LeetCode-in-Java

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

Easy

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

Solution

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

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