Longest Subarray of 1's After Deleting One Element

Identifying Problem Isomorphism

“Longest Subarray of 1’s After Deleting One Element” is approximately isomorphic to “Max Consecutive Ones II”.

In “Longest Subarray of 1’s After Deleting One Element”, the task is to find the maximum-length subarray with ones, with the option to remove one element.

“Max Consecutive Ones II” asks for the maximum number of consecutive ones you can get in an array by flipping at most one 0 to 1.

Both problems revolve around the core task of maximizing the sequence of ones, with an allowance of altering one element (either removing or changing). They both require an understanding of array manipulation and the ability to handle conditions on array elements.

“Max Consecutive Ones II” is simpler. It doesn’t involve removal of elements but simply flipping the value of an element. This makes handling the array easier as you don’t have to deal with changes in array length.

The mapping is approximate, as while the core tasks are very similar, there are differences in the constraints and requirements of the problems.

10 Prerequisite LeetCode Problems

Here are 10 problems related to arrays and subarray manipulations.

485. Max Consecutive Ones: In this problem, you need to find the maximum number of consecutive 1s in a binary array, which can help you understand how to handle subarrays of 1s.
1004. Max Consecutive Ones III: This problem extends the previous one by allowing up to K number of 0s to be changed to 1s.
209. Minimum Size Subarray Sum: This problem asks for the minimum size subarray with sum greater than a given value. It helps understand sliding window concept.
713. Subarray Product Less Than K: This problem helps in understanding the concept of a sliding window in a different context.
674. Longest Continuous Increasing Subsequence: Understanding how to track the longest increasing subsequence will help with tracking subarrays in problem 1493.
697. Degree of an Array: This problem introduces the concept of an array’s degree, and the task is to find the smallest possible length of a subarray with the same degree as that of the original array.
53. Maximum Subarray: A classic problem that helps in understanding the concept of maximum subarray, though here the subarray is not restricted to 1s only.
76. Minimum Window Substring: This problem offers practice with string manipulation and subarray problems. It involves finding the smallest window in a string which will contain all the characters of another string.
438. Find All Anagrams in a String: This problem deals with finding all the start indices of p’s anagrams in s. It can help you understand and manage multiple subarrays in a larger array.
209. Minimum Size Subarray Sum: This problem introduces the concept of finding the minimum length of a contiguous subarray of which the sum is at least a certain target, which could be useful in tracking the length of subarrays in problem 1493.

These cover subarray problems and prepare you for 1493.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
class Solution:
    def longestSubarray(self, nums: List[int]) -> int:
        k = 1
        i = 0
        for j in range(len(nums)):
            k -= nums[j] == 0
            if k < 0:
                k += nums[i] == 0
                i += 1
        return j - i

Problem Classification

This is a typical example of problems that deal with arrays and subarrays. The problem involves manipulation of binary arrays and is specifically about maximizing the length of a subarray with certain properties after performing a given operation.

The key “what” components are as follows:

We are given a binary array - nums - as input. This array contains only 0s and 1s.
We have the ability to perform a single operation: deleting one element from the array.
The main task is to return the size of the longest non-empty subarray containing only 1s after performing the deletion operation.
The subarray should exist in the array resulting from the deletion operation.
If there is no such subarray, we return 0.

This can be further classified into a few categories:

Array Manipulation: The problem involves manipulating an array (i.e., deleting an element).
Optimization Problem: The task is to optimize (maximize, in this case) the length of a certain type of subarray.
Sliding Window: This problem may be solved using a variation of the sliding window technique, which is often used in problems involving arrays and subarrays.
Greedy Algorithms: Since we’re looking for the longest subarray after one deletion, a greedy approach might be applicable where we try to keep as many 1’s as possible.

These classifications will help in identifying similar problems and may also provide hints towards potential approaches to solving the problem.

Visual Model of the Problem

Visualizing this problem involves understanding how the length of the longest subarray containing only 1s changes when we delete a single element from the binary array. Let’s take the examples provided in the problem statement to illustrate this.

Example 1: Input: nums = [1,1,0,1]
Here, you can visualize this array as a sequence of blocks where each block represents a number in the array. We can represent 1’s with filled blocks and 0’s with empty blocks:
Filled Block -> █ Empty Block -> □
The array will look like: █ █ □ █
If we remove the empty block (0), we get a new sequence as: █ █ █
So the length of the longest subarray containing only 1s is 3.
Example 2: Input: nums = [0,1,1,1,0,1,1,0,1]
The array will look like: □ █ █ █ □ █ █ □ █
If we remove the empty block (0) at position 4, we get a new sequence as: □ █ █ █ █ █ □ █
So the length of the longest subarray containing only 1s is 5.
Example 3: Input: nums = [1,1,1]
The array will look like: █ █ █
Since we must delete one element, if we remove a filled block (1) from any position, we get a new sequence as: █ █
So the length of the longest subarray containing only 1s is 2.

This visual understanding of the problem can help you devise a strategy to solve the problem. It’s clear that the goal is to choose an optimal element to delete so that the resulting array has the longest possible subarray of 1s. The optimal element to delete will usually be a 0 that is surrounded by the maximum number of 1s.

Problem Restatement

Let’s break it down:

You are given a binary array, meaning an array containing only 0s and 1s. Your task is to remove exactly one element from this array. After you remove this one element, you want to find the longest continuous streak, or subarray, of 1s that remains in the updated array. Your function should return the length of this longest streak.

If after the removal there is no subarray of 1s left, your function should return 0. If the initial array only has one element, after the removal there will be no elements left, so the function should also return 0 in this case.

The constraints specify that the length of the input array will be between 1 and 105, inclusive. And of course, each element in the array will be either 0 or 1.

In summary, the problem is asking to identify and remove a single, strategically chosen element from a binary array to create the longest possible sequence of consecutive 1s, and return the length of this sequence.

Abstract Representation of the Problem

Here is an abstract representation of the problem:

We have a sequence (S) of binary digits, where each digit is either 0 or 1. We’re allowed to perform exactly one operation: removing a single digit from S. The goal is to find the maximum length of a contiguous subsequence in the modified sequence that contains only the digit 1.

