involves identifying subarray properties, like the given problem.

10. Maximum Average Subarray I (LeetCode 643): This problem asks for the contiguous subarray of given length that has the maximum average. The sliding window technique can be employed to solve this problem efficiently.

All these problems involve finding or working with subarrays that have certain properties. The sliding window technique is commonly used to solve these problems efficiently, making them similar to the minimum size subarray problem.