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1004. Max Consecutive Ones III

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

Same as in https://leetcode.ca/2017-03-31-487-Max-Consecutive-Ones-II/, change from 1 to k for queue size check.

  • 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;
    }
}

  • 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;
    }
};

  • class Solution:
    def longestOnes(self, nums: List[int], k: int) -> int:
        res, zero, left = 0, 0, 0

        for right in range(len(nums)):
            if nums[right] == 0:
                zero += 1
            while zero > k:
                if nums[left] == 0:
                    zero -= 1
                left += 1
            res = max(res, right - left + 1)

        return res


class Solution_followup:
    def findMaxConsecutiveOnes(self, nums: List[int]) -> int:

        res, left, k = 0, 0, 1
        q = deque()

        for right in range(len(nums)):
            if nums[right] == 0:
                q.append(right)
            if len(q) > k:
                left = q.popleft() + 1
            res = max(res, right - left + 1) # max-check every for iteration

        return res

##############

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


  • 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
}

  • function longestOnes(nums: number[], k: number): number {
    const n = nums.length;
    let l = 0;
    for (const num of nums) {
        if (num === 0) {
            k--;
        }
        if (k < 0 && nums[l++] === 0) {
            k++;
        }
    }
    return n - l;
}


  • impl Solution {
    pub fn longest_ones(nums: Vec<i32>, mut k: i32) -> i32 {
        let n = nums.len();
        let mut l = 0;
        for num in nums.iter() {
            if num == &0 {
                k -= 1;
            }
            if k < 0 {
                if nums[l] == 0 {
                    k += 1;
                }
                l += 1;
            }
        }
        (n - l) as i32
    }
}

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