The constraints state that the length of S is within the range [1, 10^5]. If no subsequence of 1’s exists in the sequence after the operation, the function should return 0.

In abstract terms, this is an optimization problem that involves sequence manipulation and requires an understanding of subsequence properties. The task is to determine an optimal deletion strategy that maximizes the length of a specific subsequence.

Terminology

There are several technical concepts and specialized terms that are crucial to understanding and solving this problem. Here are some of them:

Binary Array: An array where each element is either 0 or 1. In this problem, the input is given as a binary array, which means that the array only contains 0s and 1s.
Subarray: A contiguous part of an array. In this problem, we are looking for the longest subarray consisting solely of 1s after deleting a single element from the original array.
Sliding Window Technique: A method for efficiently tracking a subset of elements (i.e., a ‘window’) in a larger dataset as the window “slides” through the data. It is commonly used in array or list-based problems. In this context, it could be used to track the longest subarray of 1s.
Array Manipulation: The act of changing an array by performing operations such as insertion, deletion, or modification of elements. In this problem, we are allowed to delete exactly one element from the binary array.
Optimization Problem: A type of problem where the goal is to find the best solution according to a particular criterion from a set of possible solutions. In this problem, we aim to find the best element to delete in order to maximize the length of the longest subarray of 1s.

Understanding these terms and concepts is crucial as they form the building blocks of the problem statement and potential solutions.

Problem Simplification and Explanation

Let’s break it down and use a metaphor to help understand this problem better.

Consider the given binary array as a long queue of people where each person is either wearing a black (representing 1) or a white hat (representing 0). Your task is to remove exactly one person from the queue to create the longest possible continuous line of people wearing black hats.

However, you can’t just remove any person. You can only remove one, so you want to make this action count. Ideally, you’d want to remove a person with a white hat who is standing between two groups of people with black hats. This way, these two groups will merge into one larger group.

Once you’ve removed one person from the queue, you want to identify the largest group of people with black hats. The size of this group is the answer you want to return.

This is essentially the problem: we have a binary array (queue of people), and we can remove one element (one person). The goal is to find the length of the longest subarray of 1s (the longest continuous line of people wearing black hats) after this operation.

The key concepts involved are array manipulation (removing a person), subarrays (groups of people with black hats), and optimization (finding the longest group of people with black hats).

I hope this analogy helps make the problem more understandable!

Constraints

Certainly, the problem statement and constraints provide some useful characteristics that can be leveraged to find an efficient solution:

Binary Array: Since the array only contains 0s and 1s, we don’t need to worry about dealing with a wide range of values or negative numbers. This can simplify our approach.
One Element Removal: The fact that we can only remove one element from the array is a limitation that can actually guide our strategy. We don’t need to consider multiple removals or swaps, just the most strategic single removal.
Subarray of 1s: The problem asks for the longest subarray of 1s. Therefore, the 0s are the break points for the 1s. Especially, a 0 with 1s on both sides is a potential candidate for removal, as removing it would join two subarrays of 1s together.
Length Constraints: The length of the array can be up to 10^5. This implies that a naive solution might not be efficient enough, pushing us towards solutions with time complexity linear to the size of the input.

In terms of patterns, a critical one is that the 0s divide the 1s into separate groups. Removing a 0 that is surrounded by the most number of 1s will generally yield the longest subarray of 1s. This observation can help us in formulating an efficient solution to the problem.

The key insights derived from analyzing the constraints are:

Binary Input: The array contains only 0s and 1s. This simplifies the problem because we only have two values to deal with, and we are specifically interested in maximizing the length of consecutive 1s.
Single Deletion: We can only delete one element from the array. This limits the number of manipulations we can make and constrains the problem to finding the single, optimal deletion that maximizes our desired outcome. It suggests that a strategy needs to be developed to identify the best candidate for removal.
Array Size: The size of the array is up to 105. This size constraint means that a solution with a time complexity worse than O(n) may not be acceptable for large inputs due to time limits. Therefore, we should aim for a solution with linear time complexity or better.
Non-Empty Subarray: We’re asked to find the longest non-empty subarray. This implies that we should return 0 if no such subarray exists, specifically if the array contains only 0s or after removal if no 1s remain.
Only 1s Subarray: The requirement of the longest subarray containing only 1s implicitly tells us that 0s are our main points of interest for the deletion operation, as they separate segments of 1s.

These insights can guide us towards developing a solution strategy that efficiently solves the problem while meeting all requirements and constraints.

Case Analysis

It’s always a good idea to cover a wide range of scenarios when developing test cases, especially edge and corner cases. Let’s go through a few:

Minimum Length Case: Input: nums = [1] In this case, we have the smallest possible array, consisting of a single element. We must delete one element, and doing so leaves us with an empty array, so the longest subarray of 1s has length 0. Expected output: 0
All Zeroes Case: Input: nums = [0,0,0,0,0] Here, all elements are 0. Deleting any element will still leave us with an array that has no 1s, so the length of the longest subarray of 1s is 0. Expected output: 0
All Ones Case: Input: nums = [1,1,1,1,1] In this case, all elements are 1. Regardless of which element we delete, we are left with a subarray of 1s of length 4. Expected output: 4
1’s separated by single 0’s: Input: nums = [1,1,0,1,1,0,1,1] Here, 1’s are separated by single 0’s. Deleting any 0 will connect two subarrays of 1s, so the longest possible subarray has length 4. Expected output: 4
1’s separated by multiple 0’s: Input: nums = [1,1,0,0,1,1,0,0,1,1] Here, the 1’s are separated by multiple 0’s. Deleting any 0 will not connect two subarrays of 1s, so the longest possible subarray remains of length 2. Expected output: 2
1’s at the ends, 0’s in the middle: Input: nums = [1,1,0,0,0,1,1] Here, 1’s are at the ends and 0’s in the middle. Deleting any 0 will not connect the two subarrays of 1s, so the longest possible subarray remains of length 2. Expected output: 2

These test cases cover a variety of scenarios that help ensure the solution handles all possible edge cases, takes into account all constraints, and is robust for a wide range of inputs.

