Maximum Subarray

Invariant

What are things that does not vary in this problem?

Search for invariant
- Contiguous subarray =>
- Largest sum =>
- At least one number in the contiguos subarray =>
Which example should we consider first?
- array of size 1 => return nums[0]
- It simplifies the problem
- array of size of 2 [-2,1] => 1 [-2,1,-3] => 1 input = [-2,1,-3,4] [-2,1,-3,4] [-2,1,-3] [1,-3,4] [-3,4] [4] ==> 4 [-2,1,-3,4,-1] [4] ==> 4
  [-2,1,-3,4,-1,2] ==> 5
  [-2,1,-3,4,-1,2,1] ==>6
  [-2,1,-3,4,-1,2,1,-5] ==>6
  [-2,1,-3,4,-1,2,1,-5,4] ==> 6
  Solution Outline
  1. sum = 0
  2. Keep a running sum
  3. Iterate through the array
  4. See if the new element will increase the sum update the sum with new sum
  5. You can also bump up the start index to drop off elements on the right of the window
  6. return sum

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# @param {Integer[]} nums
# @return {Integer}

def max_sub_array(nums)
    sum = nums[0]
    max = nums[0]

    if nums.size == 1
       return nums[0]
     end

     for i in (1..nums.size-1)
       sum = [nums[i]+sum, nums[i]].max
       max = [max, sum].max   
     end

     max
end

Invariant

What are things that does not vary in this problem?

Search for invariant
- Contiguous subarray =>
- Largest sum =>
- At least one number in the contiguos subarray =>
Which example should we consider first?
- array of size 1 => return nums[0]
- It simplifies the problem
- array of size of 2 [-2,1] => 1 [-2,1,-3] => 1 input = [-2,1,-3,4] [-2,1,-3,4] [-2,1,-3] [1,-3,4] [-3,4] [4] ==> 4 [-2,1,-3,4,-1] [4] ==> 4
  [-2,1,-3,4,-1,2] ==> 5
  [-2,1,-3,4,-1,2,1] ==>6
  [-2,1,-3,4,-1,2,1,-5] ==>6
  [-2,1,-3,4,-1,2,1,-5,4] ==> 6
  Solution Outline
  1. sum = 0
  2. Keep a running sum
  3. Iterate through the array
  4. See if the new element will increase the sum update the sum with new sum
  5. You can also bump up the start index to drop off elements on the right of the window
  6. return sum
  Running sum and max, 0 will work if all elements are positive. If negative elements, we cannot initialize to 0

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# @param {Integer[]} nums
# @return {Integer}

def max_sub_array(nums)
    sum = nums[0]
    max = nums[0]
         
     for i in (1..nums.size-1)
       sum = [nums[i]+sum, nums[i]].max
       max = [max, sum].max   
     end

     max
end

@param {Integer[]} nums

@return {Integer}

Invariant

What are things that does not vary in this problem?

Search for invariant
- Contiguous subarray =>
- Largest sum =>
- At least one number in the contiguos subarray =>
Which example should we consider first?
- array of size 1 => return nums[0]
- It simplifies the problem
- array of size of 2 [-2,1] => 1 [-2,1,-3] => 1 input = [-2,1,-3,4] [-2,1,-3,4] [-2,1,-3] [1,-3,4] [-3,4] [4] ==> 4 [-2,1,-3,4,-1] [4] ==> 4
  [-2,1,-3,4,-1,2] ==> 5
  [-2,1,-3,4,-1,2,1] ==>6
  [-2,1,-3,4,-1,2,1,-5] ==>6
  [-2,1,-3,4,-1,2,1,-5,4] ==> 6
  Solution Outline
  1. sum = 0
  2. Keep a running sum
  3. Iterate through the array
  4. See if the new element will increase the sum update the sum with new sum
  5. You can also bump up the start index to drop off elements on the right of the window
  6. return sum
  Running sum and max, 0 will work if all elements are positive. If negative elements, we cannot initialize to 0

=end

def max_sub_array(nums) sum = nums[0] max = nums[0] if nums.size == 1 return nums[0] end

 for i in (1..nums.size-1)
   sum = [nums[i]+sum, nums[i]].max
   max = [max, sum].max   
 end

 max

end

“Maximum Subarray” can be isomorphically mapped to the “Maximum Product Subarray”.

Reasoning:

Both problems require finding a contiguous subarray in a given array that has the maximum value. In the case of the “Maximum Subarray” problem, this maximum value is the sum of the elements in the subarray, while in the “Maximum Product Subarray” problem, it’s the product of the elements in the subarray.

The mapping is as follows: each number in the input array for the “Maximum Subarray” problem corresponds to the same position number in the input array for the “Maximum Product Subarray” problem. The sum of the maximum subarray maps to the product of the maximum product subarray.

This mapping isn’t exact because addition and multiplication are different operations with different properties (e.g., multiplication by zero always gives zero, which isn’t the case for addition). Therefore, the approach to solving these problems will be different, despite their isomorphism. For example, negative numbers can increase the product in the “Maximum Product Subarray” problem if there is an even number of them, whereas in the “Maximum Subarray” problem, negative numbers always decrease the sum. However, these problems are similar in their structure and how they’re approached, which is why we consider them as having an isomorphic relationship.

“Maximum Subarray” (LeetCode 53) problem is a relatively basic problem in dynamic programming. If you’re looking for even simpler problems to help build up your skills in array manipulation and basic algorithmic thinking, here are a few suggestions:

“1. Two Sum”: This problem asks for identifying two numbers in an array that add up to a specific target. It helps in understanding basic array manipulation and the use of hash maps.
“121. Best Time to Buy and Sell Stock”: This problem asks for finding the maximum profit that can be made by buying and selling a stock on different days. This helps in understanding how to track minima and maxima in an array.
“283. Move Zeroes”: This problem asks for moving all zeroes in an array to the end while maintaining the order of the other elements. It helps in understanding in-place array manipulation.
“448. Find All Numbers Disappeared in an Array”: This problem asks for identifying all elements from 1 to n that do not appear in the given array. This problem helps in understanding how to use an array to track presence or absence of elements.
“485. Max Consecutive Ones”: This problem asks for finding the maximum number of consecutive 1s in a binary array. This problem helps in understanding how to track and update a running count of elements satisfying a condition.

Once you are comfortable with these problems, you will have a better foundation to tackle the “Maximum Subarray” problem.

Problem Classification

It involves manipulation and analysis of an array, a common data structure. The sub-domain would be dynamic programming, as the problem seems to involve optimizing a certain parameter (the subarray sum) and potentially involves overlapping subproblems. Here’s how I would break it down:

Domain - Data Structures and Algorithms
- Sub-Domain - Dynamic Programming
What Components -
- Input: An integer array ’nums'.
- Output: The sum of a subarray within ’nums’, where this subarray has the largest sum compared to all other subarrays.
- Condition: A subarray is a contiguous part of an array.

The problem asks for the identification and calculation of a specific subarray with the highest sum. This falls under the category of optimization problems, where the goal is to find the best solution according to a specific criterion. Here, the criterion is the sum of the elements in the subarray.

Furthermore, this problem can also be classified as a dynamic programming problem. Dynamic programming is a strategy where we break down a problem into simpler subproblems, solve each of those subproblems just once, and store their results in case next time the same subproblem occurs. This is because a solution to this problem could involve solving for smaller subarrays and using these solutions to build up the solution for larger subarrays.

Visual Model of the Problem

Visualizing this problem can be helpful for understanding it better. Here’s how you might do it:

Number Line: The elements of the array can be thought of as changes in value along a number line (or y-axis of a graph), and each position in the array as a point on the x-axis. You’re looking for the sequence of changes (subarray) that results in the highest final value.
Graph: You could also create a running sum line graph. Plot the running total of the array values on the y-axis and the index of the array on the x-axis. The highest point on the graph represents the end of a subarray with maximum sum, and the increase from the lowest preceding point to this point represents the subarray with maximum sum.
Array: Another way is to look at the array itself. Draw the array as a sequence of boxes or cells. Then, as you consider different subarrays, draw brackets or highlight the subarrays you’re considering. The goal is to find the subarray with the highest total sum.

In all these visualizations, the key point to remember is that you’re trying to find a subarray (a continuous portion of the original array) that gives the maximum total sum. This can help you in understanding and developing algorithms like Kadane’s algorithm that solve the problem efficiently.

Case Analysis

nums = [1,2,3,4,5]

In the case where nums = [1,2,3,4,5] or any sorted array with only positive numbers, the subarray with the maximum sum is simply the entire array itself.

This is due to the key insight that a sorted array of positive numbers will always produce a higher sum when more elements are included. This property doesn’t hold true if the array includes negative numbers or is unsorted.

However, the problem of finding the subarray with the maximum sum doesn’t assume that the input array is sorted. Hence, more general solutions such as Kadane’s algorithm are needed for handling all types of input. The idea behind Kadane’s algorithm is that the maximum sum subarray ending at each position is either the maximum sum subarray ending at the previous position extended by the current element, or the current element itself (in case the sum becomes negative).

In the case of nums = [1,2,3,4,5], the maximum sum subarray at each position is simply the subarray from the start to the current position, which corresponds to the entire array.

nums = [5, 4, 3, 2, 1]

In the context of the maximum subarray problem, for any array with all positive numbers, the maximum sum will always be the sum of all elements, and the subarray will be the entire array. This is because adding any positive number to a sum will always make the sum larger.

nums = [-1, -1, -1, -1, -1]

In the case of nums = [-1, -1, -1, -1, -1], where all the numbers are negative, the maximum subarray sum would be the largest number in the array, which is -1 in this case.

nums = [-1, -5, -7, -2, -8]

The reason behind this is that for any subarray of size larger than 1, the sum would be more negative (hence smaller) than -1 because we are adding more negative numbers. Therefore, the maximum subarray sum is achieved by choosing the least negative number as a subarray of size 1.

In the case of nums = [-1, -5, -7, -2, -8], where all the numbers are negative, the maximum subarray sum would be the largest (least negative) number in the array, which is -1 in this case.

This is because the addition of any two or more negative numbers will always result in a value that is more negative (hence, smaller) than the least negative number. So no matter what subarray you choose, if it contains more than one element, the sum will be smaller than -1.

nums = [1,2,-4, -7, 5000, 5, 8, 9]

For the array nums = [1, 2, -4, -7, 5000, 5, 8, 9], the maximum subarray sum is the sum of the last four numbers: 5000 + 5 + 8 + 9 = 5022.

The maximum subarray sum in the array nums = [1,2,-4, -7, 5000, -5, 8, 9] is greatly influenced by the presence of the large positive number 5000.

No matter what subarray we consider, including 5000 in that subarray will almost certainly yield a larger sum due to its magnitude compared to other numbers in the array. This demonstrates a key aspect of Kadane’s algorithm: we are always looking for the ’local maximum’ at each step, which, in turn, informs our ‘global maximum’.

Also, the negative numbers preceding 5000 don’t diminish the value of including 5000 in the subarray, as Kadane’s algorithm effectively resets the current sum whenever a very large positive number follows a series of negatives.

However, even after encountering a large positive number like 5000, we still continue our exploration because there might be more positives ahead which could potentially increase the maximum subarray sum, like the 8 and 9 at the end.

Problem Breakdown and Solution Methodology

To solve this problem, we can use an algorithm called Kadane’s algorithm. This algorithm is particularly useful for problems involving the calculation of maximum (or minimum) sums in arrays, especially when the question involves subarrays.

Kadane’s algorithm works in the following way:

Initialize two variables, current_sum and max_sum to be the first element of the array.
Iterate over the array starting from the second element. For each element:
- Update current_sum to be the maximum of the current element and current_sum + current_element. This essentially means that at every step, we decide whether to extend the current subarray or start a new one. This is kind of like deciding whether to continue with a bad day or start afresh.
- Update max_sum to be the maximum of max_sum and current_sum. This is like keeping track of the best day we’ve had so far.
At the end of the loop, max_sum will hold the maximum subarray sum.

The time complexity of Kadane’s algorithm is O(n), as it requires a single pass through the array. It also only requires O(1) additional space, for storing current_sum and max_sum.

This approach, however, assumes that the array contains at least one positive number. If the array only contains negative numbers, the algorithm will return 0, because it will choose not to take any numbers. If we need to handle this case, we can add a check at the beginning to see if the maximum number in the array is negative. If it is, we can return that number.

Here is a sample implementation in Python:

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def maxSubArray(nums):
    if max(nums) < 0:
        return max(nums)
    
    current_sum = max_sum = nums[0]
    for num in nums[1:]:
        current_sum = max(num, current_sum + num)
        max_sum = max(max_sum, current_sum)
        
    return max_sum

And here is how it would work on a sample array [-2, 1, -3, 4, -1, 2, 1, -5, 4]:

For the first element (-2), current_sum and max_sum are both -2.
For the second element (1), current_sum is updated to 1 (since 1 > -2 + 1), and max_sum is updated to 1.
For the third element (-3), current_sum is updated to -2 (since -2 > 1 - 3), but max_sum remains 1.
For the fourth element (4), current_sum is updated to 4 (since 4 > -2 + 4), and max_sum is updated to 4.
This process continues for the rest of the array, with max_sum finally being updated to 6 when we reach the seventh element (1).

This shows that the subarray with the largest sum is [4, -1, 2, 1] with a sum of 6.

Problem Restatement

Let’s break down the problem statement:

The problem gives us an array of integers, which could have both positive and negative numbers. The task is to identify a subarray from this array, a subarray being a smaller array formed from the elements of the original array, keeping the order of elements. This subarray can be of any size, including the full size of the original array.

The goal is to find the subarray that has the largest sum of its elements compared to any other possible subarray of the given array. If multiple subarrays have the same largest sum, any one of them can be the answer. We then have to return this largest possible sum.

The constraints in this problem are that the input array will have at least one number in it, so we won’t have to deal with an empty array.

Understanding this, our main task becomes identifying a way to navigate through different subarrays and tracking the maximum sum we can obtain. This understanding is critical to approaching the problem effectively.

Abstract Representation of the Problem

Here’s how you can formulate an abstract representation of this problem:

Data: We have an array of integers. This array can be of any length and contain both positive and negative integers.
Objective: We need to find a subarray (which is a contiguous part of the original array) with the maximum sum.
Output: We will return the maximum sum we can get from any subarray.

This problem can be seen as a search problem over the space of all possible subarrays. The task is to search this space to find the subarray that yields the maximum sum.

An abstract representation might look something like this:

Array: We can visualize this array as a line of numbers, each number representing an element in the array.
Subarrays: Each subarray is a contiguous segment of this line. So the problem is to find the segment of the line (it could be the whole line, a part of it, or even a single element) which has the maximum total sum.
Sum: The sum of a subarray is the total of the numbers within the chosen segment.
Max Sum: We can imagine we’re trying to maximize the total value of the segment we pick. It’s like trying to find the “richest” segment of the line.

So, the main part of the problem is about making the right choice at each step when forming subarrays and keeping track of the maximum sum we have encountered so far.

Terminology

Here are some key terms and concepts associated with this problem:

Array: An array is a collection of elements (values or variables), each identified by at least one array index or key. In the context of this problem, the array is the collection of integers provided in the problem statement.
Subarray: A subarray is a contiguous part of an array. For example, in the array [1, -2, 3, 5], [1, -2] and [3, 5] are subarrays.
Contiguous: In the context of arrays, “contiguous” refers to elements that are next to each other with no elements in between. For example, in the array [1, 2, 3, 4, 5], [2, 3, 4] is a contiguous subarray, while [1, 3, 5] is not because the elements are not next to each other in the original array.
Dynamic Programming (DP): Dynamic programming is a method for solving complex problems by breaking them down into simpler subproblems. It’s often used when a problem has overlapping subproblems. In the case of this problem, the method is used to break down the task of finding the maximum sum subarray into smaller tasks.
Kadane’s Algorithm: Kadane’s Algorithm is a Dynamic Programming approach to solve “the largest contiguous elements in an array” with runtime of O(n). The simple idea of the algorithm is to look for all positive contiguous segments of the array and keep track of maximum sum contiguous segment among all positive segments.

