Sliding Window - Longest Subarray of 1s After Deleting One Element

Given a binary array nums, delete one element and return the size of the longest non-empty subarray containing only 1s.

Input: [1,1,0,1] → Output: 3

Input: [0,1,1,1,0,1,1,0,1] → Output: 5

Sliding window allowing at most one zero. The result is window size minus 1 (the deleted element).

Use same pattern as max consecutive ones with k=1 (allow one zero).
The answer is maxWindowSize - 1 (we must delete exactly one element).
Edge case: if no zeros in array, answer is n-1.

class Solution { public int longestSubarray(int[] nums) { int left = 0, zeros = 0, max = 0; for (int right = 0; right < nums.length; right++) { if (nums[right] == 0) zeros++; while (zeros > 1) { if (nums[left] == 0) zeros--; left++; } max = Math.max(max, right - left); // -1 for the deleted element } return max; } }

Time Complexity: O(N)
Space Complexity: O(1)