Analyzing different cases gives us several key insights into the problem and how to approach it:

Position of 0’s Matter: The position of 0’s is crucial in deciding which element to delete. A 0 surrounded by 1’s on both sides is a potential candidate for removal since this will connect two subarrays of 1s together.
All Ones or All Zeroes: In cases where the array contains all 1’s or all 0’s, the decision is straightforward. For all 1’s, regardless of which element we delete, the output will be one less than the original length of the array. For all 0’s, since there are no 1’s in the array, the output will always be 0.
1’s Separated by Multiple 0’s: If 1’s are separated by multiple 0’s, deleting a 0 won’t result in a longer subarray of 1s, as the 1’s aren’t close enough to form a continuous sequence. This means that in some cases, the deletion operation doesn’t increase the length of the longest subarray of 1’s.
Smallest and Largest Inputs: For the smallest possible input (a single-element array), the output is always 0 because we have to delete the only available element. For the largest possible input, an efficient algorithm is necessary to ensure that the solution is computed within the time constraints.

These insights help inform the strategy for developing an algorithm to solve the problem. The solution needs to strategically decide which element to delete in order to maximize the length of the longest subarray of 1’s, while also handling edge cases appropriately.

Identification of Applicable Theoretical Concepts

There are several mathematical and algorithmic concepts that can be applied to make this problem more manageable:

Sliding Window: This is an algorithmic technique commonly used for array or list-based problems, where a ‘window’ of elements is maintained and the window ‘slides’ through the data. In this problem, a sliding window can be used to keep track of the number of consecutive 1s, allowing us to identify the optimal 0 to delete.
Prefix/Suffix Sum: The concept of prefix or suffix sum can be applied in this problem. By calculating the number of 1s before and after each 0, we can easily identify the optimal 0 to delete.
Two Pointers: Another technique that could be used to solve this problem is the two-pointer technique, which involves having two pointers that can move at different speeds or directions through an array to solve a problem. Here, two pointers can be used to maintain a window that represents a valid subarray.
Greedy Algorithms: At its core, this problem can be viewed as a greedy problem where the goal is to delete the 0 that will result in the longest sequence of 1s. A greedy algorithm makes the locally optimal choice at each stage with the hope that these local choices lead to a global optimum.

These algorithmic strategies and concepts could help to design an efficient solution for the problem by minimizing unnecessary computations and taking advantage of the problem’s constraints and properties.

Problem Breakdown and Solution Methodology

Let’s approach this problem using the concept of a sliding window with the aid of prefix and suffix arrays. The idea is to find the longest sequence of 1s that can be formed by removing a single 0. We can use prefix and suffix arrays to quickly get the count of consecutive 1s before and after each 0.

Let’s break it down:

Create Prefix and Suffix Arrays: First, we’ll create two arrays - prefix and suffix - of the same length as the input array. The prefix array will store the count of consecutive 1s from the start of the array until the current position. The suffix array will store the count of consecutive 1s from the end of the array until the current position.
Populate the Prefix Array: Iterate over the input array from left to right. If the current element is 1, increment the count of consecutive 1s. If the current element is 0, reset the count to 0. Store the count in the prefix array at the current position.
Populate the Suffix Array: Similarly, iterate over the input array from right to left to populate the suffix array.
Find the Optimal 0 to Delete: Next, iterate over the input array and for each 0, calculate the sum of the corresponding values in the prefix and suffix arrays. This sum represents the length of the subarray of 1s we can get by deleting the current 0. Keep track of the maximum sum obtained.
Handle the ‘All Ones’ Case: If the input array contains all 1s, the maximum sum will be 0 because there are no 0s to delete. In this case, the longest subarray of 1s we can get by deleting a single 1 is one less than the length of the input array.

This approach works because we use the prefix and suffix arrays to quickly find the length of the subarray of 1s we can get by deleting each 0. We then find the maximum of these lengths to get our answer.

Changes in the problem’s parameters would affect the solution as follows:

If the input array becomes longer, it will require more space to store the prefix and suffix arrays, and it will take more time to populate these arrays and find the optimal 0 to delete.
If we were allowed to delete more than one 0, the problem would become significantly more complicated. We could no longer simply find the optimal single 0 to delete; we would have to consider combinations of 0s.

Let’s walk through an example:

Input: nums = [1,1,0,1,1,0,1,1] Here’s how the prefix and suffix arrays look:

nums:    1 1 0 1 1 0 1 1
prefix:  1 2 0 1 2 0 1 2
suffix:  2 1 3 2 1 2 1 0

The sum of prefix[i-1] and suffix[i+1] for each 0 gives: 2 (from prefix[1]) + 2 (from suffix[3]) = 4 and 2 (from prefix[4]) + 1 (from suffix[6]) = 3. The maximum sum is 4, so the output is 4. This corresponds to deleting the 0 at index 2 to get the subarray [1,1,1,1].

Inference of Problem-Solving Approach from the Problem Statement

The problem asks us to delete one element (which will be a ‘0’) to make the longest subarray of ‘1’s. This requires us to find a ‘0’ which has the maximum number of ‘1’s surrounding it. The count of consecutive ‘1’s before a ‘0’ and after a ‘0’ becomes a vital piece of information in making this decision.

This is where the idea of prefix and suffix arrays comes in. A prefix array can provide the count of consecutive ‘1’s before any given index, while a suffix array can provide the count of consecutive ‘1’s after any given index. These arrays can be prepared in a single pass each over the input array, giving us the required information at every index in O(N) time.

The use of prefix and suffix arrays might not be immediately obvious from the problem statement. However, upon analyzing the problem and recognizing the need to efficiently find the count of consecutive ‘1’s around each ‘0’, we can identify prefix and suffix arrays as a suitable approach.

This problem-solving pattern, where we need to efficiently compute or track properties of elements relative to their surroundings in a sequence, is fairly common in competitive programming. Therefore, with experience, one can often recognize when prefix/suffix arrays, cumulative sum arrays, or similar data structures might be useful.

