0878EasyArray

Spacing Between the Marked Sleepers

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Trains the technique from

LeetCode 1437Check If All 1's Are at Least Length K Places Away

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A track is logged as nums, where 1 marks a sleeper due for replacement and 0 one that is not.

Return true when every two marked sleepers have at least k unmarked sleepers between them, and false otherwise. A track with fewer than two marked sleepers always qualifies.

Examples

Example 1

Input: nums = [1, 0, 0, 0, 1, 0, 0, 1], k = 2
Output: true

The first pair of marks has three unmarked sleepers between them and the second pair has two, so both reach the required two.

Example 2

Input: nums = [1, 0, 1], k = 2
Output: false

The two marks have a single unmarked sleeper between them, which falls short of the required two.

Example 3

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

There is only one marked sleeper, so there is no pair to be too close.

Constraints

1 <= nums.length <= 10^5
0 <= k <= 100000
0 <= nums[i] <= 1
k is at most nums.length

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def spacing_holds(nums: list[int], k: int) -> bool:

Java

public boolean spacingHolds(int[] nums, int k)

The intended approach

Time O(n) for a track of n sleepers: one pass.Space O(1).

Only neighbouring marked sleepers need checking, because if every neighbouring pair is far enough apart then any wider pair is too. Walk the track remembering where the last marked sleeper was, and when the next one turns up, the unmarked sleepers between them number the gap in positions minus one. Report failure as soon as one pair is too close.

Pythonentry point: spacing_holds

def spacing_holds(nums, k):
    last = -1
    for i, mark in enumerate(nums):
        if mark != 1:
            continue
        if last >= 0 and i - last - 1 < k:
            return False
        last = i
    return True

The obvious version

Collect the positions of every marked sleeper, then compare every pair of them, counting the unmarked sleepers between the two. Quadratic in the number of marks, but it never assumes only neighbouring pairs matter.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_spacing_holds(nums, k):
    spots = [i for i in range(len(nums)) if nums[i] == 1]
    for a in range(len(spots)):
        for b in range(a + 1, len(spots)):
            between = 0
            for p in range(spots[a] + 1, spots[b]):
                if nums[p] == 0:
                    between += 1
            if between < k:
                return False
    return True

How this problem was checked

Reference solution executed against 12 test cases — 12 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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