README_EN.md

1004. Max Consecutive Ones III

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Description

Given a binary array nums and an integer k, return the maximum number of consecutive 1's in the array if you can flip at most k 0's.

 

Example 1:

Input: nums = [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: nums = [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.

 

Constraints:

• 1 <= nums.length <= 105
• nums[i] is either 0 or 1.
• 0 <= k <= nums.length

Solutions

Python3

class Solution:
    def longestOnes(self, nums: List[int], k: int) -> int:
        l = r = -1
        while r < len(nums) - 1:
            r += 1
            if nums[r] == 0:
                k -= 1
            if k < 0:
                l += 1
                if nums[l] == 0:
                    k += 1
        return r - l

Java

class Solution {
    public int longestOnes(int[] nums, int k) {
        int l = 0, r = 0;
        while (r < nums.length) {
            if (nums[r++] == 0) {
                --k;
            }
            if (k < 0 && nums[l++] == 0) {
                ++k;
            }
        }
        return r - l;
    }
}

C++

class Solution {
public:
    int longestOnes(vector<int>& nums, int k) {
        int l = 0, r = 0;
        while (r < nums.size())
        {
            if (nums[r++] == 0) --k;
            if (k < 0 && nums[l++] == 0) ++k;
        }
        return r - l;
    }
};

Go

func longestOnes(nums []int, k int) int {
	l, r := -1, -1
	for r < len(nums)-1 {
		r++
		if nums[r] == 0 {
			k--
		}
		if k < 0 {
			l++
			if nums[l] == 0 {
				k++
			}
		}
	}
	return r - l
}

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