1004. Max Consecutive Ones III

 Given a binary array nums and an integer k, return the maximum number of consecutive 1's in the array if you can flip at most k 0's.

 

Example 1:

Input: nums = [1,1,1,0,0,0,1,1,1,1,0], k = 2
Output: 6
Explanation: [1,1,1,0,0,1,1,1,1,1,1]
Bolded numbers were flipped from 0 to 1. The longest subarray is underlined.

Example 2:

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
Explanation: [0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,1,1,1,1]
Bolded numbers were flipped from 0 to 1. The longest subarray is underlined.

 

Constraints:

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

class Solution {

public:

    

    int longestOnes(vector<int>& A, int k) {

        // base cases about k.

        if (A.size() == 0) {

            return 0;

        }

        int ans = 0;

        int st = 0, end = 0;

        int zeros = 0;

        while (end < A.size()) {

            if (A[end] == 0) {

                zeros++;

            }

            if (zeros == k) {

                ans = max(ans, end - st + 1);

            }

            while (zeros > k && st <= end) {

                if (A[st] == 0) {

                    zeros--;

                }

                st++;

            }

            end++;

        }

        ans = max(ans, end - st);

        return ans;

    }

};