Sure, I will detail the step-by-step approach to solve this problem using prefix and suffix arrays:

Step 1 - Initialization

Initialize two arrays, prefix and suffix, with the same size as the input array. Also, initialize two variables, preCount and sufCount, to keep track of the count of consecutive 1s while populating prefix and suffix arrays.

Step 2 - Populating the Prefix Array

Iterate over the input array from left to right. If the current element is 1, increment preCount by 1. If the current element is 0, reset preCount to 0. Store preCount in the prefix array at the current position.

Step 3 - Populating the Suffix Array

Now, iterate over the input array from right to left. If the current element is 1, increment sufCount by 1. If the current element is 0, reset sufCount to 0. Store sufCount in the suffix array at the current position.

Step 4 - Finding the Optimal 0 to Delete

Next, iterate over the input array. For each 0 in the input array, calculate the sum of the corresponding value at the previous position in the prefix array and the next position in the suffix array. This sum represents the length of the subarray of 1s we can get by deleting the current 0. Keep track of the maximum sum obtained.

Step 5 - Handling the ‘All Ones’ Case

If the input array contains all 1s, the maximum sum will be 0 because there are no 0s to delete. In this case, the longest subarray of 1s we can get by deleting a single 1 is one less than the length of the input array. So, if the maximum sum is 0, return nums.length - 1.

Step 6 - Returning the Result

Return the maximum sum obtained. This represents the length of the longest subarray of 1s we can get after deleting one element.

This is the final refined step-by-step approach to solve the problem. This approach makes the problem more manageable by dividing it into smaller, more easily tackled steps.

Let’s break down our approach into more granular, actionable steps that can be directly translated into code:

Step 1: Initialization

Declare an array prefix of size nums.length, initialized to 0.
Declare an array suffix of size nums.length, initialized to 0.
Declare variables preCount and sufCount and set both to 0. These variables will be used to keep count of consecutive 1s while we populate our prefix and suffix arrays.

Step 2: Populating the Prefix Array

Iterate over the nums array from left to right (i=0 to nums.length - 1).
- If nums[i] equals 1, increment preCount.
- If nums[i] equals 0, reset preCount to 0.
- Assign preCount to prefix[i].

Step 3: Populating the Suffix Array

Iterate over the nums array from right to left (i=nums.length - 1 to 0).
- If nums[i] equals 1, increment sufCount.
- If nums[i] equals 0, reset sufCount to 0.
- Assign sufCount to suffix[i].

Step 4: Finding the Optimal 0 to Delete

Declare a variable maxSum and set it to 0.
Iterate over the nums array (i=1 to nums.length - 2).
- If nums[i] equals 0, calculate the sum of prefix[i-1] and suffix[i+1] and check if it is greater than maxSum. If it is, update maxSum.

Step 5: Handling the ‘All Ones’ Case

After the loop, if maxSum is still 0, it means that nums consists of all 1s. In that case, assign nums.length - 1 to maxSum.

Step 6: Returning the Result

Return maxSum as the result. This represents the size of the longest subarray of 1’s we can get after deleting one element from nums.

These steps are the same as our high-level approach, but they provide more details and can be directly mapped to code statements, making them ready for implementation.

Certainly, this problem can be broken down into several parts that can be solved independently, and then the results of those parts can be combined for the final solution.

1. Building Prefix Array: This can be done independently by iterating over the array from left to right and counting the number of consecutive 1s.

2. Building Suffix Array: Similarly, we can independently build the suffix array by iterating from right to left and counting the number of consecutive 1s.

3. Identifying the Best ‘0’ to Remove: Once we have the prefix and suffix arrays, we can independently process them together with the original array to identify which ‘0’ will yield the longest sequence of 1s when removed.

4. Handling Edge Case of All 1s: Finally, an independent check can be made to handle the edge case where the input array consists only of 1s. In this case, we must subtract 1 from the length of the array, as we are required to remove one element.

Each of these parts deals with a specific aspect of the problem and can be implemented and tested independently. After each part is working correctly, they can be combined to form the complete solution.

There are repeatable patterns within our solution to this problem:

1. Array Iteration: There’s a pattern of traversing an array in both directions, forward (left to right) and backward (right to left). This pattern is used when populating both the prefix and suffix arrays.

2. Counting Consecutive 1s: While populating the prefix and suffix arrays, we’re repeatedly checking the current element and updating a count based on its value (incrementing the count if it’s a 1, resetting it to 0 if it’s not). This is a pattern of counting consecutive 1s in a binary array.

3. Updating a Running Maximum: When we’re looking for the optimal 0 to remove, we maintain a running maximum length of the subarray of 1s we can get by removing a specific 0. This is a pattern of keeping track of a maximum value while iterating over an array.

Recognizing these patterns can make the solution easier to understand and implement, and can also help when solving similar problems in the future.

Solution Approach and Analysis

Our approach with metaphors and detailed examples.

Populate Prefix and Suffix Arrays: Think of the prefix array like a trail of breadcrumbs that counts how many steps we’ve taken in a forward direction. Each breadcrumb (or value in the prefix array) represents the number of continuous 1s we’ve encountered so far from the start of the array up to that point. Similarly, the suffix array is like a trail of breadcrumbs from the end back to the start, counting continuous 1s backwards. For example, for the input array [1,1,0,1,1,0,1], the prefix array will be [1,2,0,1,2,0,1] and the suffix array will be [2,1,2,1,0,1,0].
Find Optimal ‘0’ to Remove: Now imagine we are hikers who want to walk the longest path of 1s, but we are allowed to build a bridge over one ‘0’ chasm to connect two paths of 1s. We look for the ‘0’ that is surrounded by the most 1s. Using the prefix and suffix arrays, we can check how long the path of 1s would be if we built a bridge at each ‘0’. This is the equivalent of adding the previous value in the prefix array and the next value in the suffix array for every ‘0’ in the original array, and keeping track of the maximum sum.
Handle All 1s Case: If the whole terrain is just one continuous path of 1s with no ‘0’ chasms (i.e., the input array consists entirely of 1s), we cannot build a bridge. In this case, we have to remove one ‘1’, shortening our path by one step.
Return Result: We then return the length of the longest path we can walk, which is represented by the maximum sum obtained.

