1437. Check If All 1's Are at Least Length K Places Away

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Description

Given an binary array nums and an integer k, return true if all 1's are at least k places away from each other, otherwise return false.

 

Example 1:

Input: nums = [1,0,0,0,1,0,0,1], k = 2
Output: true
Explanation: Each of the 1s are at least 2 places away from each other.

Example 2:

Input: nums = [1,0,0,1,0,1], k = 2
Output: false
Explanation: The second 1 and third 1 are only one apart from each other.

 

Constraints:

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

Solutions

Solution 1

Python3

class Solution:
    def kLengthApart(self, nums: List[int], k: int) -> bool:
        j = -inf
        for i, x in enumerate(nums):
            if x:
                if i - j - 1 < k:
                    return False
                j = i
        return True

Java

class Solution {
    public boolean kLengthApart(int[] nums, int k) {
        int j = -(k + 1);
        for (int i = 0; i < nums.length; ++i) {
            if (nums[i] == 1) {
                if (i - j - 1 < k) {
                    return false;
                }
                j = i;
            }
        }
        return true;
    }
}

C++

class Solution {
public:
    bool kLengthApart(vector<int>& nums, int k) {
        int j = -(k + 1);
        for (int i = 0; i < nums.size(); ++i) {
            if (nums[i] == 1) {
                if (i - j - 1 < k) {
                    return false;
                }
                j = i;
            }
        }
        return true;
    }
};

Go

func kLengthApart(nums []int, k int) bool {
	j := -(k + 1)
	for i, x := range nums {
		if x == 1 {
			if i-j-1 < k {
				return false
			}
			j = i
		}
	}
	return true
}

TypeScript

function kLengthApart(nums: number[], k: number): boolean {
    let j = -(k + 1);
    for (let i = 0; i < nums.length; ++i) {
        if (nums[i] === 1) {
            if (i - j - 1 < k) {
                return false;
            }
            j = i;
        }
    }
    return true;
}