Problem of The Day: Max Consecutive Ones III

Problem Statement

My Solution

class Solution:
    def longestOnes(self, nums: List[int], k: int) -> int:
        N = len(nums)
        res = 0
        curr = 0
        start = 0
        for end in range(N):
            if nums[end] == 0:
                k -= 1
            if k < 0:
                while start <= end and k < 0:
                    if nums[start] == 0:
                        k += 1
                    curr -= 1
                    start += 1
            if k >= 0:
                curr += 1
            res = max(res, curr)
        return res

Editorial Solution

class Solution:
    def longestOnes(self, nums: List[int], k: int) -> int:
        left = 0
        for right in range(len(nums)):
            # If we included a zero in the window we reduce the value of k.
            # Since k is the maximum zeros allowed in a window.
            k -= 1 - nums[right]
            # A negative k denotes we have consumed all allowed flips and window has
            # more than allowed zeros, thus increment left pointer by 1 to keep the window size same.
            if k < 0:
                # If the left element to be thrown out is zero we increase k.
                k += 1 - nums[left]
                left += 1
        return right - left + 1