Let’s go over an example to illustrate the workings of the approach:

Take the input array nums = [0,1,1,1,0,1,1,0,1].

Step 1: Generate the prefix and suffix arrays:
- Prefix: [0,1,2,3,0,1,2,0,1]
- Suffix: [3,2,1,3,2,1,0,1,0]
Step 2: Iterate over nums, sum up the values from the prefix and suffix arrays at the corresponding positions for each ‘0’ (while avoiding out-of-bounds access), and find the maximum sum:
- For nums[0], there’s no prefix value, so we only add suffix[1] = 2.
- For nums[4], we add prefix[3] = 3 and suffix[5] = 1 to get 4.
- For nums[7], we add prefix[6] = 2 and suffix[8] = 0 to get 2.
- The maximum sum is 4.
Step 3: There’s no need to handle the all-1s case, as nums contains zeros.
Step 4: Return the maximum sum, which is 4.

This approach will work for different arrays and will handle variations in the problem’s parameters, as it is based on the structure of the problem rather than specific values.

Thought Process

The approach discussed so far fails to pass all test cases. Reversed engineered working C++ code:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
class Solution:
    def longestSubarray(self, nums: List[int]) -> int:
        prefix = [0] * len(nums)
        suffix = [0] * len(nums)
        ct = 0
        for i in range(len(nums)):
            prefix[i] = ct
            if nums[i] == 1:
                ct += 1
            else:
                ct = 0
        ct = 0
        for i in range(len(nums) - 1, -1, -1):
            suffix[i] = ct
            if nums[i] == 1:
                ct += 1
            else:
                ct = 0
        ans = 0
        for i in range(len(nums)):
            ans = max(ans, prefix[i] + suffix[i])
        return ans

Sliding Window Approach

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
class Solution:
    def longestSubarray(self, nums: List[int]) -> int:
        k = 1
        i = 0
        for j in range(len(nums)):
            k -= nums[j] == 0
            if k < 0:
                k += nums[i] == 0
                i += 1
        return j - i

This code solution uses a sliding window approach, which is a common technique in problems that deal with subarrays or substrings. This approach can maintain a window of elements that fulfill a certain condition. When the condition is violated, it changes the window until the condition is fulfilled again.

Here’s a breakdown of the code:

Initialize k as 1. k represents the number of zeros we can still delete.
Initialize two pointers i and j. i points to the start of the current subarray (the left end of the sliding window), and j traverses the array (the right end of the sliding window).
The for loop traverses the array. If the current element is a zero, k is decremented. This reflects that we have used our chance to delete a zero.
Inside the loop, if k becomes negative, it means we’ve encountered more than one zero in our subarray, which violates our condition. To rectify this, we check if the element at i is zero. If it is, we effectively “add back” the zero to k by incrementing it, meaning we are ignoring the zero at position i. Then, we increment i to move the start of our subarray to the right. This process continues until k is no longer negative, i.e., until we have at most one zero in our subarray.
The loop continues, adjusting i and j as necessary to maintain at most one zero in our subarray.
Finally, we return j - i, which is the length of the longest subarray of 1’s after deleting one element. The subtraction is used here because j and i are indices, so subtracting them gives the length of the subarray.

In terms of concepts, this solution uses the sliding window technique for subarray problems, combined with a counter (k) to track the condition that needs to be fulfilled. It’s a more optimized solution compared to the prefix-suffix approach, as it only requires a single pass through the array, making it more efficient.

From Brute Force to Optimal Solution

Brute Force Approach:

Generate all possible subarrays.
For each subarray, if it contains a 0, then delete it and check if the remaining subarray contains only 1’s.
Keep track of the maximum length subarray that satisfies the condition.
Return this maximum length.

The main problem with the brute force approach is that it’s extremely inefficient. For an array of size n, there are n*(n+1)/2 subarrays, so we would need to perform that many checks. In each check, we may need to scan up to n elements. Therefore, the time complexity would be O(n^3), which is not feasible for large n.

Optimized Approach:

To optimize the solution, we can use a technique called “Sliding Window”. This approach maintains a “window” of elements that satisfy our conditions, and “slides” it along the array.

Initialize a left pointer i and a right pointer j to 0. These pointers represent the window’s edges.
Initialize k as 1, representing the number of zeros we can delete.
Traverse the array. If we encounter a zero, decrement k.
If k is negative, increment i and adjust k accordingly until k is no longer negative.
The maximum length of subarray we can have is j - i.

This solution is significantly more efficient. It requires a single pass through the array, so the time complexity is O(n). Additionally, it only requires a constant amount of extra space (for i, j, and k), so the space complexity is O(1).

This optimization is a significant improvement over the brute force approach. It takes advantage of the problem’s specific properties (we can delete at most one zero) to maintain a running window of valid elements, rather than having to generate and check all possible subarrays.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
class Solution:
    def longestSubarray(self, nums: List[int]) -> int:
        k = 1  # this is our deletion budget, initially we can delete 1 element
        i = 0  # 'i' is the starting index of our sliding window
        
        # 'j' is the ending index of our sliding window; 
        # we run a loop where 'j' goes from the first element to the last one in the array
        for j in range(len(nums)):
            
            # if the current number is 0, we subtract it from our budget 'k'
            k -= nums[j] == 0
            
            # if 'k' falls below 0, it means we've encountered more zeros than we can delete;
            # hence, we need to adjust our window by moving the start index 'i' forward
            if k < 0:
                # if the number at index 'i' was 0, we add it back to our budget 'k',
                # as we are no longer considering this number in our current window
                k += nums[i] == 0
                
                # increment the start index 'i', thereby shrinking our window
                i += 1
        
        # return the length of the longest subarray with only 1's after deleting one element
        # this is obtained by subtracting the final starting index 'i' from the final ending index 'j'
        return j - i

In the given solution, the right boundary of the window is defined by the variable j. As we iterate over the array with the for loop for j in range(len(nums)), j moves from the start to the end of the array. This acts as the right boundary of our sliding window. Whenever j moves, we are effectively expanding the window to the right.

