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

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Question

Approach

I track the index of the previous 1. Each time I see a new 1, I check the distance to the previous one. If the gap is smaller than k, I return False. If all 1s satisfy the distance constraint, I return True

Implementation

class Solution:
    def kLengthApart(self, nums: List[int], k: int) -> bool:
        prev = -1
        
        for i, val in enumerate(nums):
            if val == 1:
                if prev != -1 and i - prev - 1 < k:
                    return False
                prev = i
        
        return True

Complexities

• Time: O(n)
• Space: O(1)