Longest Subarray of 1's After Deleting One Element

Problem Statement

In this problem, we are given an array nums consisting entirely of binary digits (0s and 1s). Our task is to remove exactly one element from the array (regardless of its value) and find the length of the longest contiguous subarray made entirely of 1s in the resulting array. If no such subarray exists, the function should return 0. This problem must efficiently handle arrays with up to 100,000 elements, making it crucial to consider efficient algorithms instead of brute-force solutions.

Examples

Example 1

Input:

nums = [1,1,0,1]

Output:

3

Explanation:

After deleting the number in position 2, [1,1,1] contains 3 numbers with value of 1's.

Example 2

Input:

nums = [0,1,1,1,0,1,1,0,1]

Output:

5

Explanation:

After deleting the number in position 4, [0,1,1,1,1,1,0,1] longest subarray with value of 1's is [1,1,1,1,1].

Example 3

Input:

nums = [1,1,1]

Output:

2

Explanation:

You must delete one element.

Constraints

• 1 <= nums.length <= 105
• nums[i] is either 0 or 1.

Approach and Intuition

The challenge is to identify the longest sequence of 1s that can be achieved by removing a single element from the binary sequence. To approach this problem, consider the following steps:

1. Understanding the significance of removing a zero versus a one:

• Removing a zero potentially connects two separate sequences of ones, potentially increasing the length of a contiguous subarray of ones.
• Removing a one only decreases the length of a contiguous sequence.

2. Keeping track of sequence lengths:

• Traverse the array while maintaining information about the sequences of ones and where the zeroes are located. You could use variables to store the length of the current sequence of 1s, the previous sequence of 1s, and the number of 1s since the last zero.
• You also need to handle the edge cases like start and end sequences of 1s specially since removing elements at the ends has different implications.

3. Calculating the possible maximum:

• As you identify a zero (which breaks a sequence of ones), calculate what the maximum length of a contiguous sequence would be if that zero were removed. This is done by adding the length of 1s sequences before and after the zero.
• Keep track of the global maximum length for all identified potential removals.

4. Iterating efficiently:

• Utilize a single traversal of the array (O(n) complexity), updating sequences and counting lengths as you proceed.
• This is efficient and avoids the unnecessary re-computation that a naïve method might employ, like considering each possible array resulting from each possible removal.

By systematically tracking sequences of 1s and carefully determining the impact of removing zeroes, this approach efficiently determines the optimal length of a sequence of 1s after one deletion. This strategy directly leverages the structure of contiguous sequences disrupted by zeroes, making it well-suited to large inputs as described in the problem's constraints.

Solutions

class Solution {
public:
    int longestSubarray(vector<int>& arr) {
        int numZeros = 0;
        int maxLen = 0;
        int leftIdx = 0;
            
        for (int rightIdx = 0; rightIdx < arr.size(); rightIdx++) {
            numZeros += (arr[rightIdx] == 0);
                              
            // Adjust the window when more than one zero is inside
            while (numZeros > 1) {
                numZeros -= (arr[leftIdx] == 0);
                leftIdx++;
            }
                  
            maxLen = max(maxLen, rightIdx - leftIdx);
        }
    
        return maxLen;
    }
};