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

## 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

• 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;
}
}

• 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;
}
};

• 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


• 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
}

• 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;
}