The condition if k < 0 helps in controlling this right boundary. If we encounter more zeroes than allowed (i.e., k becomes less than zero), we need to move our left boundary to the right to maintain the limit of one zero in our window. So, while j defines the right boundary, the balance of k controls how far this boundary can extend before needing to adjust the window.

In the provided solution, the window is shrunk when the condition if k < 0 is met. This occurs when the count of zeros (k) in the window exceeds the allowable limit, which is 1 in this case.

The lines of code that execute under this condition are:

1
2
k += nums[i] == 0
i += 1

Here, k += nums[i] == 0 effectively increases k if the left-most element of the window (nums[i]) is a 0, as we are excluding this element from the window.

Then, i += 1 moves the left boundary of the window one step to the right, thus shrinking the window.

In summary, whenever the number of zeros in the window exceeds 1, the left boundary of the window is moved to the right, which shrinks the window.

The window in the solution is expanded by the for loop for j in range(len(nums)):. Each iteration of the loop corresponds to moving the right boundary of the window one step to the right, thus expanding the window.

In particular, the variable j represents the right boundary of the window. As j increases with each iteration of the loop, the window effectively expands to include more elements from the array nums. This expansion continues until we encounter more than one zero in the window, at which point the window will need to be shrunk.

Here is the revised code with comments, highlighting where the invariant is maintained:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
class Solution:
    def longestSubarray(self, nums: List[int]) -> int:
        # 'k' is a counter representing how many zeros we can still remove from our subarray. 
        # We're allowed to remove one zero from the subarray according to the problem statement, hence we initialize it to 1.
        k = 1 

        # 'i' is the starting index of our sliding window/subarray. We start from the beginning of the array.
        i = 0 

        # We begin iterating over the array. 'j' is the ending index of our sliding window.
        for j in range(len(nums)):
            # We encounter a zero at the current position, 'j'. This reduces our capacity 'k' to remove zeros.
            # We are virtually 'removing' a zero from our subarray by decreasing 'k'.
            k -= nums[j] == 0 

            # If 'k' becomes negative, it means we have more than one zero in our current subarray which is not allowed.
            # So, we need to adjust our subarray to make it valid again. This is where we maintain the invariant.
            if k < 0:
                # We increase 'k' if the element being removed from the subarray is zero. 
                # Essentially, we're 'adding back' the capacity to remove a zero because we're excluding it from the subarray.
                k += nums[i] == 0 

                # We move the starting index of our subarray one step to the right to exclude the current starting element.
                # This step is shrinking our subarray (window) to exclude excess zeros and make it valid again.
                i += 1 

        # After traversing the entire array, we return the length of the longest valid subarray we found.
        # This is calculated as the difference between our last ending index 'j' and the last starting index 'i'.
        return j - i 

The invariant here is that at any point in our sliding window, we will have at most one zero. We maintain this invariant by adjusting the start index i and updating k whenever we have more than one zero in our subarray. The implementation of this invariant using a sliding window approach allows us to keep track of the longest subarray satisfying this condition throughout the iteration.

This code leverages the concept of a sliding window to find the longest subarray of 1’s after deleting one element (zero) from the binary array. The window is expanded by moving its right boundary and is shrunk when it contains more than one zero, by moving its left boundary. The length of the longest such window encountered gives the solution.

Coding Constructs

High-level problem-solving strategies: The code uses the “Sliding Window” strategy to solve this problem. This strategy is particularly useful in problems dealing with subarrays or substrings. It creates a ‘window’ over the array and moves this window such that it always satisfies the problem’s constraints.
Explaining to a non-programmer: Imagine you are reading a long line of text, looking for the longest sentence that contains at most one comma. You could do this by keeping track of the longest sentence you’ve read so far that meets this criteria. As you move along the text, if you encounter another comma, you’d start the sentence after the previous comma. This is the same idea the code is using, but instead of sentences and commas, it’s looking for the longest subarray with at most one zero in a binary array.
Logical elements or constructs: The primary logical constructs in this code are conditional statements (if statements) and a loop. The loop traverses through the array, and the if statements check the conditions on the current window (whether we’ve used up our chance to delete a zero).
Algorithmic approach in plain English: The code scans the binary array from left to right. It maintains a record of the longest sequence of 1’s it has seen that contains at most one zero. If it sees more than one zero, it effectively skips over the first zero it encountered, and continues scanning.
Key steps or operations: The key operations in the code are updating the window’s boundaries and maintaining the count of zeros in the window. When it sees a zero, it decrements the counter. If the counter goes below zero, it increments the counter and the start of the window, effectively ignoring the zero at the window’s start.
Algorithmic patterns or strategies: The key algorithmic pattern used by this code is the Sliding Window pattern. It also uses a counter to keep track of the number of zeros in the current window. The conditional logic then adjusts the window based on the value of this counter. This strategy helps us maintain a window that always satisfies the problem’s constraints.

Language Agnostic Coding Drills

Dissecting the Code into Distinct Concepts:
a. Looping Over an Array: The ability to iterate over an array from start to finish is an essential programming concept. It allows the algorithm to inspect or manipulate each element in the array.
b. Conditional Statements (if): The ability to perform different actions depending on whether a particular condition is true or false. In this case, it allows the algorithm to decide how to adjust the window based on the current number in the array (1 or 0).
c. Pointer Manipulation: The use of pointers (or indices) to represent a range or window in the array. Here, it is used to keep track of the current subarray the algorithm is considering.
d. Updating Variables: The ability to modify the values of variables based on new information. In this case, it is used to track the number of zeros in the current window (via the variable ‘k’) and to move the window (via the variables ‘i’ and ‘j’).
List of Coding Concepts in Order of Increasing Difficulty:
a. Looping Over an Array: This is a fundamental concept and typically one of the first things a new programmer learns.
b. Conditional Statements (if): Also a basic concept, but requires understanding of boolean expressions and control flow.
c. Updating Variables: This concept requires understanding of variables and assignments, but it’s slightly more advanced than basic looping and conditionals because the updates can be based on more complex logic.
d. Pointer Manipulation: This is a more advanced concept that requires understanding of how arrays are structured and how to use indices to access and represent portions of an array.
Problem-solving Approach:
a. Start by understanding the problem requirements and constraints. We need to find the longest subarray containing only 1’s after deleting at most one element.
b. Recognize that we can use a sliding window to track the current subarray we’re considering. Initiate two pointers, ‘i’ and ‘j’, to represent the start and end of the window, and a counter ‘k’ to represent the number of zeros we can still delete.
c. Loop over the array. For each element, adjust ‘k’ and the window as necessary. If the element is a 0, decrement ‘k’. If ‘k’ becomes negative, increment ‘i’ and adjust ‘k’ until ‘k’ is no longer negative.
d. After each adjustment, note the length of the current window. The longest window we see during this process will be our answer.

