Sliding Window - Max Consecutive Ones III

Given a binary array nums and integer k, return the maximum number of consecutive 1s if you can flip at most k zeros.

Input: nums=[1,1,1,0,0,0,1,1,1,1,0], k=2 → Output: 6

Input: nums=[0,0,1,1,0,0,1,1,1,0,1,1,0,0,0,1,1,1,1], k=3 → Output: 10

Sliding window with a count of zeros in current window. Expand right; if zeros in window > k, shrink from left.

Maintain left pointer and zeros count.
Expand right: if nums[right]==0, increment zeros.
While zeros > k: if nums[left]==0 decrement zeros; increment left.
Answer = max(right - left + 1).

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

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