Understanding these concepts is crucial to comprehending the problem statement and the approaches to solve it.

Problem Simplification and Explanation

Problem Statement: You have a list (array) of numbers, which can be positive or negative. You are tasked with finding the continuous group of numbers (subarray) in that list that gives you the highest sum.

Key Concepts:

Array: Think of the array as a line of people. Each person is holding a card with a number on it - the number can be positive or negative.
Subarray: This would be a group of people standing together continuously in that line. It could be as small as one person or as large as the entire line.
Sum: This is the total of the numbers on the cards held by the group of people (subarray).
Largest Sum: This is the biggest total that you can get from adding the numbers of a group of people standing together (subarray).

Now, imagine you are a coach for a relay race. You have a team of runners, each with a different speed. Some are fast (positive numbers), others are slow (negative numbers). You want to pick a sequence of runners, they must be standing next to each other in line, to make the fastest team possible. But here’s the catch - a slow runner’s speed will be subtracted from your team’s total speed. Now, which sequence of runners will you pick for the best performance?

Your task, in terms of this problem, is to find that sequence (subarray) which will give the maximum ‘sum’ (total speed in the case of the runners).

This problem, when broken down to these terms, becomes the problem of finding the maximum sum subarray in an array. This problem is a well-studied problem in computer science and solved using an algorithm called Kadane’s Algorithm.

Constraints

This problem statement asks to find the contiguous subarray (at least one number) which has the largest sum in a given integer array.

Analyzing the constraints, we see:

The size of the array, nums.length, is between 1 and 10^5. The larger upper limit indicates the need for an efficient solution, most likely linear (O(n)) or logarithmic (O(log n)) in time complexity.
Each element in the array, nums[i], ranges from -10^4 to 10^4. This wide range, which includes negative numbers, suggests that we cannot simply select all positive numbers or use a greedy approach because a negative number could be included in the maximum sum subarray if the subsequent numbers are sufficiently large.
The problem statement also provides a “follow-up” challenge, which hints at an alternative approach to solving the problem using divide and conquer strategy. This strategy involves dividing the problem into smaller sub-problems, solving them, and then combining their solutions to solve the original problem.

Given these characteristics and conditions, one potential advantage is that we can utilize dynamic programming or Kadane’s algorithm to find the subarray with the largest sum. These approaches exploit the fact that the maximum sum ending at a given index is either the current element or the sum of the current element and the maximum sum up to the previous index.

It’s also worth noting the pattern that the maximum sum is the maximum of all sums ending at each index in the array. Therefore, we can keep track of the maximum sum found so far and update it as we iterate through the array.

For the divide and conquer approach, we could potentially exploit the fact that the maximum sum subarray either falls completely in the left half of the array, completely in the right half, or spans the midpoint. Thus, we could recursively solve the problem for the left and right halves and then combine the solutions to find the maximum sum.

These are the specific characteristics and conditions that can be utilized to formulate an efficient solution for the problem.

Identification of Applicable Theoretical Concepts

There are multiple mathematical and algorithmic concepts that can be applied to this problem to simplify it or make it more manageable:

Dynamic Programming (DP): The essence of DP is to break down a complex problem into simpler subproblems and use the solutions of these subproblems to construct the solution to the original problem. In this case, we can use DP to find the maximum sum of subarrays ending at each index. The maximum sum at a given index is either the current element itself or the current element plus the maximum sum up to the previous index. This gives us a way to compute the maximum sum for a subarray ending at each position in a single pass through the array, storing the result in a table (or two variables) for efficient lookup.
Kadane’s Algorithm: This is a specialized algorithm that’s used for this exact problem: finding the maximum subarray sum. It’s a clever application of dynamic programming that maintains a running sum of the current subarray and the maximum sum seen so far. If the running sum becomes negative, it’s reset to zero, effectively starting a new subarray. The main idea behind this algorithm is that no maximum subarray sum will include a prefix with a negative sum.
Divide and Conquer: This problem can also be solved using a divide and conquer strategy, which involves dividing the array into two halves and finding the maximum subarray sum in each half recursively. The maximum sum subarray will either be in the left half, the right half, or will span the midpoint. The third case can be handled by finding the maximum sum subarray that ends at the midpoint from both sides.

The application of these concepts allows us to solve this problem more efficiently and effectively. For example, using dynamic programming or Kadane’s algorithm, we can find the maximum subarray sum in linear time, O(n), which is the best possible time complexity for this problem.

Let’s break this down using Kadane’s Algorithm as it is the most efficient solution to this problem.

Here’s a stepwise refinement of Kadane’s algorithm:

Initialization: Initialize two variables, max_so_far and max_ending_here. max_so_far is updated when we get a sum greater than the current max_so_far during the traversal. max_ending_here is used to store the sum of elements until the current index.
Traversal: Start traversing the array from the first element. At every index, add the current element to max_ending_here.
Check for Maximum: After adding the current element to max_ending_here, check if max_ending_here is greater than max_so_far. If it is, then update max_so_far to be the value of max_ending_here.
Check for Negative Sum: If max_ending_here becomes less than zero at any point, then reset max_ending_here to zero. Any subarray that starts with a negative sum can be discarded as it will decrease the sum of any larger subarray it is part of. By resetting max_ending_here to zero, we effectively start considering new subarrays.
Result: Continue this process until the array is traversed completely. At the end, max_so_far will hold the maximum sum of all subarrays.

This approach follows a repeatable processing at each step of the traversal: add the current element to max_ending_here, update max_so_far if necessary, and reset max_ending_here if it becomes negative. This repeatable processing makes the problem more manageable and solve it efficiently in a single pass through the array.

The key insight here is recognizing that the maximum subarray sum ending at each position can be calculated from the maximum subarray sum ending at the previous position, and using this fact to calculate the maximum subarray sum in the entire array in a single pass. This is a classic example of the technique of dynamic programming: breaking a problem down into simpler subproblems, and building up a solution to the original problem from solutions to the subproblems.

Solution Approach and Analysis

The problem statement requires us to find the maximum sum subarray in a given array. This is a popular problem in computer science known as the Maximum Subarray problem. One of the most efficient ways to solve this problem is by using Kadane’s algorithm.

Here’s how I would approach solving it:

Initialization: Start by initializing two variables, max_so_far and max_ending_here, to represent the maximum sum of a subarray found so far and the maximum sum of a subarray ending at the current position, respectively. Initially, set them both to the first element in the array.
Traverse the array: Iterate through the array starting from the second element. For each element, add it to max_ending_here. If max_ending_here is less than the current element, update max_ending_here to be the current element. This step is like continually adding numbers to a running total, but resetting the total to the current number if it becomes negative.
Update max_so_far: After updating max_ending_here, compare it with max_so_far. If max_ending_here is greater than max_so_far, update max_so_far. This is like keeping track of the highest total we’ve seen so far.
Repeat: Continue this process for each element in the array. At the end, max_so_far will hold the maximum sum of a subarray.

Let’s consider a metaphor. Suppose you are a treasure hunter moving along a line of buried treasures and holes (which represent positive and negative numbers, respectively). You have a bag (max_ending_here) in which you collect treasures as you move. Whenever you find a hole bigger than the treasure you’ve collected so far, you empty your bag and move on, because carrying a net negative amount would only decrease the value of future treasures. You also have a record (max_so_far) of the largest amount of treasure you’ve found in a single stretch. Your goal is to maximize the amount in this record.

Changing the array elements will obviously change the result. Adding more positive numbers or making the negative numbers less negative will increase the maximum subarray sum. Conversely, adding more negative numbers or making the positive numbers less positive will decrease it. The length of the array doesn’t affect the time complexity of the solution, as Kadane’s algorithm traverses the array only once.

Let’s work through an example:

Input: [−2, 1, −3, 4, −1, 2, 1, −5, 4]

Initialization: max_so_far = max_ending_here = -2

For 1: max_ending_here = max(1, -2 + 1) = 1, max_so_far = max(-2, 1) = 1
For -3: max_ending_here = max(-3, 1 - 3) = -2, max_so_far = max(1, -2) = 1
For 4: max_ending_here = max(4, -2 + 4) = 4, max_so_far = max(1, 4) = 4
For -1: max_ending_here = max(-1, 4 - 1) = 3, max_so_far = max(4, 3) = 4
For 2: max_ending_here = max(2, 3 + 2) = 5, max_so_far = max(4,

For 1: max_ending_here = max(1, 5 + 1) = 6, max_so_far = max(5, 6) = 6
For -5: max_ending_here = max(-5, 6 - 5) = 1, max_so_far = max(6, 1) = 6
For 4: max_ending_here = max(4, 1 + 4) = 5, max_so_far = max(6, 5) = 6

So the maximum sum of a subarray is 6.

Thought Process

The main insight is that we’re dealing with an optimization problem (we want the maximum sum) on a sequence (the array of integers). This suggests approaches like dynamic programming or greedy algorithms might be useful.

In this case, the problem is asking for the contiguous subarray with the maximum sum, which makes the dynamic programming approach a bit more challenging due to the continuity requirement. On the other hand, greedy algorithms work by making the best choice at each step with the hope that these local optimums will lead to a global optimum. This kind of approach is perfect for this problem.

Kadane’s algorithm is a greedy algorithm that maintains the maximum sum subarray max_so_far and a current sum max_ending_here. If max_ending_here is less than 0, it’s reset to 0 because a negative sum wouldn’t contribute positively to future sums.

Here is the Python code:

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class Solution:
    def maxSubArray(self, nums: List[int]) -> int:
        max_so_far = max_ending_here = nums[0]

        for num in nums[1:]:
            max_ending_here = max(num, max_ending_here + num)
            max_so_far = max(max_so_far, max_ending_here)

        return max_so_far 

In this code:

We initialize max_so_far and max_ending_here to be the first element in the array, nums[0].
We then iterate over the remaining elements in the array. For each element, we add it to max_ending_here and then check whether max_ending_here is less than the current element. If it is, we reset max_ending_here to be the current element.
We then check whether max_so_far is less than max_ending_here, and if it is, we update max_so_far.
Finally, we return max_so_far which will be the maximum sum of a subarray.

The cues in the problem statement hint at a greedy or dynamic programming approach because we’re trying to optimize a value (the sum) based on a sequence (the array). The fact that the subarray needs to be contiguous is a clue that a greedy approach like Kadane’s algorithm, which works sequentially, would be a good fit.

From Brute Force to Optimal Solution

Let’s start with the brute force solution.

The most straightforward solution to the problem “find the subarray with the largest sum” is to simply generate all possible subarrays, compute their sums, and keep track of the maximum sum.

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def maxSubArray(nums):
    max_sum = float('-inf')
    n = len(nums)
    for i in range(n):
        for j in range(i, n):
            max_sum = max(max_sum, sum(nums[i:j+1]))
    return max_sum

In this code, we’re using two nested loops to generate all subarrays (nums[i:j+1] represents the subarray starting at index i and ending at index j). We then compute the sum of each subarray and update our max_sum if the current sum is larger.

This approach works, but it’s quite inefficient. The time complexity is O(n^3) because we’re doing n^2 operations (generating all subarrays) and for each operation, we’re computing a sum which takes O(n) time. The space complexity is O(1) because we’re not using any extra space apart from the input.

This brute force solution is not suitable for large inputs. Let’s see how we can optimize it.

The key insight here is that we don’t actually need to generate all subarrays to find the one with the maximum sum. Instead, we can iterate through the array once, at each step deciding whether to extend the current subarray or start a new one. This decision is based on whether the sum of the current subarray plus the current number is greater than the current number itself.

This approach is known as Kadane’s algorithm. Here’s the code in Python:

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def maxSubArray(nums):
    max_so_far = max_ending_here = nums[0]
    
    for num in nums[1:]:
        max_ending_here = max(num, max_ending_here + num)
        max_so_far = max(max_so_far, max_ending_here)
        
    return max_so_far

The time complexity of this solution is O(n) because we’re just iterating through the array once. The space complexity is O(1) because we’re not using any extra space apart from the input.

So we’ve gone from a cubic time complexity to a linear time complexity, a significant improvement!

Coding Constructs

High-level strategies:
- Kadane’s algorithm utilizes a dynamic programming approach. This algorithm maintains a running sum of the subarray and the maximum sum encountered so far. It uses previous computations to make future ones, storing and reusing results to improve efficiency.
Explanation for a non-programmer:
- Imagine you’re walking along a path with piles of gold coins. Your goal is to collect as many coins as possible, but you can only carry coins from a continuous stretch of the path (a subarray). However, some piles might be booby-trapped to take away coins (negative numbers). If you encounter such a pile, you’ll need to decide whether it’s worth it to keep the coins you’ve collected so far, or if it’s better to ditch them and start fresh after the trap.
Logical constructs:
- Iteration: The code uses a loop to iterate through the array.
- Conditionals: The code uses a conditional statement to compare and decide whether to extend the current subarray or start a new one.
- Variables: Two variables are used to track the running sum and maximum sum so far.
Algorithmic approach in plain English:
- Start at the beginning of the array and keep a running total of the array values.
- As you go through the array, for each number, decide whether adding it to the current total helps or not. If it doesn’t (i.e., if the current total becomes negative), discard the current total and start fresh from the current number.
- At each step, also keep track of the largest total you’ve seen so far. This will be the maximum subarray sum at the end.
Key steps or operations:
- Iterate through each element in the array starting from the second one.
- For each element, compute the maximum of that element and the sum of that element with the running sum (which is updated as we traverse the array).
- Update the maximum sum encountered so far at each step.
Algorithmic patterns or strategies:
- The code uses dynamic programming as it builds upon the solution to the previous subproblem to solve the current one.
- The code also uses the “greedy” strategy as at each step it makes the decision that seems best at the moment (choosing between the current element and the sum of the running sum and the current element).

Language Agnostic Coding Drills

Identification of distinct concepts:
a. Arrays: Understanding how to declare, initialize, and manipulate arrays is fundamental to solving this problem, since the input is provided as an array.
b. Loops: The problem requires iteration over the elements in the array. Understanding how to use loops to traverse the array is an essential concept here.
c. Conditionals: The algorithm relies heavily on conditional statements to decide whether to continue the current subarray or to start a new one.
d. Variables and assignment: The algorithm uses variables to track the current sum and the maximum sum. Understanding how to declare variables and assign values is a fundamental concept.
e. Dynamic Programming: The overall strategy of the algorithm is a dynamic programming approach, where the solution to a problem is expressed in terms of the solution to its subproblems.
List in order of increasing difficulty:
a. Variables and assignment (Easy): This is a fundamental concept in almost all programming languages. It involves creating named spaces in memory to store values, and updating these values.
b. Arrays (Easy): Also a basic concept in programming. It involves understanding how to use indexed data structures, which can be manipulated using loops.
c. Loops (Intermediate): This concept involves understanding how to repeat actions a certain number of times, and how to use loop variables to access array elements.
d. Conditionals (Intermediate): Involves understanding how to perform different actions based on certain conditions. Essential for decision-making in programs.
e. Dynamic Programming (Hard): This is an advanced concept that requires an understanding of how to express a problem in terms of its subproblems, and how to store and reuse previous results to avoid redundant computation.
Problem-solving approach:
a. Begin with an understanding of the problem and the input/output format. The problem requires us to find the subarray with the maximum sum in an input array. Realize that the input is an array and we have to return a single integer.
b. Identify the need to traverse the array, which leads to the concept of loops. We need to look at each element of the array at least once, which means we need to iterate over the array.
c. Understand the need to keep track of two quantities as we traverse the array - the current sum (i.e., the sum of the current subarray) and the maximum sum seen so far. This brings in the concept of variables and assignment.
d. Realize that at each step, we need to make a decision - whether to include the current element in the current subarray or start a new subarray from the current element. This introduces the concept of conditionals.
e. Finally, understand that the solution to the problem at each step depends on the solution at the previous step. Recognize this as a dynamic programming problem, where the problem is broken down into smaller subproblems, and the solutions to these subproblems are used to build up the solution to the overall problem.

