README.md

1004. Max Consecutive Ones III (Medium)

Given an array A of 0s and 1s, we may change up to K values from 0 to 1.

Return the length of the longest (contiguous) subarray that contains only 1s. 

 

Example 1:

Input: A = [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: A = [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.

 

Note:

1. 1 <= A.length <= 20000
2. 0 <= K <= A.length
3. A[i] is 0 or 1 

Solution 1. Sliding Window

// OJ: https://leetcode.com/problems/max-consecutive-ones-iii/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int longestOnes(vector<int>& A, int K) {
        int ans = 0, zero = 0, i = 0, j = 0, N = A.size();
        while (j < N) {
            zero += A[j++] == 0;
            if (zero <= K) ans = max(ans, j - i);
            while (zero > K) zero -= A[i++] == 0;
        }
        return ans;
    }
};