All Divisions With The Highest Score Of A Binary Array - Leetcode Solution

Problem Statement:

Given a binary array nums of size n, you are allowed to replace at most k elements of this array with any other element in the array (both 0 and 1).

Find the maximum length of a contiguous subarray containing only 1's after performing at most this many replacements.

Example 1: Input: nums = [1,1,0,0,1,1,1,0,1], k = 2 Output: 6 Explanation: [1,1,0,0,1,1,1,0,1] One possible longest contiguous subarray is [1,1,0,0,1,1,1,1,1].

Solution:

Approach:

We can use the sliding window technique to solve this problem. We can maintain a window of contiguous 1's and expand the window until we hit k zeroes, then shrink the window until we can replace k zeroes again. We can keep track of the maximum size of this window as we expand and shrink it.

Algorithm:

1. Initialize left and right pointers to 0 and a zeros counter to 0.
2. While right pointer is less than length of array: a. If nums[right] is 0, increment zeros counter. b. If zeros counter is greater than k, decrement zeros counter and move the left pointer. c. Increment the right pointer. d. Update the maximum length of the window.
3. Return the maximum length of the window.

Code:

class Solution: def longestOnes(self, nums: List[int], k: int) -> int: left = right = zeros = maxLength = 0 while right < len(nums): if nums[right] == 0: zeros += 1 while zeros > k: if nums[left] == 0: zeros -= 1 left += 1 maxLength = max(maxLength, right - left + 1) right += 1 return maxLength

Time Complexity:

The time complexity of this algorithm is O(n), where n is the length of the input array, since we traverse the array once from left to right.

Space Complexity:

The space complexity of this algorithm is O(1), since we are only using a constant amount of extra space to store variables.

1