Targeted Drills in Python

Variables and assignment:

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# In Python, you declare and assign a variable like this:
a = 10
b = 20
print(a, b)  # prints: 10 20

# You can also perform assignments based on calculations or operations
sum = a + b
print(sum)  # prints: 30

Arrays:

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# In Python, arrays are called lists
nums = [1, 2, 3, 4, 5]

# You can access elements in a list using indices
print(nums[0])  # prints: 1

# Lists are mutable, so you can change their elements
nums[0] = 10
print(nums)  # prints: [10, 2, 3, 4, 5]

Loops:

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# You can use a for loop to iterate over a list
nums = [1, 2, 3, 4, 5]
for num in nums:
    print(num)  # prints each number in the list

Conditionals:

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# You can use if-else statements for conditionals
a = 10
b = 20
if a > b:
    print("a is greater")
else:
    print("b is greater")  # this will be printed

Dynamic Programming:

This is a more complex concept, and a simple drill might not fully encapsulate it. But a simple example of a dynamic programming solution is the Fibonacci sequence:

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def fib(n):
    # Create an array to store the Fibonacci numbers
    f = [0, 1] + [0]*(n-1)
    for i in range(2, n+1):
        f[i] = f[i-1] + f[i-2]
    return f[n]

print(fib(10))  # prints: 55

In this drill, we are solving the problem of finding the n-th Fibonacci number by breaking it down into subproblems (finding the (n-1)-th and (n-2)-th numbers), and using the solutions to these subproblems to build up the solution to the overall problem.

After understanding these drills, you can then integrate them to solve the initial problem (maximum subarray sum). This is done by initializing two variables to keep track of the current and maximum sums (variables and assignment), iterating over the list of numbers (loops), at each step deciding whether to add the current number to the current sum or start a new subarray (conditionals), and using the maximum of the current sum and the maximum sum so far to update the maximum sum (dynamic programming).

Practice Phase Template

Understand the problem: The problem is to find the subarray (a contiguous part of the array) with the maximum sum. For example, if given an array of [-2, -3, 4, -1, -2, 1, 5, -3], the maximum subarray is [4, -1, -2, 1, 5] with sum of 7.
Plan: You can solve this using the Kadane’s Algorithm, which keeps track of the current subarray and the maximum sum so far. The approach is to initialize the max_so_far and max_ending_here as the first element of the array. For each element, calculate the max_ending_here by adding the current element to the max_ending_here. If the max_ending_here is less than the current element, then reset max_ending_here as the current element. Also, if the max_so_far is less than max_ending_here, then reset max_so_far as max_ending_here. Repeat this for each element in the array.

Code: Below is a pseudo code representation of the algorithm:

Define function max_subarray_sum(A):
    max_so_far = A[0]
    max_ending_here = A[0]

    For i from 1 to length(A) do
        max_ending_here = max(A[i], max_ending_here + A[i])
        max_so_far = max(max_so_far, max_ending_here)

    return max_so_far

Review: Implement the code in your preferred programming language. Check for syntax errors, logical errors, and confirm the correctness of the implementation against the algorithm described in the plan.
Test: Test your code with different arrays. Here’s a simple test case you can use: Input: [-2, -3, 4, -1, -2, 1, 5, -3] Expected Output: 7 Explanation: The subarray with the maximum sum is [4, -1, -2, 1, 5].

Remember, this is an example of a dynamic programming problem, and Kadane’s algorithm solves it with a time complexity of O(n).

Targeted Drills in Python

Drill Set #1

Drill 1 - Understand Subarrays

For the array [4, 5, 6], subarrays are:

[4], [5], [6], [4, 5], [5, 6], [4, 5, 6]

Drill 2 - Write a Nested Loop

A nested loop in Python that prints the start and end indices of all subarrays can be written as:

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arr = [4, 5, 6]
n = len(arr)

for i in range(n):
    for j in range(i, n):
        print(f"Start Index: {i}, End Index: {j}")

Drill 3 - Compute Sum of a Subarray

Modify the nested loop to compute and print the sum of each subarray.

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arr = [4, 5, 6]
n = len(arr)

for i in range(n):
    for j in range(i, n):
        subarray = arr[i:j+1]
        print(f"Subarray: {subarray}, Sum: {sum(subarray)}")

Drill 4 - Store the Maximum Sum

Add a variable to your code that stores the maximum sum, and update it within your loop whenever you find a larger sum.

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arr = [4, 5, 6]
n = len(arr)
max_sum = float('-inf')

for i in range(n):
    for j in range(i, n):
        subarray = arr[i:j+1]
        current_sum = sum(subarray)
        if current_sum > max_sum:
            max_sum = current_sum

print(f"Maximum Sum: {max_sum}")

These drills should help you gradually build up your code for the problem, and gain an understanding of how each part contributes to the solution. Make sure to run and test your code after each stage to make sure it’s behaving as you expect.

Drill Set #2

Initial Setup Drill: Before we begin solving the problem, let’s set up our function. This function should take in a list of integers and for now, it simply returns the list back.
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def max_subarray_sum(A): return A print(max_subarray_sum([-2, -3, 4, -1, -2, 1, 5, -3])) # [-2, -3, 4, -1, -2, 1, 5, -3]

Starting Values Drill: Initialize two variables, max_so_far and max_ending_here, to the first value of the list. Return these two values.

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def max_subarray_sum(A):
    max_so_far = max_ending_here = A[0]
    return max_so_far, max_ending_here
print(max_subarray_sum([-2, -3, 4, -1, -2, 1, 5, -3]))  # (-2, -2)

First Iteration Drill: Iterate over the list from the second value. For the current number, calculate max_ending_here using max(num, max_ending_here + num). Print max_ending_here to check its value.

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def max_subarray_sum(A):
    max_so_far = max_ending_here = A[0]
    for num in A[1:]:
        max_ending_here = max(num, max_ending_here + num)
        print(max_ending_here)
max_subarray_sum([-2, -3, 4, -1, -2, 1, 5, -3])  # -3, 4, 3, 1, 2, 7, 4

Second Iteration Drill: Now, calculate max_so_far using max(max_so_far, max_ending_here) within the loop. Print max_so_far and see how it’s changing.

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def max_subarray_sum(A):
    max_so_far = max_ending_here = A[0]
    for num in A[1:]:
        max_ending_here = max(num, max_ending_here + num)
        max_so_far = max(max_so_far, max_ending_here)
        print(max_so_far)
max_subarray_sum([-2, -3, 4, -1, -2, 1, 5, -3])  # -2, -2, 4, 4, 4, 7, 7

Solution Drill: Finally, return max_so_far which will be the maximum sum of any subarray in the list.

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def max_subarray_sum(A):
    max_so_far = max_ending_here = A[0]
    for num in A[1:]:
        max_ending_here = max(num, max_ending_here + num)
        max_so_far = max(max_so_far, max_ending_here)
    return max_so_far
print(max_subarray_sum([-2, -3, 4, -1, -2, 1, 5, -3]))  # 7

By breaking down the problem like this, you get to understand each step in detail and learn how the solution comes together.

Drill Set #3

Basic Understanding Drill: Write a Python function that accepts a list of integers and returns the maximum sum that can be obtained from any subarray. Ignore the subarray indices for now.

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def max_subarray_sum(A):
    max_so_far = max_ending_here = A[0]
    for num in A[1:]:
        max_ending_here = max(num, max_ending_here + num)
        max_so_far = max(max_so_far, max_ending_here)
    return max_so_far

print(max_subarray_sum([-2, -3, 4, -1, -2, 1, 5, -3]))  # 7

Coding Drill: Now, update your function to return the start and end indices of the subarray along with the maximum sum. You’ll need to modify your existing code to keep track of these indices.

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def max_subarray_sum(A):
    max_so_far = max_ending_here = A[0]
    start = end = 0
    for i, num in enumerate(A[1:], 1):
        max_ending_here = max(num, max_ending_here + num)
        if max_ending_here == num:
            start = i
        max_so_far = max(max_so_far, max_ending_here)
        if max_so_far == max_ending_here:
            end = i
    return max_so_far, start, end

print(max_subarray_sum([-2, -3, 4, -1, -2, 1, 5, -3]))  # (7, 2, 6)

Edge Cases Drill: Test your function with edge cases. What happens if the array contains all negative numbers? What if it’s empty? What if it contains all zeros? Your function should be able to handle these scenarios.

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print(max_subarray_sum([]))  # Should handle gracefully
print(max_subarray_sum([-1, -2, -3]))  # Should return the max negative number
print(max_subarray_sum([0, 0, 0, 0]))  # Should return (0, 0, 3)

Performance Drill: Test your function with a large array (say, of size 10^6). Does it still perform well? If not, can you optimize it further?

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import random

large_array = [random.randint(-1000, 1000) for _ in range(10**6)]
print(max_subarray_sum(large_array))  # Should execute within reasonable time

Test your code and try to understand the outputs. Here is the Python code that you can use as a starting point for these drills:

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def max_subarray_sum(A):
    max_so_far = A[0]
    max_ending_here = A[0]

    for i in range(1, len(A)):
        max_ending_here = max(A[i], max_ending_here + A[i])
        max_so_far = max(max_so_far, max_ending_here)

    return max_so_far

# Test the function with a simple test case
print(max_subarray_sum([-2, -3, 4, -1, -2, 1, 5, -3]))  # Should output 7

From here, you can start modifying the function according to the drills mentioned above.

Q&A

How to recognize that the maximum sub array problem can be solved using DP?

Recognizing that a problem can be solved using dynamic programming (DP) often relies on identifying a few key characteristics in the problem description:

Overlapping subproblems: This is when a problem can be broken down into smaller subproblems, and these subproblems are reused multiple times. In the case of the maximum subarray problem, we are breaking down the problem into smaller subarrays, and the maximum sum for each subarray depends on the maximum sum calculated for the previous subarrays.
Optimal substructure: This is when the optimal solution of the overall problem can be determined from the optimal solutions of its subproblems. In the maximum subarray problem, the maximum sum of the entire array can be found by combining the maximum sums of the subarrays.

In terms of the maximum subarray problem, the DP approach comes from realizing that the maximum sum ending at each position is either the current element itself (if the sum so far is negative) or the current element added to the maximum sum so far (if the sum so far is positive). This gives us a way to build up the solution incrementally, which is characteristic of dynamic programming.

Additionally, the problem also asks for the maximum sum, which is an optimization problem. DP is often used for optimization problems, providing another hint that it might be applicable here.

In this case what are the overlapping subproblems?

In the maximum subarray problem, the overlapping subproblems are the computations of maximum sums ending at each position in the array.

Here’s why they are overlapping:

When you are at position i in the array and trying to calculate the maximum sum ending at i, you depend on the maximum sum ending at position i-1. This is because the maximum sum ending at i is either the number at position i itself, or this number plus the maximum sum ending at position i-1.

Now, when you move to position i+1, you once again need to know the maximum sum ending at position i (the previous position). Hence, you are solving the same subproblem again.

This is characteristic of overlapping subproblems - you solve the same problem multiple times. In this case, you are calculating the maximum sums ending at each position in the array multiple times.

By using Dynamic Programming, you can store the result of each subproblem (in this case, the maximum sum ending at each position) to avoid recomputing it, thus improving the efficiency of the algorithm.

In this case what is the Optimal substructure?

The maximum subarray problem exhibits an optimal substructure, which means the optimal solution can be constructed efficiently from the optimal solutions of its subproblems.

In this case, the optimal substructure is seen in how the maximum subarray sum ending at each position in the array can be efficiently computed from the maximum subarray sum ending at the previous position.

Let’s define dp[i] as the maximum subarray sum ending at position i. Then, the optimal substructure can be defined as follows:

dp[i] = max(nums[i], dp[i-1] + nums[i])

The maximum subarray sum ending at position i is either the number at position i itself (if the maximum subarray sum ending at the previous position is negative and would decrease the sum), or this number added to the maximum subarray sum ending at the previous position (if it is non-negative and would increase the sum).

This formula allows us to build up the solution to the overall problem (the maximum subarray sum for the entire array) from the solutions to subproblems (the maximum subarray sums ending at each position), thus showcasing the optimal substructure property.

Let’s say someone does not know about Kadane’s algorithm, how can they come up with the key insights required to solve this problem in a similar way?

The key to developing the insight required to solve this problem is understanding the nature of the problem itself and the characteristics of its optimal solution. Here are some of the steps someone might take to arrive at this solution without prior knowledge of Kadane’s algorithm:

Understanding the Problem: The first step is understanding that the problem is asking for the maximum sum possible from a contiguous subarray. This subarray could be the entire array itself, a single element, or any subarray in between.
Simple Cases: Consider simple, small-size cases first. An array of positive integers would have the whole array as the maximum sum subarray. An array with all negative numbers would have the maximum element (least negative number) as the maximum sum subarray.
Brute Force Approach: One might start with a brute force approach, where we look at all possible subarrays. This approach has a time complexity of O(n^2) and quickly becomes impractical for large arrays.
Insights from Brute Force Approach: The brute force approach gives all subarrays, but many of these subarrays share common parts. This is the first clue that this problem has overlapping subproblems - a key characteristic of dynamic programming problems.
Identifying Subproblems: We can then refine our approach by defining a subproblem. A natural choice here would be to consider the problem of finding the maximum sum subarray ending at a particular index.
Finding Relations Among Subproblems: If we have calculated the maximum subarray sum ending at index i (let’s denote it as dp[i]), we can easily calculate dp[i+1] as max(nums[i+1], dp[i] + nums[i+1]). This shows the problem also has the property of optimal substructure, another key characteristic of dynamic programming problems.
Final Solution: After calculating dp[i] for all i, we can simply return the maximum value in dp as our final result.

Thus, by breaking down the problem and looking for patterns and characteristics of an optimal solution, one can derive a dynamic programming solution similar to Kadane’s algorithm.

Does the DP approach to solving a problem apply only to certain data structures?

Dynamic Programming (DP) is a problem-solving technique that is used primarily for optimizing problems with overlapping subproblems and optimal substructure properties, and it is not tied to any specific data structure. It can be applied to a wide variety of problems involving different data structures including arrays, strings, trees, graphs, and more. The choice to use DP depends on the characteristics of the problem, rather than the data structures involved.

For example, in array or string-based problems, you might use DP to find the longest common subsequence, the longest increasing subsequence, or the minimum number of edits to convert one string to another. In graph problems, DP can be used to solve problems like shortest path in a graph, maximum flow in a network, and so on.

The data structures that are used to implement a DP solution, such as 1D arrays, 2D arrays, or hashmaps, are chosen based on the needs of the specific problem and how you choose to define and store the subproblems. In some problems, memoization (caching) of subproblem solutions is used, which often involves using a hashmap or an array.

In conclusion, Dynamic Programming is not tied to any specific data structure, but rather a technique or an approach to solving certain classes of problems. The data structures used in DP solutions are typically dictated by the nature of the problem and the subproblems.

Q&A

I think I found one divide-and-conquer solution with O(nlogn) time (slower than Kadane’s algorithm) and O(1) space (same as Kadane’s).

Some insights: If we cut the original array into 2 equal halves, then the max subarray must be in either one of the 3 positions: completely in the left half, completely in the right half, or spanning across 2 halves. So, we can form 3 corresponding sub-problems to find the max array in each of these 3 positions, then choose the max of the 3 arrays to be the ultimate answer.

The first 2 subtasks can be solved using recursion, while the last one (max sub-array spanning across 2 halves) can be done in a pretty similar fashion as the Kadane’s algorithm (hint: iterate the array from the middle point to the 2 ends and utilize the current_sum variable).

