1004. Max Consecutive Ones III #

Problem #

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

Problem Summary #

This problem tests the sliding window technique.

Given an array whose elements contain only 0 and 1. Also given K, representing the number of times 0 can be changed to 1. Find the longest consecutive length of 1 after the transformations.

Solution Approach #

Just process it according to the sliding window approach, continuously updating and maintaining the maximum length.

Code #

package leetcode

func longestOnes(A []int, K int) int {
	res, left, right := 0, 0, 0
	for left < len(A) {
		if right < len(A) && ((A[right] == 0 && K > 0) || A[right] == 1) {
			if A[right] == 0 {
				K--
			}
			right++
		} else {
			if K == 0 || (right == len(A) && K > 0) {
				res = max(res, right-left)
			}
			if A[left] == 0 {
				K++
			}
			left++
		}
	}
	return res
}

func max(a int, b int) int {
	if a > b {
		return a
	}
	return b
}