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

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:
class Solution:
    def kLengthApart(self, nums: List[int], k: int) -> bool:
        minDist = float('inf')
        prev = None
        numOnes = 0
        for i, num in enumerate(nums):
            if num == 1:
                numOnes += 1
                if prev:
                    minDist = min(minDist, i - prev - 1)
                prev = i
        if numOnes <= 1:
            return True
        if minDist == float('inf'):
            return False
        return minDist >= k

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