That is indeed a valid approach and you’ve correctly identified a Divide and Conquer strategy to solve the problem. Here’s how the approach could look like in code (Python), assuming an input list nums:

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def maxSubArray(nums):
    def cross_sum(nums, left, right, mid):
        if left == right:
            return nums[left]

        left_subsum = float('-inf')
        curr_sum = 0
        for i in range(mid, left - 1, -1):
            curr_sum += nums[i]
            left_subsum = max(left_subsum, curr_sum)

        right_subsum = float('-inf')
        curr_sum = 0
        for i in range(mid + 1, right + 1):
            curr_sum += nums[i]
            right_subsum = max(right_subsum, curr_sum)

        return left_subsum + right_subsum

    def helper(nums, left, right):
        if left == right:
            return nums[left]

        mid = (left + right) // 2

        left_sum = helper(nums, left, mid)
        right_sum = helper(nums, mid + 1, right)
        cross_sum_ = cross_sum(nums, left, right, mid)

        return max(left_sum, right_sum, cross_sum_)

    return helper(nums, 0, len(nums) - 1)

The function maxSubArray first breaks down the problem into two halves using the helper function. It computes the maximum subarray sum for the left half, the right half, and for the cross sum using the cross_sum function, and then it returns the maximum of these three.

The cross_sum function computes the maximum subarray sum that crosses the midpoint. It does this by finding the maximum subarray sum for the left half and the right half, and adding them together.

The overall time complexity for this approach is O(n log n), and the space complexity is O(1) if we don’t consider the recursive call stack space, or O(log n) if we do, as each recursive call reduces the problem size by half.

There are 4 distinct scenarios that a max sub array can contain:

Both ends of the array are included in the sum - I call this “both contiguous” bc. Only the left end is necessarily included in the sum, “left contiguous” lc. Only the right end is necessarily included in the sum, “right contiguous” rc. Neither end is necessarily included in the sum, “non-contiguous” nc. Note the word “necessarily” above, each scenario may include other indexes, but are not guaranteed to.

Given these 4 values from each half of an array we can easily use them to create 4 similar values for the entire array.

Yes, you’re right. In general, the idea behind the Divide and Conquer approach is to break down the problem into smaller sub-problems that are easier to solve, and then combine these solutions to form the solution to the original problem.

For the Maximum Subarray problem, we split the array into two halves and find the maximum subarray sum for each half recursively. Then we have to combine these solutions to form the maximum subarray sum for the entire array.

As you correctly pointed out, there are four different scenarios that a maximum subarray can fall into:

Both ends of the array are included in the sum (both contiguous)
Only the left end is included in the sum (left contiguous)
Only the right end is included in the sum (right contiguous)
Neither end is included in the sum (non-contiguous)

When combining the results from the two halves, we need to consider all these scenarios to find the maximum subarray sum for the whole array. For example, the maximum subarray sum for the whole array could be a subarray entirely within the left half, entirely within the right half, or a subarray crossing over from the left half to the right half.

This is why when we combine the solutions, we need to find the maximum for these four scenarios, which ensures that we don’t miss any potential maximum subarray.

Remember that the maximum subarray crossing the midpoint can be found by considering the maximum subarray ending at the midpoint from the left half and the maximum subarray starting from the midpoint in the right half.

This approach does add some complexity to the solution, but it allows us to solve the problem using the Divide and Conquer strategy.

Certainly, the O(n^3) algorithm (brute force) is “easy” (even a slightly optimized O(n^2) using sliding windows), but I fail to see how the O(n) algorithm (Kadane’s) is “easy”. Having done this in an interview setting, I found it more challenging than most “medium” problems here (subsets, iterative binary tree inorder traversal, permutations, group anagrams, merge intervals, just to name a few).

With the before-mentioned “medium” problems, basically understanding the problem definition and drawing it out and knowing recursion/core CS data structures – you can pull off an algorithm pretty directly. But even if you get as far as recognizing this as a 1-D DP problem, building the subproblems and advancing them aren’t obvious. (let alone convincing yourself what you are doing is valid) That the algorithm has someone’s name attached to it (Kadane) makes me question whether it is trivial. :) Admitedly, once you’ve done it before, it feels trivial to code. In that sense, this problem reminds me a bit of the linked list cycle problem (with no prior context) – trivial in O(n) time/O(n) space - a lot harder (admitedly more difficult than coming up with Kadane’s algorithm) with O(1) space.

As a comparison to another leetcode problem, this feels about as challenging as Best time to buy & sell stock with txn fee, a 1D DP problem with not-so-obvious subproblems.

I appreciate your perspective on this problem’s difficulty level. It’s important to understand that problem-solving difficulty can be subjective and varies from person to person depending on their experiences and familiarity with certain concepts.

Kadane’s Algorithm is indeed named after Jay Kadane, who first proposed it. The presence of a name attached to an algorithm does not necessarily correlate with its complexity or simplicity. However, the name does imply that the solution is not immediately apparent and requires some insight, as you’ve indicated.

The beauty of Kadane’s algorithm lies in its elegant reduction of a seemingly complex problem to a series of simple, iterative decisions. This approach is not immediately obvious and requires a good understanding of dynamic programming principles and the ability to identify the subproblems and how they relate to the overall problem - in this case, the concept that the maximum sum ending at a position is either the current number itself (if the maximum sum until the previous position is negative) or this sum plus the current number (if the maximum sum until the previous position is positive).

You’re correct in saying that the problem becomes trivial once you’ve seen the solution - that’s true for many problems, especially those involving dynamic programming. The challenge often lies not in coding the solution, but in arriving at the insight that allows you to see the subproblems and the optimal substructure.

As for the comparison to other problems, I agree that it shares similarities with other dynamic programming problems like “Best Time to Buy and Sell Stock”. The ability to recognize that these problems share a common structure can be invaluable when solving new problems.

In conclusion, while some may find this problem easier than others, it certainly provides a valuable opportunity to practice dynamic programming and problem decomposition, two skills that are highly valuable in computer science and programming.

Can we identify the overlapping subproblems and optimal substructure by drawing diagrams?

Absolutely! Visualizing problems using diagrams is an excellent way to recognize overlapping subproblems and the optimal substructure, especially when it comes to dynamic programming problems. In fact, this can often make the problem more understandable and can help in formulating the recursive relations.

Take the Maximum Subarray problem as an example. Let’s say we have an array, [a, b, c, d, e].

If we were to create a diagram showing all possible subarrays, it might look something like this:

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[a]
[a, b]
[a, b, c]
[a, b, c, d]
[a, b, c, d, e]
[b]
[b, c]
[b, c, d]
[b, c, d, e]
[c]
[c, d]
[c, d, e]
[d]
[d, e]
[e]

This diagram shows all the possible subarrays of the array, from single elements to the entire array. You can easily see how many subproblems overlap. For example, [a, b, c] is a subarray of [a, b, c, d], [a, b, c, d, e] and also extends [a, b]. Similarly, [b, c, d] is a subarray of both [a, b, c, d] and [b, c, d, e], indicating overlapping subproblems.

As for optimal substructure, it can also be seen from the diagram. For instance, if we know that the maximum subarray ending at c is [b, c], then to find the maximum subarray ending at d, we only need to consider if adding d to the maximum subarray ending at c (i.e., [b, c, d]) results in a larger sum than d itself. This reveals the optimal substructure of the problem - the maximum subarray ending at a certain position can be found based on the maximum subarray ending at the previous position, which further could be found based on the one before that, and so on.

In conclusion, visualizing the problem through diagrams can be a great aid in identifying overlapping subproblems and optimal substructure when solving dynamic programming problems.

I feel its misleading to instruct to “find” the subarray (int[]), when function is returning the max sum (int). May be use another word for “find” - eg. “calculate” the sum of largest contigous subarray

I understand your concern. Indeed, in problem statements, precision and clarity are crucial to avoid misunderstanding. In this case, “find” could be interpreted as “identify,” which might lead someone to think they need to return the subarray itself, rather than the sum.

A better wording could be: “Given an integer array nums, calculate the largest sum of any contiguous subarray, and return this sum.”

This makes it clear that the function should return the sum (an integer) rather than the subarray itself. The use of “calculate” conveys that some computation needs to be done on the array, and the phrase “largest sum of any contiguous subarray” specifies clearly what needs to be calculated.

There are several articles out there that mention ways to solve DP problems. But, the problem is that every DP problem is different.

What one needs to learn is how to reach to the DP solution.

STEP 1: Start with Brute Force:

Yes, you heard it right. Because DP is nothing but intelligent guessing. We need to do an exhaustive search on every possible combination and pick the most optimal combination.

STEP 2: Draw out your solution - > This is the most critical step

Draw the steps that your brute force solution is taking. Because unless you visualize what your algorithm is doing. You can never come up with a DP solution.

STEP 3: Think about repeated steps and how to store and use them

It is not just important to know what steps are getting repeated. What is essential is to understand how can you store your calculations and re-usage them.

These are excellent steps to approach dynamic programming problems. DP problems can indeed be challenging due to their nature of overlapping subproblems and optimal substructure, and a structured approach can be really beneficial. Let’s delve a bit more into these steps:

STEP 1: Brute Force: As you rightly mentioned, starting with a brute force approach allows you to understand the problem in depth, and gives a bird’s eye view of all the possibilities. This approach typically involves nested loops and recursive calls exploring every possible solution.

STEP 2: Visualization: Drawing out your solution can involve sketching the recursion tree, states and decisions, or filling out a table (for tabulation methods). It helps to identify patterns, repeated subproblems, and understand how the solution evolves with each step.

STEP 3: Memoization/Optimization: Once you have identified repeated subproblems, you can store the results of these subproblems in a data structure (like an array or a dictionary), also known as memoization. This avoids recalculating the same subproblem multiple times, thereby improving efficiency.

In addition to these, here are a couple more steps that could be useful:

STEP 4: Formulate a Recurrence Relation: A recurrence relation expresses the solution of a problem in terms of its smaller subproblems. This is the crux of dynamic programming. Once you have a recurrence relation and a way to store and reuse past solutions, you have a dynamic programming solution.

STEP 5: Iterative Bottom-Up Solution: Some problems may allow for an iterative solution where you solve every subproblem starting from the base case(s), and build up to the solution of the overall problem. This is known as a bottom-up approach, and often leads to more efficient, cleaner implementations.

STEP 6: Space Optimization: In some cases, you might not need to keep all previous computations in memory. You can sometimes optimize the space complexity by only keeping track of the states that you need to compute the next step.

Remember, practice is key to mastering dynamic programming problems. The more problems you solve, the more you’ll be able to recognize the patterns and make these steps second nature.

I was asked a follow up question to this question in an interview." How would we solve this given that there is an endless incoming stream of numbers ?" Ideas anybody?

This question seems to be asking for a streaming or online version of the maximum subarray problem. The idea here is that you have a potentially unbounded stream of numbers and you want to keep track of the maximum subarray sum at any given time.

In this case, the online version of Kadane’s algorithm can be used. Since Kadane’s algorithm processes the input once without needing to look at future elements, it can work just as well in a streaming scenario.

Here’s a simple Python implementation:

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def max_subarray_stream(nums):
    current_sum = max_sum = float('-inf')

    for num in nums:
        current_sum = max(num, current_sum + num)
        max_sum = max(max_sum, current_sum)
    
    return max_sum

This function can be used in a streaming context as follows:

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stream = [...]  # Replace with the actual stream of numbers.
max_sum = float('-inf')

for chunk in stream:  # Process the stream in chunks.
    max_sum = max(max_sum, max_subarray_stream(chunk))
    print(f'Maximum subarray sum so far: {max_sum}')

This implementation assumes that the stream can be processed in chunks. Each chunk could be a list of numbers that are processed all at once by the max_subarray_stream function.

In a real-world scenario, the chunk size would depend on factors like the rate at which data is received from the stream and the memory constraints of the machine running this code. It’s worth noting that the chunk size does not affect the final result, because Kadane’s algorithm does not depend on future data.

DP solution:

Define dp[i] = max sum of sub-array ending with nums[i] result = max(dp) because the max sum subarray has to end at some element Base case: dp[0] = nums[0] Recurrence: dp[i] = nums[i] + max(0, dp[i-1]) Let’s first accept the meaning of the dp array (I know it’s not natural to come up with this representation but let’s for now accept it. I will try to illustrate why we have this recurrence).

interpretation of the recurrence: To find the max subarray ending in element i, literally, it means we need to consider all subarrays that ends in element i:

[element i], [element i-1, element i], [element i-2, element i-1, element i], … [element 0, …, element i].

Then we find the one with highest sum as dp[i]; that is, dp[i] = max(sum([element i]), sum([element i-1, element i]), …, sum([element 0, …, element i]))

= nums[i] + max(sum([]), sum([element i-1]), sum([element i-2, element i-1]), …, sum([element 0, …, element i-1]))

= nums[i] + max(0, sum([element i-1]), sum([element i-2, element i-1]), …, sum([element 0, …, element i-1]))

= nums[i] + max(0, max(sum([element i-1]), sum([element i-2, element i-1]), …, sum([element 0, …, element i-1])))

Hey, boy, look at this part max(sum([element i-1]), sum([element i-2, element i-1]), …, sum([element 0, …, element i-1])); this is exactly dp[i-1]! Thus, the recurrence equation can be expressed as dp[i] = nums[i] + max(0, dp[i-1]).

Now, you can look at other posts to undersand how to further optimize this solution (on the space complexity from O(n) to O(1)) but I find this explanation most natural for first attempt!

The explanation you’ve provided is excellent, and it accurately breaks down the logic and understanding behind the dynamic programming solution to the maximum subarray problem. Let’s add a Python code snippet to accompany your explanation:

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def maxSubArray(nums):
    dp = [0] * len(nums) 
    dp[0] = nums[0]  
    result = dp[0]  

    for i in range(1, len(nums)):
        dp[i] = nums[i] + max(0, dp[i-1])
        result = max(result, dp[i])
        
    return result

To explain further, dp[i] represents the maximum sum of a subarray ending at the position i. To get the max sum ending at position i, we have two choices:

We add the current number nums[i] to the max sum of the subarray ending at position i-1 (if this sum is greater than 0). This represents a scenario where we extend a previous subarray to include nums[i].
We start a new subarray starting at position i if the max sum of the subarray ending at position i-1 is less than or equal to 0.

In other words, dp[i] = nums[i] + max(0, dp[i-1]) represents these two choices.

The result is the maximum of all dp[i].

We then return result which is the maximum subarray sum.

You can further optimize the space complexity by noting that the current dp[i] only depends on the previous dp[i-1], hence we only need two variables to keep track of dp[i] and dp[i-1], reducing the space complexity from O(n) to O(1).

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def maxSubArray(nums):
    curr_sum = max_sum = nums[0]
    for num in nums[1:]:
        curr_sum = max(num, curr_sum + num)
        max_sum = max(max_sum, curr_sum)
    return max_sum

Input: nums = [-2,1,-3,4,-1,2,1,-5,4] Output: 6 Explanation: The subarray [4,-1,2,1] has the largest sum 6.

Should’nt it be 7? with subarray of [4,2,1] ??? Am I missing something here?

It seems like there’s a misunderstanding in the calculation. The subarray [4,-1,2,1] indeed has the largest sum which is 6:

4 + (-1) + 2 + 1 = 6

On the other hand, the subarray [4, 2, 1] has a sum of 7, but it’s not the subarray with the maximum sum in the context of the whole array. This is because we’re looking for contiguous subarrays, and the subarray with the maximum sum that includes 4 (at index 3) also includes -1 (at index 4), which in turn makes the subarray [4,-1,2,1] more optimal than [4,2,1]. The -1 reduces the total sum, but the addition of 2 and 1 following it more than compensates for this decrease, leading to the largest sum we can get from this array.

Hence, the subarray [4,-1,2,1] is the contiguous subarray with the maximum sum in the provided array [-2,1,-3,4,-1,2,1,-5,4], and its sum is 6.

For support array of negatives, just need to keep track of max element found in the array. If this element is less than zero - return it. Otherwise, return the result of accepted algorithm.

