Longest Subarray of 1's After Deleting One Element

Given a binary array nums, you should delete one element from it.

Return the size of the longest non-empty subarray containing only 1's in the resulting array. Return 0 if there is no such subarray.

Sliding Window#

Because we must delete exactly one element, the best we can do is keep a window that contains at most one zero — that zero is the element we "delete". Use two pointers (start, end) to maintain this window: expand end freely, and whenever the window accumulates more than one zero, shrink from the left by advancing start until only one zero remains. The length of the window minus 1 (to account for the mandatory deletion) gives the count of 1s; tracking the maximum across all valid windows gives the answer. Note that end - start (not end - start + 1) already accounts for the deletion since we never count the deleted zero.

• One zero is allowed in window, since we can delete it.
• As soon as you discover second zero, shift the zero

class Solution {
public:
    int longestSubarray(vector<int>& nums) {
        int n = nums.size();

        int start = 0;
        int zeroCount = 0;
        int maxLen = INT_MIN;

        for(int end = 0; end < n; end++){
            if(nums[end] == 0){
                zeroCount++;
            }

            while(zeroCount > 1){
                if(nums[start]==0)
                    zeroCount--;
                start++;
            }

            maxLen = max(maxLen, end-start);
        }
        if(maxLen == INT_MIN)
            return 0;
        return maxLen;
    }
};

Time Complexity: O(n)

Space Complexity: O(1)