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

Solution: Two Pointers

• Time complexity: O(n)
• Space complexity: O(1)

 

JavaScript

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {boolean}
 */
const kLengthApart = function (nums, k) {
  const n = nums.length;
  let prevOne = -1;

  for (let index = 0; index < n; index++) {
    if (!nums[index]) continue;

    if (prevOne !== -1 && index - prevOne - 1 < k) {
      return false;
    }

    prevOne = index;
  }

  return true;
};