This process uses each of the coding drills to gradually build up to the final solution. We start with the simpler concepts of looping and conditionals to iterate over the array and make decisions based on the current element. We then use variable updates to adjust our window and count of zeros based on these decisions. Finally, we use pointer manipulation to maintain and adjust our window as we loop over the array. Each of these drills plays a crucial role in the overall solution.

Targeted Drills in Python

Sure, let’s create Python-based coding drills for the concepts we have identified.

Looping over an array:
1 2 3
nums = [1, 0, 1, 1, 0] for num in nums: print(num)
This basic code iterates over each element in the array and prints it out.

Conditional statements (if):

1
2
3
4
5
6
nums = [1, 0, 1, 1, 0]
for num in nums:
    if num == 1:
        print("One")
    else:
        print("Zero")

This code checks if each element in the array is a 1 or a 0, and prints out “One” or “Zero” respectively.

Updating variables:

1
2
3
4
5
6
nums = [1, 0, 1, 1, 0]
count_of_ones = 0
for num in nums:
    if num == 1:
        count_of_ones += 1
print(count_of_ones)

This code counts the number of ones in the array and prints out the final count.

Pointer manipulation:
1 2 3 4 5
nums = [1, 0, 1, 1, 0] i = 0 j = 2 subarray = nums[i:j+1] print(subarray)
This code uses two pointers (i and j) to get a subarray from the array, and prints out the subarray.

Now, for problem-specific concepts:

Sliding window with deletions:
This is a specialized form of pointer manipulation and variable updating. We need to keep track of a sliding window and the number of deletions allowed (which decrease each time we encounter a 0). Here’s how to implement it:
1 2 3 4 5 6 7 8 9
nums = [1, 0, 1, 1, 0] i = 0 k = 1 for j in range(len(nums)): k -= nums[j] == 0 if k < 0: k += nums[i] == 0 i += 1 print(i, j)
This code implements a sliding window that slides through the array. When it encounters a 0, it decreases k (the number of allowed deletions). If k becomes negative, it shifts the start of the window forward and adjusts k until it’s no longer negative.

After mastering each of these drills, you can combine them in order to solve the original problem. You’d start by creating the sliding window, and then, while iterating over the array, you’d use your conditional checks and variable updating to adjust the window as necessary. Finally, after each adjustment, you’d use your understanding of arrays and subarrays to calculate and track the length of the current window, which is your ultimate solution.

Q&A

Sliding window consists of two pointers and traversing the array. Did I miss anything

Yes, you’re correct that a sliding window algorithm typically involves two pointers that traverse the array. However, there are a few more details that are commonly associated with sliding window algorithms:

Window Size: The window size can either be fixed or variable. In fixed-size sliding window problems, the window size is constant, and the window slides one position at a time. In variable-sized sliding window problems, the size of the window can expand or contract based on certain conditions.
Conditions for Moving Pointers: The conditions to move the start and end pointers of the window vary from problem to problem. Usually, the end pointer moves forward unconditionally (i.e., traverses the array), while the start pointer moves forward based on some specific condition (e.g., the window size exceeding a certain limit, or the elements within the window violating some constraint).
Maintaining Window State: As the window slides, the algorithm maintains some kind of state or information about the elements within the window. This might be the sum of the elements, the count of a specific element, the maximum or minimum element, etc.
Result Calculation: The algorithm typically needs to calculate some kind of result based on the state of the sliding windows, such as the maximum sum, the longest contiguous sequence satisfying some condition, etc.

Remember that these are just general characteristics of sliding window algorithms, and the specifics can vary based on the problem at hand.

Just Maintain a sliding window where there is at most one zero on it.

Yes, you are correct. In the context of the problem at hand, the invariant for the sliding window is that it should contain at most one zero. In other words, the window can contain as many ones as possible, but at most one zero.

This invariant is maintained as follows:

When a new element is included in the window (i.e., when the window expands), if the element is a zero and there is already a zero in the window, then the window shrinks from the left until the first zero is excluded.
If the new element is a one, it is simply included in the window, and the window expands without needing to shrink.

Remember, an invariant is a condition that remains true throughout the execution of an algorithm. In the case of this sliding window algorithm, ensuring that the window always satisfies this invariant condition is essential for finding the solution to the problem.

It’s a good one to practice sliding window.

Why would you delete 1’s?

In the context of this problem, we don’t want to deliberately delete 1’s. The action of deleting or removing 1’s from our window in this problem only occurs as a byproduct of maintaining our invariant: that the window contains at most one zero.

When we encounter a second zero in our array traversal, we need to shrink our window from the left until the first zero is no longer in the window. During this process, if there are 1’s to the left of this first zero, they will also be removed from the window.

It’s important to note that “removing” or “deleting” from the window doesn’t mean that these elements are erased or forgotten. They’re simply not in the current window under consideration. As our window slides through the array, any 1’s removed in an earlier step could be included again if they meet the condition of the problem statement.

They are not removed from the given array? We are not modifiying it?

No, we are not modifying the given array. The terms “remove” and “delete” are used in the context of our sliding window, not the original array.

