Editorial

Solution walkthrough

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Video explanation

Solution

Intuition

Only consecutive occurrences of one need checking: any more distant pair has at least as much separation. The number of positions between ones at indices previous and i is i - previous - 1, not the raw index difference.

Approach

Scan the array while remembering the last index containing one. On each new one, reject if a previous one exists and the number of intervening positions is below k; then update the remembered index. Return true after the scan. Zero or one occurrence of one satisfies the condition automatically, and k = 0 allows neighboring ones.

class Solution:
    def kLengthApart(self, nums: List[int], k: int) -> bool:
        n = len(nums)
        prev = -1

        for i in range(n):
            if nums[i] == 1:
                if prev != -1 and i - prev - 1 < k:
                    return False
                prev = i

        return True

Complexity analysis

The single index scan takes O(n) time and O(1) auxiliary space.