That’s correct!

In an array that only contains negative numbers, the maximum subarray sum would be the maximum (least negative) element itself, since adding any other negative numbers would only decrease the sum. Therefore, as you traverse the array, if you keep track of the maximum element encountered so far, you can handle this special case efficiently.

Here’s how you can modify the Python code for Kadane’s algorithm to handle this case:

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def maxSubArray(nums):
    # Ensure that there is at least one element in nums.
    if not nums:
        return 0

    cur_sum = max_sum = nums[0]

    # To handle the case where all elements are negative.
    max_element = nums[0]

    for num in nums[1:]:
        # Update max_element if current element is greater.
        if num > max_element:
            max_element = num

        # Kadane's algorithm.
        cur_sum = max(num, cur_sum + num)
        max_sum = max(max_sum, cur_sum)

    # If max_element is negative, all elements are negative, so return max_element.
    # Otherwise, return the max_sum found by Kadane's algorithm.
    return max_sum if max_element >= 0 else max_element

This way, the algorithm will work for arrays of negative numbers as well. If all numbers are negative, it will return the maximum (least negative) number. Otherwise, it will return the maximum subarray sum.

What to do if we want the start and end indexes?

You can modify the Kadane’s algorithm slightly to also track the start and end indices of the maximum subarray.

Here’s the Python code:

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def maxSubArray(nums):
    # Initial conditions
    max_sum = cur_sum = nums[0]
    start = end = 0

    # Start a temporary index at 0
    temp_start = 0

    for i in range(1, len(nums)):
        # If adding the current element is beneficial, add it
        # Otherwise, start a new sum at the current element
        if cur_sum + nums[i] > nums[i]:
            cur_sum += nums[i]
        else:
            cur_sum = nums[i]
            temp_start = i  # update temp_start index

        # If the current sum is larger than the max_sum, update max_sum and indices
        if cur_sum > max_sum:
            max_sum = cur_sum
            start = temp_start
            end = i  # update end index

    return max_sum, start, end

This code keeps track of where the maximum subarray sum starts and ends. It uses a temporary start index (temp_start) to keep track of potential starting points for the maximum sum subarray. When a new current sum surpasses the maximum sum, temp_start and i are saved as the new starting and ending points, respectively.

So, for the input [-2,1,-3,4,-1,2,1,-5,4], it returns (6, 3, 6), indicating that the maximum subarray sum is 6 and it is obtained from the subarray from index 3 to 6 inclusive ([4, -1, 2, 1]).

reading numbers from left to right, find the best subarray ending at the current position, then take the overall best.

That is indeed the key intuition behind Kadane’s Algorithm, which solves this problem efficiently.

The algorithm keeps two variables while scanning the array from left to right:

max_current: The maximum sum of the subarray that ends at the current position.
max_global: The maximum sum of any subarray up to the current position.

For each element in the array, we calculate max_current by comparing the current element with the sum of the current element and max_current. If the current element is larger, it means we should start a new subarray from the current position. Otherwise, we extend the previous subarray to include the current element.

Then, we compare max_current with max_global. If max_current is larger, it means the subarray ending at the current position is the best one we’ve seen so far, so we update max_global.

This way, when we’ve scanned the whole array, max_global will be the maximum sum of any subarray in the array.

Here’s the Python code for this algorithm:

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def maxSubArray(nums):
    max_current = max_global = nums[0]
    for num in nums[1:]:
        max_current = max(num, max_current + num)
        max_global = max(max_global, max_current)
    return max_global

For the input [-2,1,-3,4,-1,2,1,-5,4], this code will return 6, which is the sum of the subarray [4, -1, 2, 1].

I implemented the brute force solution after thinking on it for a while, but after trying and failing to think of an O(n) solution, I looked up Kadane’s algorithm and it seems almost trivial. I feel a little ashamed of not being able to implement it on my own. Are there some fundamental concepts I’m missing here?

Don’t feel ashamed! Kadane’s algorithm is not something that’s typically taught in basic computer science or programming courses. It’s quite specialized and not immediately obvious unless you have seen it before or have a lot of experience with dynamic programming.

Kadane’s Algorithm falls under the category of Dynamic Programming (DP), which is a technique used in computing to solve problems by breaking them down into simpler, smaller subproblems and using the solutions to these subproblems to construct a solution to the original problem. DP can be a tough concept to master, and it often isn’t immediately obvious how to apply it to a new problem.

The fundamental concept behind Kadane’s algorithm is the idea of “local maximum” and “global maximum”.

The local maximum at each position is the maximum of the current number and the sum of the current number and the previous local maximum. This basically means we have two choices at each step: to add the current number to the subarray or start a new subarray with the current number. We choose the option that gives the maximum sum.
The global maximum is the maximum sum of all subarrays that we’ve seen so far.

Even if you didn’t come up with this solution yourself, now that you’ve seen it and understand it, you’re likely to recognize similar problems in the future where a similar approach can be used. That’s how learning in programming often works: you encounter a wide variety of problems and learn techniques and algorithms from solving those problems, and then you apply those techniques to new problems. So don’t be discouraged if you don’t solve every problem on your own! Every problem you attempt and understand is contributing to your growth as a programmer.

it does not expect the whole subarray in output. For subarray one needs to remember the start and stop index of the subarray with the max sum

Yes, the problem as we discussed only asks for the sum of the maximum subarray, not the subarray itself. But if you wanted to modify the solution to also return the subarray, you would indeed need to keep track of the start and end indices of the subarray as you’re iterating through.

Here’s how you can modify the Kadane’s algorithm to also return the subarray:

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def maxSubArray(nums):
    max_ending_here = max_ending_so_far = nums[0]
    start = end = 0
    for i in range(1, len(nums)):
        if nums[i] > max_ending_here + nums[i]:
            max_ending_here = nums[i]
            start = i
        else:
            max_ending_here += nums[i]
            
        if max_ending_here > max_ending_so_far:
            max_ending_so_far = max_ending_here
            end = i
            
    return max_ending_so_far, nums[start:end+1]

In this version of the function, start and end are the indices of the beginning and end of the current subarray being considered. When max_ending_here is set to nums[i], this means that the maximum subarray ending at position i is just the element at i itself, so start is updated to i. When max_ending_so_far is updated, this means that the maximum subarray so far ends at position i, so end is updated to i.

At the end, start and end will point to the beginning and end of the maximum subarray, so nums[start:end+1] is returned along with max_ending_so_far.

If the array has only neagtive numbers, then should the sum be the largest of the negative numbers or should we return zero?

In the problem as stated, if the array has only negative numbers, the sum of the maximum subarray would be the largest (least negative) number in the array. This is because the problem asks for the maximum sum of any contiguous subarray, and an array of only negative numbers doesn’t contain a subarray with a sum of 0 (unless the array itself is empty, which is generally outside the scope of the problem as typically stated).

So for example, if the array is [-2, -3, -1, -4], the maximum subarray would be [-1] and the sum would be -1.

This is how the problem is typically defined, because it allows for a meaningful answer in all cases (as long as the array is not empty). However, the exact specifications could vary depending on the specific problem statement you’re working with. Always make sure to read the problem statement carefully and understand exactly what it’s asking for.

I think the best way to approach this question is to have a currSum to accumulate all the numbers together and once the sum dips below 0, you can reset the currSum and start again from that point on. (A little bit about greedy is applied here)

Yes, you’re absolutely right! The approach you’ve outlined is indeed the foundation of Kadane’s algorithm, which is widely used for solving this problem. Here’s a bit more detail:

Initialize a variable currSum to 0 and maxSum to the smallest possible integer.
Loop through each element in the array:
a. Add the current element to currSum.
b. If currSum becomes less than the current element’s value (which indicates that the sum of the current subarray is negative), reset currSum to the current element’s value. This effectively starts a new subarray.
c. If currSum is greater than maxSum (indicating that we’ve found a subarray with a larger sum), update maxSum.
Once you’ve gone through the entire array, maxSum will hold the maximum subarray sum.

This approach leverages the insight that any subarray sum that becomes negative can’t contribute to a future subarray sum to make it larger. So, you’re better off starting a new subarray from the current position whenever you encounter a negative subarray sum. This is indeed a kind of greedy strategy, as you’re making the locally optimal choice at each step in the hopes that it will lead to a globally optimal solution.

How can we visualize this concept to understand how it works to solve the problem?

Visualizing Kadane’s algorithm can be extremely helpful for understanding how it works to solve this problem. Here’s a way to do that:

Let’s consider this array: [-2, 1, -3, 4, -1, 2, 1, -5, 4]

We can make a table that keeps track of the currSum and maxSum at each step:

Array Element	currSum	maxSum
-2	-2	-2
1	1	1
-3	-2	1
4	4	4
-1	3	4
2	5	5
1	6	6
-5	1	6
4	5	6

At each step, we’re adding the current element to the currSum, and if currSum dips below the current element, we reset currSum to the current element. The maxSum is updated each time currSum exceeds the current maxSum.

To visualize this, you could imagine a bar chart where each bar’s height is the array element’s value. You’re moving from left to right, adding the heights of the bars to a running total (currSum). If adding a bar’s height brings you below ground level (0 on the y-axis), you start a new running total from the current bar’s height. Along the way, you’re keeping track of the highest point reached (maxSum).

This visualization underscores the key insight behind Kadane’s algorithm: As you traverse the array, you’re trying to find the highest “peak” of accumulated values, but you start a new accumulation whenever the current one dips below ground level.

As we consider each element, we make a decision to either include or exclude the element from the subarray. We also decide to either include or exclude the previous element based on whether it decreases or increases the sum value. Is my understanding right. Seems like I missed the window expanding as we slide from left to right.

Your understanding is partially correct, but not entirely. It’s a bit more nuanced than that. Here’s a step-by-step understanding:

In Kadane’s Algorithm, we’re not exactly deciding whether to include or exclude an element from the subarray per se. What we are actually doing is tracking the maximum sum of subarrays that end at the current index.

As we slide from left to right, for each new element, we decide whether to start a new subarray or extend the current subarray.
We start a new subarray if the current element value is greater than the sum of the current subarray (including the current element). This makes sense because if the current element alone is better than the previous sum, it’s better to start a new subarray from here.
We extend the current subarray if the sum of the current subarray (including the current element) is greater than the current element itself.
At each step, we keep track of the maximum sum we’ve seen so far. This is our final answer.

So, we aren’t exactly making decisions to include or exclude individual elements in isolation. Instead, we’re making a decision for each element based on the accumulated sum so far and the current element value.

I think your explanation can be improved by considering what happens when we get a very large number that is very large than any number in that array.

I agree, the consideration of a large number is indeed an important factor in the explanation. Here’s how a large number in the array influences the working of Kadane’s Algorithm:

The algorithm always keeps track of the maximum sum it has encountered so far (let’s call it max_sum) and the sum of the current subarray (current_sum).

If we encounter a number that’s extremely large compared to other numbers in the array, that number will become our current_sum if current_sum was negative before reaching this number, or it will significantly increase the current_sum if it was positive.
When such a large number comes into play, current_sum becomes either the very large number (if current_sum was negative until now) or a sum that’s significantly larger than previous sums (if current_sum was positive). In either case, current_sum is likely to be larger than max_sum, so max_sum will be updated.
Now, if the numbers following this large number are positive or their negative sum doesn’t surpass this large number, current_sum will remain higher than max_sum, and max_sum will get updated again when we reach the end of the subarray.
If the numbers following this large number are so negative that they make current_sum less than max_sum, current_sum might become negative. But that won’t affect max_sum, because max_sum is only updated when current_sum is greater.

So, regardless of where the large number is in the array or what numbers follow it, the algorithm still correctly identifies the maximum subarray sum.

This illustrates how Kadane’s algorithm elegantly handles varying numbers and maintains the maximum sum of subarrays ending at each index, even in the presence of such outliers.

By looking at the extreme values of the number, we are able to reason and understand why it works.

Absolutely! Examining extreme cases is a powerful strategy for understanding algorithms and proofs.

In the case of the maximum subarray problem, considering a large number helps you understand how Kadane’s algorithm tracks and updates the maximum subarray sum. The algorithm doesn’t care about the absolute values of the numbers in the array, but it does pay attention to how each number affects the sum of the current subarray. Whether the number is extremely large, small, or even negative, the algorithm handles it correctly.

This demonstrates the robustness of Kadane’s algorithm and highlights its ability to deliver the correct solution in various scenarios. This sort of reasoning can also be applied to understand and validate other algorithms, making it a crucial skill in algorithm design and analysis.

In the case that the array contains all negative values, wouldn’t the dynamic programming solution given fall apart?

Kadane’s algorithm will work even if all the values in the array are negative. The dynamic programming approach works by keeping track of the current subarray sum and the maximum subarray sum found so far. If all values are negative, the maximum subarray sum would be the largest (or least negative) number in the array.

Here’s a brief explanation:

Initialize current_sum and max_sum to the first element of the array.
For each element in the array (from the second element onwards):
- Update current_sum to be the maximum of the current element and the sum of the current element and the current_sum from the previous step.
- If current_sum is greater than max_sum, update max_sum with the value of current_sum.
The max_sum at the end of the array will be the maximum subarray sum.

So, in the case of all negative numbers, during each iteration, current_sum will choose the current number if it’s larger (less negative) than the sum of the current number and the previous current_sum (which would be more negative). Hence, current_sum will always hold the least negative number encountered so far, and max_sum will be updated accordingly.

In conclusion, Kadane’s algorithm can handle arrays with all negative numbers correctly.

2 things to take care:

When temp sum is positive, if next one make it negative, get the new starting point When temp sum is negative, if next one is bigger, start the new temp sum with the next value

Yes, your understanding aligns with the concept behind Kadane’s algorithm. To be more specific:

When the temporary sum (current sum of the subarray) is positive, and adding the next element makes it negative, it indicates that starting a new subarray from the next element might lead to a larger sum subarray (because a positive subarray followed by a negative number results in a smaller sum). So, we update the starting point to be the next element.
If the temporary sum is negative, and the next element is larger than this sum, it’s better to start a new subarray from this next element (since it provides a larger sum than the current negative sum). In this case, we start the new temporary sum with the next value.

The crux of the algorithm is to always keep track of the maximum subarray sum found so far, and update it whenever we find a larger sum. This allows the algorithm to effectively identify the maximum sum subarray in a single pass through the array, achieving a time complexity of O(n).

So how can we assume that the max is either in the left right or cross? What if it’s a straight up positive array filled with positive numbers and all of them add up to the maximum sum?

The scenarios of maximum subarray sum being entirely in the left half, entirely in the right half, or spanning across the two halves are not mutually exclusive. They are merely possibilities to consider when you are breaking down the problem into smaller parts, as done in the divide and conquer approach.

When you say “a straight-up positive array filled with positive numbers and all of them add up to the maximum sum”, that’s perfectly valid and can be captured in any of the three scenarios:

Left Half: If the entire array is the “left half”, then the maximum sum subarray is the entire array.
Right Half: If the entire array is the “right half”, then the maximum sum subarray is the entire array.
Spanning Across: If we consider the mid point to be any index in the array, then the maximum subarray sum spans across the “left half” and “right half”.

So, even in the scenario you provided, these three cases cover it. The key is to understand that the three scenarios are not limiting but are designed to capture all possible configurations of the maximum subarray sum.

I was reading CLRS “divide and conquer” section and this problem was the example, but the example was introduced as a “best time to buy and sell stock” problem. So I just realized that actually in this problem every element in the list can be considered as the difference between two stock prices in two consecutive days. The O(n) solution is the same idea as the solution to the stock price problem, and it is easier to understand for some people, I guess.

Indeed, you’re correct. The “Maximum Subarray” problem and the “Best Time to Buy and Sell Stock” problem can essentially be seen as two sides of the same coin. Both of these problems involve finding the maximum difference in values within a sequence of numbers.

In the “Maximum Subarray” problem, you’re tasked with finding a contiguous subarray that results in the maximum sum. This can be thought of as adding up the differences between consecutive numbers.

