Max Consecutive Ones III

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:

Example 2:

Note:

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

Solution

Solution 1: Sliding window (dynamic)

class Solution:
    def longestOnes(self, A: List[int], K: int) -> int:
        # Time complexity: O(n).
        # Space complexity: O(1).
        m, beg = len(A), 0
        
        for end in range(m):
            
            if A[end] == 0:
                K -= 1
            
            if K < 0:
                if A[beg] == 0:
                    K += 1
                beg += 1
        
        return end - beg + 1