The array itself remains unchanged. The sliding window is just a conceptual tool used in our algorithm to examine a subset of the array. As we move the window through the array (by adjusting our pointers), some elements will fall outside the window, but they are not removed from the array.

keep track of the count of 1’s in the current and the previous substring of 1’s to solve without sliding window

While it’s true that we need to keep track of the count of 1s in the current and previous substring of 1s, it’s important to realize that this is essentially the same as managing a sliding window. Here’s how you might go about it:

Initialize two counters, cur and prev, to 0. These represent the counts of 1s in the current and previous substrings of 1s, respectively. Also initialize maxLen to 0, which will hold our result.
Iterate over the array. For each element:
- If it’s a 1, increment cur. Update maxLen with the maximum of its current value and cur + prev.
- If it’s a 0, set prev to cur (since the current substring of 1s now becomes the previous substring), and reset cur to 0.
Return maxLen.

This strategy gives the same result as the sliding window method, but it approaches the problem from a slightly different angle. We’re still effectively maintaining a window on the array, but we’re doing so by updating our counts rather than by explicitly defining window boundaries. It’s another way to visualize and tackle the problem.

Remember that the main goal is to find the longest contiguous subarray that would become all 1s if we change exactly one 0 to 1. In both methods, we’re essentially looking for the longest sequence of 1s that are separated by a single 0.

It’s basically an application of Kadane’s Algorithm. We could easily solve the problem, if no elements are allowed to delete. So just maintain two counters curr_cnt and prev_cnt. curr_cnt stores the maximum length of a subarray of contiguous 1’s till the current index, while prev_cnt stores the previous value of curr_cnt.

Always update your result as res = max(res,prev_cnt+curr_cnt). Just have to look for subarrays containing 0’s but are of length 1, the rest are useless.

One last edge comes in when all the elements are 1 in the array. To handle it just check if res==len(nums). If yes return res-1, otherwise just return res.

Yes, indeed! You’ve distilled the problem down to its core. It’s essentially a variant of Kadane’s Algorithm where we’re looking to maximize the length of a contiguous subarray of 1s with the flexibility to flip exactly one 0 to 1. The two counters, curr_cnt and prev_cnt, are used to track the lengths of the current and previous subarrays of contiguous 1s.

However, we should remember that in the special case where all the elements in the input array are 1s, we can’t increase the length of the subarray by flipping a 0 to 1, as there are no 0s. This is why we must decrease the result by one in this edge case.

Here’s how this might look in Python:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
class Solution:
    def longestSubarray(self, nums: List[int]) -> int:
        prev_cnt = curr_cnt = res = 0

        for num in nums:
            if num == 1:
                curr_cnt += 1
            else:
                prev_cnt, curr_cnt = curr_cnt, 0

            res = max(res, prev_cnt + curr_cnt)

        # Handle the all-1s case
        if res == len(nums):
            return res - 1

        return res

This code follows the approach we just discussed. It keeps track of the current and previous counts of contiguous 1s, updates the maximum subarray length accordingly, and handles the special case where all the elements are 1s.

What’s wrong with this code:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
class Solution:
    def longestSubarray(self, nums: List[int]) -> int:
        i = 0
        j = 0
        a = []
        zeroCnt = 0
        longWindow = 0
        while j < len(nums):
            if nums[j] == 0:
                zeroCnt+=1
            a.append(nums[j])

            if zeroCnt == 1:
                longWindow = max(longWindow,(j-i))
                j+=1
            elif zeroCnt>1:
                while zeroCnt>1:
                    a.pop(0)
                    zeroCnt-=1
                    i+=1
                j+=1
            else:
                j+=1
        return longWindow 

The issue with this code is in how it treats the presence of zeros in the array. Specifically, it starts by assuming that there are no zeros, then when it encounters a zero, it increments the zeroCnt counter. However, it only checks for windows of maximum size when zeroCnt == 1.

The problem arises when the array has consecutive zeros. If zeroCnt > 1, the elif condition starts to remove elements from the beginning of the window to adjust the zeroCnt to 1. This leads to the array a and zeroCnt being modified, but the length of the window (j - i) is not updated immediately, resulting in a wrong value for longWindow.

In other words, the code does not properly handle scenarios where there are multiple zeros together. For example, in an input like [1, 1, 0, 0, 1], the longest subarray should be 3 ([1, 1, 0]) but the code incorrectly outputs 2 ([1, 1]).

The solution should handle cases with more than one zero by considering windows with a single zero and updating longWindow accordingly before moving the window past the extra zeros. Here’s a fixed version:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
class Solution:
    def longestSubarray(self, nums: List[int]) -> int:
        i = 0
        j = 0
        zeroCnt = 0
        longWindow = 0
        while j < len(nums):
            if nums[j] == 0:
                zeroCnt+=1
            if zeroCnt > 1:
                while zeroCnt > 1:
                    if nums[i] == 0:
                        zeroCnt-=1
                    i+=1
            longWindow = max(longWindow, j-i)
            j+=1
        return longWindow

There is an incorrect testcase, the output is supposed to be 13, and it is marked incorrectly as 14

[1,0,0,1,1,1,1,1,0,0,0,0,0,0,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,1,1,1,0,1,1,0,0,0,1,1,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,1,1,1,1,0,1,1]

I believe there may be a misunderstanding with this problem. The prompt states that we can remove at most one zero from the array to maximize the length of the contiguous subarray consisting of only 1s.

Looking at your example, the longest subarray of 1s, with at most one 0 removed, would be of length 14:

[1,1,1,1,1,1,1,1,1,1,1,1,1,1]

This is from index 8 to index 21 (0-indexed) in the provided array, with one 0 removed at index 14. So the expected output should indeed be 14.

Please review the problem statement and the expected output. If the problem indeed allows removing more than one 0, the solution will need to be adjusted accordingly. However, as far as I understand, only one 0 can be removed.

In this problem, we have to return the size of the longest non-empty subarray containing only 1’s in the resulting array and return 0 if there is no such subarray.

The idea behind the solution is to maintain a sliding window that tracks the longest subarray with at most one zero. We keep track of the number of zeroes encountered in the current window and adjust the window boundaries accordingly.