In the “Best Time to Buy and Sell Stock” problem, you’re given a series of stock prices (for each day), and you need to find the maximum profit you can achieve by buying on one day and selling on another. This is essentially like finding a pair of numbers in the array such that their difference (price on the day of selling - price on the day of buying) is maximum, with the constraint that the selling day comes after the buying day.

Thus, if you transform the “Best Time to Buy and Sell Stock” problem by replacing each price with the difference between that price and the price of the previous day, then the problem reduces to finding a subarray with the maximum sum, which is exactly the “Maximum Subarray” problem.

Understanding these two problems in parallel can be beneficial, as the insights you gain from one can directly apply to the other. And as you mentioned, depending on a person’s background and preferred way of thinking, one problem may be easier to understand than the other.

One of the ways to solve this problem is sliding window , Keep on adding elements starting from left, whenever the sum becomes negative start removing elements from the beginning until it becomes non negative again. return the max of all these sums.

While sliding window is indeed a common technique for array problems, in this case it would not yield the correct answer. The sliding window approach generally assumes that adding more elements (expanding the window) could potentially improve the answer, which is not necessarily true for this problem.

Consider a scenario where we have an array of [-2, 1, -3, 4, -1, 2, 1, -5, 4]. The maximum subarray sum is 6 (from subarray [4, -1, 2, 1]). If we follow your approach, we start with -2, then add 1 to get -1, then add -3 to get -4. At this point, we’ve hit a negative sum, so we start removing elements from the beginning. After removing -2, we’re left with -2, but the next element is 4. If we had kept the -2, we could have reached a sum of 2 (4 - 2), which is greater than the sum of 4 alone.

In the Maximum Subarray problem, whether adding a new element would improve the solution depends not only on the new element itself, but also on the accumulated sum so far. The Kadane’s algorithm effectively implements this idea by keeping track of the maximum subarray ending at each position. If the sum of the subarray ending at the previous position is negative, then we start a new subarray at the current position; otherwise, we extend the previous subarray to include the current position.

This dynamic programming approach works because it exhaustively considers all subarrays ending at each position, and thus ensures that the maximum sum is found. The sliding window technique, on the other hand, could potentially miss some subarrays that yield a higher sum.

I tried submitting the divide and conquer approach which has complexity O(nlogn), but it is giving me TLE. Is it because my code has errors or?

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class Solution {
public:
	int callsum(int i, int j, vector&nums){

		if(i > j) return -1e9;
		if(i == j) return nums[i];
		int mid = (i+j)/2;
		int z = 0;
		int maxm = -1e9;
		for(int k = mid; k >= 0; k--) 
			z += nums[k],maxm = max(z, maxm);
		z = 0;
		int hi = maxm;
		for(int k = mid + 1; k < nums.size(); k++) 
			z += nums[k], maxm = max(z + hi, maxm);
		int x = callsum(i, mid, nums);
		int y = callsum(mid + 1, j, nums);
		
		return max({x, y, maxm});
	}
	int maxSubArray(vector& nums) {
		return callsum(0 ,nums.size()-1, nums);
	}
};

Your code looks fine for the divide-and-conquer approach. It’s true that the divide-and-conquer approach has a time complexity of O(n log n), which is higher than the O(n) time complexity of Kadane’s algorithm.

However, “Time Limit Exceeded” (TLE) errors often indicate that your code may be doing unnecessary work or has an inefficient implementation that causes it to take longer than the maximum allowed time.

There are a few things that could potentially be improved in your code:

Recursion overhead: While recursion can make the code more readable, it does add overhead, especially in languages like C++.
Unnecessary computations: The loops from mid to 0 and mid+1 to nums.size() will go out of the bounds of the current subarray when the subarray size is small, resulting in unnecessary computation.
Multiple calls to max function: You can store the maximums in variables to reduce the number of calls to the max function.

One more thing to consider, some online coding platforms could have tighter time limits, and algorithms with higher time complexity, even if correctly implemented, might still exceed the time limit. So the divide-and-conquer approach might still not pass even after optimizing the above points.

Also remember, it’s common practice in coding interviews to discuss the time complexity of your solution. If your interviewer asks for a more efficient solution, that’s your cue to discuss or implement the O(n) Kadane’s algorithm.

Example 3: Input: nums = [5,4,-1,7,8] Output: 23 Explanation: The subarray [5,4,-1,7,8] has the largest sum 23. How is Example 3 a subarray? Shouldn’t the answer be 15

In the context of this problem, a “subarray” of an array is a contiguous portion of the array. A subarray could consist of any number of elements, from a single element up to the entire array. So, in Example 3, [5,4,-1,7,8] is considered a subarray of the original array [5,4,-1,7,8] – in fact, it’s the entire array.

The subarray with the largest sum is [5,4,-1,7,8], and the sum of this subarray is 5+4+(-1)+7+8 = 23, not 15. The algorithm is trying to find the maximum sum that can be obtained from a subarray, not the maximum length or number of positive numbers in a subarray.

So the output 23 is correct for this example.

I have question. the problem indicate the nums element constraint can be from -10^4 to 10^4 which means the whole array can be negative nums even tho there is not a single test case that all array contains negative nums

Just because the constraints allow for elements to be in the range -10^4 to 10^4, it doesn’t mean that all the test cases will have all negative numbers. The constraints are there to provide the boundaries within which your solution must work, but the actual test cases used to evaluate your solution can be anything within those boundaries, including cases where all elements are negative, all elements are positive, or a mix of positive and negative elements.

For the Maximum Subarray problem, if all the elements in the input array are negative, then the maximum subarray is simply the element with the least negative value (i.e., the element closest to zero). This is because adding negative numbers together only makes the sum more negative.

For example, if you have an input array like [-3, -2, -1], the maximum subarray is simply [-1], and the maximum sum is -1.

Here’s an example solution using Kadane’s algorithm:

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class Solution:
    def maxSubArray(self, nums: List[int]) -> int:
        if max(nums) < 0:
            return max(nums)
        else:
            max_sum = 0
            curr_sum = 0
            for num in nums:
                if curr_sum < 0:
                    curr_sum = 0
                curr_sum += num
                max_sum = max(max_sum, curr_sum)
            return max_sum

With this code, if you input [-3, -2, -1], the output will indeed be -1. Even though there is not a single test case where all numbers are negative, your solution should still be able to handle that situation.

Here, what if all the integers are negative? for example, [-1,-9, -2,-5] then largest sum subarray might be [-1], but how can we assume empty subarray equals to zero.

In the context of the Maximum Subarray problem, an empty subarray is generally not considered, as the problem constraints specify that the size of the input array nums is at least 1.

If all the numbers in the array are negative, then the maximum subarray would be the subarray containing the single number with the least negative value (i.e., the number closest to zero). So, for the example [-1,-9, -2,-5], the maximum subarray is indeed [-1].

There’s no need to consider an empty subarray in this case, because an empty subarray would have a sum of zero, which would be less than any non-negative number and only more than negative numbers. Since the array always contains at least one number, the maximum subarray sum will always be at least as large as the maximum individual number.

If you are using Kadane’s algorithm to solve this problem, it essentially does the same thing - it maintains a running sum of the array, and if the sum becomes negative at any point, it resets the sum to zero (effectively “discarding” the previous numbers), because a negative sum would only decrease the total sum if we were to add more positive numbers to it. However, the algorithm also keeps track of the maximum sum it has seen so far, so even if all the numbers are negative, it will still return the least negative number as the maximum subarray sum.

Here’s the algorithm for reference:

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class Solution:
    def maxSubArray(self, nums: List[int]) -> int:
        max_sum = nums[0]
        curr_sum = 0
        for num in nums:
            if curr_sum < 0:
                curr_sum = 0
            curr_sum += num
            max_sum = max(max_sum, curr_sum)
        return max_sum

In the case of all negative numbers, curr_sum would always be reset to 0, but max_sum would keep track of the maximum (least negative) number seen.

For example nums = [-1, -2, -3] Why subarray can’t be empty and maxSubArray = 0?

In the context of the Maximum Subarray problem, the definition of a subarray typically requires that it contains at least one element. An empty set does not constitute a subarray as per this definition. Therefore, in the case of nums = [-1, -2, -3], the maximum subarray cannot be empty and the maximum subarray sum cannot be 0.

If you consider an empty subarray as valid and having a sum of 0, then for any array that only has negative numbers, the maximum subarray would always be an empty set and the maximum sum would always be 0. But by requiring that a subarray contains at least one number, the problem becomes more interesting, as you then have to consider which number (or numbers) maximizes the sum.

So for nums = [-1, -2, -3], the maximum subarray would be [-1], and the maximum subarray sum would be -1, because -1 is greater than -2 and -3.

Largest sum right? why are we required to get a subarray with negative values for eg -1 in requirements?

The idea is to find the subarray that has the “maximum sum”. In your example of an array with all negative values (e.g., [-1, -2, -3]), a subarray with a single element of -1 does provide the maximum sum. Although it’s a negative value, -1 is greater than -2 and -3.

The purpose of this requirement is to find the maximum sum of a continuous sub-sequence in the array, even if the maximum sum is negative. It adds more complexity to the problem as compared to simply discarding all negative numbers.

So even if the input array only has negative numbers, we don’t consider the maximum sum to be 0 by choosing an empty subarray; instead, we consider the maximum negative number as the maximum sum. If we allowed an empty subarray to be the maximum, then the solution for any input array with all negative numbers would be 0, which simplifies the problem and ignores the fact that a higher (less negative) number is actually “greater” than a lower (more negative) number.

Hint: if the the subarray has the largest sum, then the remaining elements have the minimum sum(either negative or no elements).

Actually, that hint isn’t entirely correct. It’s possible for a subarray with the maximum sum to not leave the remaining elements with the minimum sum. Consider the array [2, 1, -3, 4, -1, 2, 1, -5, 4]. The subarray with the largest sum is [4, -1, 2, 1] with a sum of 6. However, the remaining elements [2, 1, -3, -5, 4] don’t necessarily sum to the minimum possible sum in this array.

What we can say is that if we’re trying to find the subarray with the maximum sum, it doesn’t make sense to include a prefix or suffix of negative numbers in that subarray. That’s because removing that prefix or suffix would lead to a higher sum.

So in the array [2, 1, -3, 4, -1, 2, 1, -5, 4], the subarray [4, -1, 2, 1] does not include the prefix [-3] or the suffix [-5], because doing so would lead to a smaller sum.

A better hint might be:

When examining the elements in the array, if the current element plus the running total (subarray sum up to this point) is less than the current element itself, then the running total should be discarded (set to zero) and start a new subarray from the current element. This is because a maximum subarray sum will never include a previous sum that is negative, it is better to start a new sum from the current element. This is the core idea behind Kadane’s algorithm.

what do you guys think of such an algorithm

condense the array(in the sense that you combine all contiguous positive and contiguous negative numbers together so that you end up with an array with alternating positive and negative numbers) find the largest number in the condensed array pair up positive and negative numbers together starting from the largest number and excluding the largest number(so if the largest number is index 3 in an array of length 7, index 2 and 1 will be paired, 0 will be alone, 4 and 5 will be paired, 6 will be alone. recur over and over again until 2 end cases: a. length of the array becomes 2(guaranteed to be a positive and negative number, so just take the positive number) b. length of array becomes 3(alternates between positive and negative number e.g. pos neg pos or neg pos neg, so just take the largest number of the 3) im not sure what the time complexity and space complexity of such an algorithm would be, would anyone be so kind as to critique such an algorithm thanks

i am using divide and conqure still getting error in [-2,1,-3,4,-1,2,1,-5,4] testcase kindly see my code and help me

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class Solution {
public:
	int maxsum(vector&nums, int n){
		if(n == 1){
			return nums[0];
		}
		int m = n/2;
		int leftmaxsum = maxsum(nums, m);
		int rightmaxsum = maxsum(nums, n-m);
		int leftsum = -1000000;
		int rightsum = -100000000;
		int sum = 0;
		for(int i = m-1; i >= 0; i--){
			sum += nums[i];
			leftsum = max(leftsum, sum);

		}
		sum = 0;
		for(int i = m; i < n-m-1; i++){
			sum += nums[i];
			rightsum = max(rightsum, sum);
		}
		int ans = max(leftmaxsum, rightmaxsum);
		int temp= leftsum + rightsum;
		ans = max(ans, temp);

		return ans;
	}
	int maxSubArray(vector<int>& nums) {
		int n = nums.size();
		return maxsum(nums, n);
	}
};

The divide and conquer approach you’ve implemented for the problem seems to be the correct idea. However, there seems to be an issue with how the array is split in the recursive calls.

In your code, when you call maxsum(nums, m) and maxsum(nums, n-m), you’re always starting from the beginning of the array, instead of dividing it into two parts. That’s why it’s causing the error.

Instead of passing the length of the subarray, you should pass the starting and ending indices of the subarray. And when you’re making recursive calls, you should properly specify the start and end indices of the left and right subarrays.

Here’s a corrected version of your function:

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class Solution {
public:
    int maxsum(vector<int>& nums, int start, int end) {
        if(start == end)
            return nums[start];

        int mid = start + (end - start) / 2;
        int leftmaxsum = maxsum(nums, start, mid);
        int rightmaxsum = maxsum(nums, mid + 1, end);

        int leftsum = INT_MIN, rightsum = INT_MIN, sum = 0;
        for(int i = mid; i >= start; i--) {
            sum += nums[i];
            leftsum = max(leftsum, sum);
        }

        sum = 0;
        for(int i = mid + 1; i <= end; i++) {
            sum += nums[i];
            rightsum = max(rightsum, sum);
        }

        int crosssum = leftsum + rightsum;
        return max(max(leftmaxsum, rightmaxsum), crosssum);
    }

    int maxSubArray(vector<int>& nums) {
        int n = nums.size();
        return maxsum(nums, 0, n - 1);
    }
};

The divide and conquer approach works by dividing the array into two halves and solving the problem for each half. It then combines the solutions from each half by considering the maximum subarray sum that could cross the midpoint. The maximum of these three is returned as the final result. The time complexity is O(n log n) because it’s a divide and conquer strategy similar to merge sort. The space complexity is O(log n) due to the recursion stack.

(186/210) [2,-1,-1,2,0,-3,3] with expected 3. How?

The maximum subarray in your example [2,-1,-1,2,0,-3,3] is [2,-1,-1,2,0,-3,3] itself, and it sums to 3. Here’s how you calculate it:

2 - 1 - 1 + 2 + 0 - 3 + 3 = 2

The subarray sums up to 2, not 3. It seems there might be a mistake in the expected result. If the maximum subarray sum was indeed 3, then it seems the input array or expected result is not correct according to the rules of the maximum subarray problem.

i want to know how we fiind a subarray .In my point of view positive to positve ex 4,-1,2,1 -> and one more example nums = [5,4,-1,7,8] whole index ..Whether my understanding is right?

Your understanding is partly right, but there’s more to the problem. We’re looking for a contiguous subarray (a sub-section of the array) with the largest sum. This doesn’t necessarily mean we only add the positive numbers. Negative numbers can also be part of the maximum subarray if they’re surrounded by large enough positive numbers.

Let’s go through the array you provided to illustrate this:

Given array nums = [2,-1,-1,2,0,-3,3]

We start with the first number, 2. This is our current maximum.

Next, we add -1, getting a sum of 1. Even though -1 is negative, adding it to our sum still results in a positive number, so we keep this subarray.

The next number is -1 again, our sum becomes 0. It’s not beneficial, but it’s not reducing our maximum found so far (2), so we keep going.

The next number is 2, so our sum is now 2. This is equal to our maximum found so far.

Then we have 0, -3, and 3. Adding these, we end up with a sum of 2, which is equal to our maximum.

The maximum subarray can thus be [2, -1, -1, 2] or [2, -1, -1, 2, 0, -3, 3] both resulting in a sum of 2.

