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

Description

Given an array nums of 0s and 1s 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.

Example 3:

Input: nums = [1,1,1,1,1], k = 0
Output: true

Example 4:

Input: nums = [0,1,0,1], k = 1
Output: true

 

Constraints:

• 1 <= nums.length <= 10^5
• 0 <= k <= nums.length
• nums[i] is 0 or 1

Solution(javascript)

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {boolean}
 */
const kLengthApart = function (nums, k) {
  const start = nums.findIndex(x => x === 1)
  if (start === -1) {
    return true
  }
  let prev = start
  for (let i = start + 1; i < nums.length; i++) {
    if (nums[i] === 1) {
      if (i - prev <= k) {
        return false
      }
      prev = i
    }
  }
  return true
}