You can see that we have included negative numbers in our subarray because, even though they momentarily decrease the sum, they may be part of a larger sum subarray. The algorithm essentially tries different “parts” of the array to find the combination that yields the highest sum.

I’m new to divide and conquer can someone tell me where I am wrong in this code? (maximum subarray problem )

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class Solution {
public:
	int maxSubArray(vector& nums,int s,int e) {
		int mid;
		s = nums.front();
		e = nums.back();
		if(s == e){
			return nums.at(s);
		}
		mid = e+(s-e)/2;
		int leftpart = maxSubArray(nums, s, mid);
		int rightpart=maxSubArray(nums, mid + 1, e);
		int rmax =0;
		int lmax =0;
		int lsum =0;
		int rsum =0;
		
		for(int i = mid;i >= s; i--){
			lsum = lsum + nums[i];
			if(lsum > lmax){
				lmax = lsum;
			}
		}
		for(int i= mid + 1; i <= e; i++){
			rsum = rsum + nums[i];
			if(rsum > rmax){
				rmax = rsum;
			}
		}
		int overlapping = rmax + lmax;
		int maxisubsum = max(max(rmax, lmax), overlapping);
		return maxisubsum;

	}
};

Your code seems to be implementing the divide and conquer approach for the maximum subarray problem. It tries to find the maximum subarray sum in the left half of the array, right half of the array, and the maximum crossing the midpoint. However, there are some issues in your code:

Incorrect Initialization: You’re initializing s and e with nums.front() and nums.back(), which represent the first and the last elements of the array, not the indices.
Misused .at(): In the base case, you’re using nums.at(s) which could cause out-of-bounds errors because s is initialized with a value from the array, not an index.
Not Calling maxSubArray from Outside: You have not called maxSubArray method from anywhere. You may want to define a helper method which is called from a public method maxSubArray(vector<int>& nums) to start the recursion.

Here’s your corrected code:

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class Solution {
public:
    int maxCrossingSum(vector<int>& nums, int s, int mid, int e) {
        int sum = 0;
        int left_sum = INT_MIN;
        for (int i = mid; i >= s; i--) {
            sum = sum + nums[i];
            if (sum > left_sum)
              left_sum = sum;
        }

        sum = 0;
        int right_sum = INT_MIN;
        for (int i = mid + 1; i <= e; i++) {
            sum = sum + nums[i];
            if (sum > right_sum)
              right_sum = sum;
        }

        return max(left_sum + right_sum, max(left_sum, right_sum));
    }

    int maxSubArraySum(vector<int>& nums, int s, int e) {
        if (s == e)
            return nums[s];

        int mid = s + (e -s) / 2;

        return max(max(maxSubArraySum(nums, s, mid),
                    maxSubArraySum(nums, mid+1, e)),
                    maxCrossingSum(nums, s, mid, e));
    }

    int maxSubArray(vector<int>& nums) {
        return maxSubArraySum(nums, 0, nums.size()-1);
    }
};

In this corrected version, we define three functions: maxCrossingSum finds the maximum subarray sum that crosses the midpoint, maxSubArraySum is a helper function that implements the divide-and-conquer approach, and maxSubArray is the main function that calls maxSubArraySum. Also, the base case in maxSubArraySum checks if s == e and returns the value at index s (or e, they’re the same in this case) of the array.

Use sum as 0 and maximum as the first element of the array. use a for loop from 0 to last element of array and add it as sum + the element, and check whether the sum is greater than maximum or not if yes then maximum = sum and also check whether the sum is less than 0 or not if yes then sum =0 and return the maximum value;

Yes, that’s a correct summary of the Kadane’s algorithm, which is an efficient way to solve the maximum subarray problem. Here’s the implementation of what you just described in C++:

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class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        int max_so_far = nums[0];
        int curr_max = 0;

        for (int i = 0; i < nums.size(); i++) {
            curr_max = curr_max + nums[i];
            if (max_so_far < curr_max)
                max_so_far = curr_max;

            if (curr_max < 0)
                curr_max = 0;
        }
        return max_so_far;
    }
};

The max_so_far variable tracks the maximum subarray sum found so far, and curr_max keeps track of the current sum of elements. If the curr_max becomes negative at any point, it is reset to 0, because a negative sum would not contribute to a larger sum of a future subarray. If curr_max exceeds max_so_far, then max_so_far is updated to be the value of curr_max.

This algorithm works by essentially looking for all positive continuous sections of the array (curr_max is used for this). And keep track of maximum sum contiguous segment among all positive segments (max_so_far is used for this). Each time we get a positive sum compare it with max_so_far and update max_so_far if it is greater than max_so_far. This algorithm has a linear time complexity, O(n), making it efficient for large inputs.

I know kadane algo and solved but can anyone tell me if question was that print the maximum subarray it actuall subarray and return type is vector than what changes we need to made in this code

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int maxSubArray(vector& nums) {
	int ans = INT_MIN;
	int n = nums.size();
	int CS = 0;
	for(int i = 0; i < n; i++)
	{
		CS += nums[i];
		ans = max(ans, CS);
		if(CS < 0)
			CS = 0;
	}
	return ans;
}

The code you provided finds the maximum subarray sum using Kadane’s alg`orithm, but doesn’t keep track of the indices where the maximum sum subarray starts and ends. If we modify your code to remember the beginning and ending indices of the maximum subarray, we can use those indices to return the maximum subarray itself. Here’s how you could do it:

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vector<int> maxSubArray(vector<int>& nums) {
    int maxSum = INT_MIN, currSum = 0;
    int start = 0, end = 0, tempStart = 0;
    int n = nums.size();

    for(int i = 0; i < n; i++) {
        currSum += nums[i];

        // If new sum is greater, update maxSum and indices
        if (maxSum < currSum) {
            maxSum = currSum;
            start = tempStart;
            end = i;
        }

        // If current sum is negative, start a new sum
        if (currSum < 0) {
            currSum = 0;
            tempStart = i + 1;
        }
    }

    // Using the start and end indices, construct the max subarray
    vector<int> maxSub(nums.begin() + start, nums.begin() + end + 1);

    return maxSub;
}

In this updated code, tempStart remembers the starting index of a potential maximum subarray. When we find a larger sum, we update start and end to be the current boundaries of the maximum subarray. After the loop, we use the start and end indices to create a new vector that contains the maximum subarray, and return that vector.

//im getting error is this write approach to do this problem

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class Solution {
public:
	int maxSubArray(vector& nums) {
		vectora(nums.size(),0);
		int n = nums.size();
		for(int i = 1; i < n; i++){
			nums[i] = nums[i] + nums[n-i];
		}

		a = nums;
		for(int i = 1; i < n; i++){
			a[i] = a[i] - a[n-1];
		}
    
		int max = a[0];

		for (int j = 1; j < n; j++)
			if (a[j] > max)
				max = a[j];

		return max;
	}
};

There seem to be a few issues in the code you posted.

Syntax Error: You need to declare the data type of your vector a. So, it should be vector<int> a(nums.size(), 0); instead of vector a(nums.size(), 0);.
Invalid Indexing: The line nums[i] = nums[i] + nums[n-i]; could result in accessing elements beyond the array when i = 0. Because n-i when i = 0 would give n, and nums[n] is out of bounds.
Incorrect Logic: The logic you are using doesn’t seem to follow the Kadane’s algorithm, which is the recommended approach for the maximum subarray problem. The operations you’re performing on the array don’t align with finding the maximum sum of a subarray.

Here’s an example of a correct implementation using Kadane’s algorithm:

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class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        int max_sum = INT_MIN, curr_sum = 0;
        int n = nums.size();

        for (int i = 0; i < n; i++) {
            curr_sum += nums[i];
            if (max_sum < curr_sum)
                max_sum = curr_sum;
            if (curr_sum < 0)
                curr_sum = 0;
        }

        return max_sum;
    }
};

This implementation starts with max_sum set to the smallest possible integer and curr_sum set to 0. It iterates through the array, adding the current number to curr_sum. If curr_sum is larger than max_sum, max_sum is updated. If curr_sum becomes negative, it is reset to 0. At the end of the iteration, max_sum contains the maximum sum of a subarray in nums.

how could i solve this with divide and conquer?? if anyone know please explain it.

The divide and conquer approach for the maximum subarray problem is a bit more complex than the iterative approach like Kadane’s algorithm. It is based on the principle of breaking down the problem into smaller subproblems, solving them, and combining their results to find the solution to the original problem. Here’s a step by step breakdown:

Divide the given array in two halves.
Return the maximum of the following three:
- Maximum subarray sum in the left half (recursive call).
- Maximum subarray sum in the right half (recursive call).
- Maximum subarray sum such that the subarray crosses the midpoint.

The third case is handled differently. We first find the maximum sum starting from the mid point and ending at some point on the left half, then find the maximum sum starting from mid + 1 and ending at some point on the right half and finally combine the two.

Here’s the C++ code that implements this:

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class Solution {
public:
    int crossSum(vector<int>& nums, int left, int right, int mid) {
        if (left == right) return nums[left];

        int leftSubsum = INT_MIN;
        int currSum = 0;
        for (int i = mid; i > left - 1; --i) {
            currSum += nums[i];
            leftSubsum = max(leftSubsum, currSum);
        }

        int rightSubsum = INT_MIN;
        currSum = 0;
        for (int i = mid + 1; i < right + 1; ++i) {
            currSum += nums[i];
            rightSubsum = max(rightSubsum, currSum);
        }

        return leftSubsum + rightSubsum;
    }

    int helper(vector<int>& nums, int left, int right) {
        if (left == right) return nums[left];

        int mid = left + (right - left) / 2;

        int leftSum = helper(nums, left, mid);
        int rightSum = helper(nums, mid + 1, right);
        int crossSum = this->crossSum(nums, left, right, mid);

        return max(max(leftSum, rightSum), crossSum);
    }

    int maxSubArray(vector<int>& nums) {
        return helper(nums, 0, nums.size() - 1);
    }
};

In the code above, crossSum calculates the maximum subarray sum that crosses the midpoint, helper performs the recursive division of the array and calculates the maximum subarray sums, and maxSubArray is the main function that calls helper to start the recursion. The result of maxSubArray is the maximum subarray sum in the input array nums.

Why my solution is beating only 41% of answers? Any help please to improve my working code.

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class Solution {
	public int maxSubArray(int[] nums) {
		int maxSoFar = Integer.MIN_VALUE;
		int maxTillYet = 0;

		for(int i = 0; i < nums.length; i++) {
			maxTillYet += nums[i];
			if (maxSoFar < maxTillYet) {
				maxSoFar = maxTillYet;
			}
			if (maxTillYet < 0) {
				maxTillYet = 0;
			}
		}
		return maxSoFar;
	}
}

Your code is using Kadane’s algorithm, which is the most optimal solution for the maximum subarray problem. It already runs in O(n) time complexity and uses O(1) space complexity, meaning you can’t really optimize it further from a time or space complexity standpoint.

The percentage you see on online judge platforms like LeetCode (41% in your case) is not entirely dependent on the efficiency of your algorithm. The speed of your solution can also be affected by factors that aren’t really related to the algorithm’s efficiency, like:

The programming language you use: Some languages are inherently faster than others. Solutions in languages like C or C++ often run faster than equivalent solutions in languages like Python or Java.
System-level factors: These include things like the load on the online judge servers at the time of submission, the system’s I/O processing speed, etc.

Therefore, while it’s always a good idea to aim for efficient code, don’t be too concerned about the percentage reported by the online judge system. It’s more important to make sure that your solution is correct, readable, and as efficient as it can be in terms of time and space complexity.

Intuition behind kadane’s algo At index i what it the max subarray sum?

Can previous sum, nums[i-1], help maximize my current nums[i] Given nums=[-2, 1] , i=1

Relationship between previous sum and current number: In order to maximize it

prev sum cur max why +1 +1 1+1 prev+ cur -1 +1 1 cur +1 -1 0 prev+cur, cur must be included always! -1 -1 -1 cur Options:

nums[i] is positive, then nums[i] is part of the solution nums[i-1] + nums[i] : previous + i nums[i] : previous discarted, perhaps sum to 0 or negative number nums[i] is negative nums[i-1]+nums[i]: previous being is positive will help increasing nums[i] nums[i]: if both previous is also negative it should not be added

Yes, your understanding of Kadane’s algorithm is correct.

The main intuition behind Kadane’s algorithm is that any maximum subarray sum ending at index i will be either the maximum subarray sum ending at i-1 plus nums[i], or just nums[i], whichever is larger. In other words, the maximum subarray sum ending at an index i is either a continuation of the maximum subarray sum at i-1 or it starts from the current element nums[i] itself.

This is based on the idea that as we iterate through the array, we are interested in finding the maximum sum up to the current position. We only want to include previous sum if it’s positive because that would increase our current sum. If previous sum is negative, we discard it and start a new subarray from the current position.

As per the given options and your understanding:

For nums[i] is positive:

nums[i-1] + nums[i]: If the previous sum is positive, it would increase our current sum. Hence, we continue with the sum.
nums[i]: If the previous sum is negative, we would start a new sum from current position.

For nums[i] is negative:

nums[i-1] + nums[i]: Here, the previous sum being positive might help in increasing the current negative number. Hence, we continue with the sum.
nums[i]: If the previous sum is also negative, adding it would not help. Hence, we start a new sum from current position.

Remember that the core of Kadane’s algorithm is to look for all positive continuous sections of the array and keep track of maximum sum contiguous segment among all positive segments (max_so_far is used for this). Each time we get a positive sum, compare it with max_so_far and update max_so_far if it is greater than max_so_far.

In the problem they mentioned maximum sum of subarray So what If I add only positive numbers so that it will make maximum number..? Can I do that Anyone clarify my doubt

The Maximum Subarray problem asks for the maximum sum of a contiguous subarray, meaning the elements in the subarray have to be next to each other in the original array. This is different from simply adding up all the positive numbers in the array.

For example, consider an array like [-2, -3, 4, -1, -2, 1, 5, -3]. If you add up all the positive numbers, you’ll get 4 + 1 + 5 = 10. However, the maximum sum of a contiguous subarray is 4 - 1 - 2 + 1 + 5 = 7, which includes some negative numbers.

Adding up all positive numbers in the array would not give you the correct answer to this problem, because it ignores the requirement that the subarray be contiguous. If the problem instead asked for the maximum sum of a non-contiguous subarray, then yes, simply adding up all the positive numbers would give you the correct answer. But in the Maximum Subarray problem, the subarray must be contiguous.

this is my solution and pass the all test case but not pass the only nums[] = {-1}

please help anyone ;

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class Solution {
	public int maxSubArray(int[] nums) {
		int ms = Integer.MIN_VALUE ;
		int cs = 0 ;
		for(int i = 0 ; i < nums.length; i++){
			cs += nums[i];
			if(cs < 0){
				cs = 0;
			}
			ms = Math.max(cs, ms) ;
		}
		return ms ;
	}
}

The issue with your code lies in the check if(cs < 0), which resets the current sum cs to 0 if it falls below 0. This approach works fine for arrays that have at least one positive number, because the maximum sum will always be greater than 0. However, if the array contains only negative numbers (like in your case nums[] = {-1}), then the maximum sum will be the maximum (or least negative) number in the array.

To handle this case, you should check whether cs is greater than ms before resetting cs to 0. Here is the corrected code:

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class Solution {
    public int maxSubArray(int[] nums) {
        int ms = Integer.MIN_VALUE ;
        int cs = 0;
        for(int i = 0; i < nums.length; i++){
            cs += nums[i] ;
            ms = Math.max(cs, ms);
            if(cs < 0){
                cs = 0 ;
            }
        }
        return ms ;
    }
}

This code works by first updating ms with the maximum of cs and ms, and then checking if cs has fallen below 0. If it has, cs is reset to 0. By updating ms before the check, you ensure that ms receives the highest sum encountered, even if that sum is negative.

To solve this problem in O(n) time complexity, you can imagine it by using the concept of sliding window.

The problem can indeed be solved in O(n) time complexity, but it’s not exactly a sliding window approach. The most common solution is Kadane’s algorithm, which is an example of dynamic programming.

In the sliding window approach, the window size changes based on certain conditions while moving through the array. However, in this problem, we are not changing the window size based on a condition while iterating through the array. Instead, we are just keeping track of the current maximum sum and the overall maximum sum at each step.

Here is how you can solve the problem using Kadane’s algorithm (in Python):

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def maxSubArray(nums):
    current_sum = max_sum = nums[0]
    for num in nums[1:]:
        current_sum = max(num, current_sum + num)
        max_sum = max(max_sum, current_sum)
    return max_sum

This algorithm works by iterating through the array and at each step, it decides whether to add the current element to the previous maximum sum or start a new sum from the current element. It also keeps track of the maximum sum seen so far.

So while the time complexity is the same as the sliding window approach (both are O(n)), the methodology is different.

Why it is not a sliding window? What are the reasons?

In a sliding window problem, we are given an array or a list of elements, and a “window” of a fixed size is defined. The goal is often to find or calculate something in that window, and then the window “slides” over one or more places, and the calculation is repeated until the window reaches the end of the array.

The main characteristics of the sliding window problems are:

The window size is defined and it either stays constant or changes according to certain conditions as we traverse the array.
There is often a condition that needs to be satisfied for all elements in the window (for example, the sum of elements in the window must be less than a given value).

In the maximum subarray problem:

There is no fixed window size. Instead, we have a variable-length subarray that could include any number of contiguous elements from the array.
There is no condition that all elements in the subarray need to satisfy. Instead, we’re looking for the subarray (of any size) that has the maximum sum.

So, while you can solve the maximum subarray problem in O(n) time (which is common in sliding window problems), it’s not accurate to categorize it as a sliding window problem because it doesn’t have the key characteristics of a sliding window problem. Instead, it’s usually categorized as a dynamic programming problem, because it involves making an optimal decision at each step based on the decisions made in previous steps.

Here is my solution with O(n) tc but its wrong answer for one case that is {-1}. if anyone know plz help me

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class Solution {
public:
	int maxSubArray(vector& nums) {
		int maxsum = INT_MIN;
		int currsum = 0;

		for(int i = 0; i < nums.size(); i++){
			currsum += nums[i];
        
			if(currsum < 0){
				currsum = 0;
			}
			maxsum = max(maxsum, currsum);
		}
		return maxsum;

	}
};

Your solution is almost correct. The problem occurs when all the numbers in the array are negative. In this situation, your current implementation will return 0, because it resets currsum to 0 whenever currsum becomes negative. However, for an array of all negative numbers, the maximum subarray sum is the maximum (or least negative) number in the array.

To fix this, you should update maxsum before checking whether currsum is negative. Here is the corrected code:

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class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        int maxsum = INT_MIN;
        int currsum = 0;

        for(int i = 0; i<nums.size(); i++){
            currsum += nums[i];
            maxsum = max(maxsum, currsum);

            if(currsum < 0){
                currsum = 0;
            }
        }
        return maxsum;
    }
};

Now, maxsum is updated with the maximum sum found so far before currsum is reset to 0. This way, even if all numbers in the array are negative, maxsum will hold the maximum (least negative) number.

Hey why it’s not working

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

    max_sum = -float('inf')
    sum_ = 0
    start = 0
    end = len(nums)-1
    prefix_sum = [nums[0]]
    for i in range(1,len(nums)):
        prefix_sum.append(nums[i] + prefix_sum[-1])
    for start in range(len(nums)):
        if start==0:
            sum_ = prefix_sum[end]
        else:
            sum_ = prefix_sum[end] - prefix_sum[start-1]

        if sum_ > max_sum:
            max_sum = sum_
    return max_sum                

Your current implementation does not correctly solve the problem, because it only calculates the sum from the beginning of the array to an end index, and it does not consider all possible subarrays.

In other words, your current implementation only calculates the prefix sum of the array and returns the maximum prefix sum. This is not the same as finding the maximum sum of any subarray, which could start and end anywhere in the array.

For example, consider the array [-2,1,-3,4,-1,2,1,-5,4]. The maximum sum of any subarray is 6, from the subarray [4,-1,2,1]. However, your current implementation would return 7, which is the sum of the entire array from the beginning to the second-to-last element.

To solve this problem, you could use Kadane’s algorithm, which involves keeping track of the current subarray sum and the maximum subarray sum seen so far:

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class Solution:
    def maxSubArray(self, nums: List[int]) -> int:
        max_sum = nums[0]
        curr_sum = nums[0]
        for num in nums[1:]:
            curr_sum = max(num, curr_sum + num)
            max_sum = max(max_sum, curr_sum)
        return max_sum

This algorithm works by iterating through the array, and at each step, it decides whether to add the current element to the previous subarray or start a new subarray from the current element. It also keeps track of the maximum subarray sum seen so far.

There is no need to find the subarray we can just traverse the original array and find the maximum sum of any subpart.

Yes, that’s correct. The problem only asks for the maximum sum, not for the actual subarray that produces this sum.

In this case, you can use Kadane’s algorithm, which is an efficient way to solve this problem with a time complexity of O(n). Here’s how it works:

Initialize two variables, max_sum and current_sum, both set to the first element of the array.
Iterate through the array, starting from the second element. For each element, update current_sum to be the maximum of the current element and current_sum + current_element. This step means that current_sum is the maximum sum ending at the current element.
Update max_sum to be the maximum of max_sum and current_sum. This step means that max_sum is the maximum sum of any subarray seen so far.
Repeat steps 2-3 until you’ve processed the entire array. At the end, max_sum will be the maximum sum of any subarray in the array.

In Python, this could look like:

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def maxSubArray(nums):
    current_sum = max_sum = nums[0]
    for num in nums[1:]:
        current_sum = max(num, current_sum + num)
        max_sum = max(max_sum, current_sum)
    return max_sum

Even though this solution doesn’t find the actual subarray, it still correctly solves the problem as it is asked.

I have assigned max sum as the first value of the array. But as soon as the value is -ve, my code does not take the next +ve element as the largest value. Can someone help?

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class Solution
{
	public int maxSubArray(int[] nums)
	{

		int max_sum = nums[0];
		int c_sum = nums[0];
		for(int i=1;i<nums.length;i++)
		{
        
			c_sum = c_sum + nums[i];

			if(c_sum > max_sum)
			{
				max_sum = c_sum;
			}

			if(c_sum < 0)
			{
				c_sum = 0;
			}

		}
		return max_sum;
	}
}

Your current implementation is almost correct. However, the order of your conditions is causing the code to not consider a positive number immediately after the c_sum becomes negative.

In your implementation, you’re adding nums[i] to c_sum and then checking if c_sum is greater than max_sum. If c_sum becomes negative after the addition, you’re resetting c_sum to 0. But in the next iteration, you’re still adding nums[i] to c_sum before checking if c_sum is negative, which could cause you to miss a positive number.

You should first check if c_sum is negative, reset it if necessary, and then add nums[i] to c_sum. Here is the corrected code:

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class Solution
{
    public int maxSubArray(int[] nums)
    {
        int max_sum = nums[0];
        int c_sum = 0;

        for(int i=0; i<nums.length; i++)
        {
            if(c_sum < 0)
            {
                c_sum = 0;
            }

            c_sum = c_sum + nums[i];

            if(c_sum > max_sum)
            {
                max_sum = c_sum;
            }
        }
        return max_sum;
    }
}

Now, you’re correctly considering every positive number in the array. Also, by setting c_sum = 0 before the loop, you no longer need to initialize c_sum to nums[0], which makes the code slightly more efficient.

Can someone help me with this. My code is working fine but consider a case of [-2,-1]. Since my maxSum is set to 0. My output is coming 0 instead of-1. How should I modify my code?

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class Solution {
	public int maxSubArray(int[] nums) {
		int sum = 0, maxSum = 0, i = 0;
		if(nums.length == 1) return nums[0];
		for(i = 0; i < nums.length-1; i++){
			sum += nums[i];
			for(int j = i+1; j < nums.length; j++){
				sum += nums[j];
				if(sum > maxSum){
					maxSum = sum;
				}
			}
			sum = 0;
		}
		return (nums[i] > maxSum) ? nums[i] : maxSum;
	}
}

Your current implementation checks the sum of all subarrays in the array and it is not efficient as it runs in O(n^2) time complexity. A better way is to use Kadane’s algorithm as mentioned before.

In addition, your current approach uses maxSum=0 as a starting value, which means it will not work correctly if all numbers in the array are negative, because the maximum sum would be less than zero.

Instead of starting maxSum at 0, you should start it at the smallest possible integer value, Integer.MIN_VALUE, which guarantees that any sum of subarray will be larger than the initial maxSum. Here’s how you can modify your code:

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class Solution {
    public int maxSubArray(int[] nums) {
        int sum = 0;
        int maxSum = Integer.MIN_VALUE;
        for(int i = 0; i < nums.length; i++){
            sum += nums[i];
            if(sum > maxSum){
                maxSum = sum;
            }
            if(sum < 0){
                sum = 0;
            }
        }
        return maxSum;
    }
}

This code initializes maxSum to Integer.MIN_VALUE and updates maxSum whenever sum is greater than maxSum. If sum becomes negative, it resets sum to 0 because no max subarray sum would include a negative sum.

is a list of one element considered to be a subarray? E.g. for [-1, 100, -1] will a correct answer be sum([100])==100 or sum([-1,100])==99?

Yes, a list with just one element is considered a subarray. In your example, the subarray with the maximum sum would indeed be [100], which results in a sum of 100.

The term “subarray” just means a contiguous (i.e., unbroken, with no gaps) subset of the array. It doesn’t specify a minimum or maximum size. So a subarray could have as few as one element or as many as the entire array.

So in your example, both [100] and [-1, 100] are valid subarrays, but the one with the maximum sum is [100].

What if there are only negatives in the array?

I believe the regular Kadane’s Algorithm will take care of it if we initialize the maxSum variable correctly. By this I mean, for Python, I did maxSum = -float(‘inf’). Even if all the values are negative, this still works.

Yes, you are correct. The key is to initialize maxSum as negative infinity or the smallest possible number (for languages that don’t support negative infinity). This ensures that maxSum gets updated even if all numbers in the array are negative.

When all elements of the array are negative, the maximum subarray sum would be the least negative number. This is because adding any additional negative numbers would only decrease the sum. Kadane’s Algorithm correctly handles this situation by considering each element as a potential starting point of a new subarray when the sum of the current subarray becomes negative.

Here’s the Python implementation of Kadane’s Algorithm with maxSum initialized to negative infinity:

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def maxSubArray(nums):
    maxSum = -float('inf')
    currSum = 0

    for num in nums:
        currSum = max(num, currSum + num)
        maxSum = max(maxSum, currSum)

    return maxSum

In this code, currSum is the maximum sum ending at the current position, and maxSum is the maximum sum of any subarray seen so far. For each number, it considers either starting a new subarray from the current position or extending the current subarray. This ensures that maxSum is updated correctly even when all numbers are negative.

What should be the length of the sub array which has largest sum?

The length of the subarray with the largest sum could be any length between 1 (if a single element is the maximum sum) and the length of the entire array (if all elements contribute to a maximum sum). It’s entirely dependent on the array and isn’t a fixed length.

For example, consider the array [5, -2, 3]. Here, the subarray with the maximum sum is [5, -2, 3], which has a length of 3.

Now, consider the array [-2, 1, -3, 4, -1, 2, 1, -5, 4]. Here, the subarray with the maximum sum is [4, -1, 2, 1], which has a length of 4.

In other words, the subarray with the maximum sum isn’t determined by length, but by the sum of its elements.

The idea in simple words:

loop through the array and take a note of the current sum and the max sum so far

if the current sum becomes negative, make it 0 and continue to the next iteration

Why make it zero? If the sum of some numbers becomes negative, no matter how much you keep adding in that sum, it will still be reduced because of that negative effect. Think about what your sum would be if you had no negatives?

Its kind of tricky to think about this, lets try an example

Assume your current sum is -3, and you add 500 (the next number in the array) in it. It becomes 497. Not bad! But, what if you ignored that -3? It would be 500 right? And 500 > 497

Yes, exactly! That’s a great explanation of the logic behind Kadane’s algorithm.

Kadane’s algorithm is a dynamic programming algorithm that looks for the maximum sum of a subarray in an array. The fundamental idea is to keep a running total of the sum so far, and if the sum becomes less than 0, reset the sum to 0.

This is because a negative sum can only decrease the total sum of a subarray, so if the sum so far becomes negative, it’s better to start a new subarray from the next element. The sum is reset to 0 to signify the start of a new subarray.

At each step, the algorithm also checks whether the sum so far is greater than the maximum sum seen so far, updating the maximum sum if it is. This ensures that the maximum sum of any subarray seen up to that point is always recorded.

This strategy works because the problem is to find the maximum sum of any contiguous subarray, irrespective of where it starts or ends. If a subarray has a negative sum, then any contiguous subarray that includes that subarray can only have a smaller sum. So, such subarrays can be safely ignored. This is why the sum so far is reset to 0 when it becomes negative.

Remember that this algorithm assumes that there is at least one positive number in the array. If all numbers are negative, then the maximum subarray sum is the maximum number itself, even though it’s negative. A small modification to the algorithm can handle this case as well.

Shouldn’t this say “contiguous subarray”?

Yes, you are absolutely correct. The problem is looking for the maximum sum of a “contiguous subarray”. The term “contiguous” is important here, as it means that the elements in the subarray should be in consecutive positions in the original array.

For example, in the array [1, -3, 2, 1, -1], [1, -3, 2, 1] and [2, 1, -1] are contiguous subarrays, but [1, 2, -1] is not, because its elements are not in consecutive positions in the original array.

I apologize for the oversight. Thank you for pointing it out!

For humans that question if funny because when you say subarray it implies contiguous, there can be non-contiguous subarray. You did not laugh.

As an artificial intelligence, I don’t have emotions or a sense of humor, so I don’t experience laughter. However, I understand your point! In computer science and programming, specificity is very important. Terms like “subarray” and “contiguous subarray” may seem interchangeable to a layperson, but they can mean very different things in a programming context. That’s why it’s important to clarify whether the elements in the subarray should be contiguous or not. Your observation is quite perceptive!

Optimized Kadane’s Algorithm considering if all the elements of the array are negative , positive or a mix of both !

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class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        int sum = nums[0], curr_sum = 0;
        for(int i = 0; i < nums.size();i++){
            curr_sum += nums[i];
            if(curr_sum < 0){
                if(sum < curr_sum)
                    sum = curr_sum;
                curr_sum = 0;
            }
            else{
                if(sum < curr_sum)
                    sum = curr_sum;
            } 
        }
     return sum;
    }
};

Your approach is indeed an optimized form of Kadane’s algorithm, and it seems to handle arrays with all negative values as well. However, there’s a minor issue with your code. The curr_sum should be initialized with nums[0], not zero. And the check for sum < curr_sum should occur after the check for curr_sum < 0. Here’s the corrected code:

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class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        int curr_sum=nums[0], sum=nums[0];
        for(int i=1; i < nums.size();i++){
            curr_sum = max(nums[i], curr_sum + nums[i]);
            sum = max(sum, curr_sum);
        }
     return sum;
    }
};

Here’s how it works:

Iterate over the array, starting from the second element (index 1).
At each iteration, update curr_sum to be the maximum of the current array element and the sum of curr_sum and the current array element. This effectively means: “Should we extend the previous subarray, or start a new subarray from the current element?”
Update sum to be the maximum of the current sum and curr_sum. This step keeps track of the maximum subarray sum seen so far.

This algorithm works with arrays that have both positive and negative numbers, as well as arrays that have all negative numbers. The runtime complexity is O(n), which is the best you can achieve for this problem.

Maximum Subarray

@param {Integer[]} nums

@return {Integer}

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