0384MediumArray

Farthest Free Locker

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Trains the technique from

LeetCode 849Maximize Distance to Closest Person

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 gym corridor has one row of lockers. lockers[i] is 1 when locker i is already assigned to somebody and 0 when it is free. At least one locker is free and at least one is assigned.

You get to claim one free locker. The quiet score of a locker is the number of doors between it and the closest assigned locker, that is |i - j| for the closest index j with lockers[j] == 1.

Return the largest quiet score you can claim.

Examples

Example 1

Input: lockers = [1, 0, 0, 0, 0, 1, 0, 1]
Output: 2

Claiming locker 2 puts two doors between it and locker 0, and two doors between it and locker 5.

Example 2

Input: lockers = [0, 0, 0, 1]
Output: 3

Claiming locker 0 puts three doors between it and locker 3, the only assigned one.

Example 3

Input: lockers = [1, 0, 0]
Output: 2

Claiming locker 2 puts two doors between it and locker 0.

Example 4

Input: lockers = [1, 0, 1, 0, 1]
Output: 1

Both free lockers sit next to an assigned locker on either side.

Constraints

2 <= lockers.length <= 2 * 10^4
lockers[i] is 0 or 1.
At least one locker is free.
At least one locker is assigned.

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 max_dist_to_closest(lockers: list[int]) -> int:

Java

public int maxDistToClosest(int[] lockers)

The intended approach

Time O(n)Space O(1)

A free locker's quiet score depends only on the run of free lockers it belongs to. Walk the corridor once, remembering the index of the previous assigned locker. A run that starts at the wall scores its own length, because there is no assigned locker on that side; the same holds for a run that ends at the far wall. A run bounded by assigned lockers on both sides scores half the distance between them, rounded down. Report the largest score seen.

Pythonentry point: max_dist_to_closest

def max_dist_to_closest(lockers):
    # Time O(n) -- one pass over the corridor.
    # Space O(1) -- three counters.
    n = len(lockers)
    best = 0
    prev = -1
    for i in range(n):
        if lockers[i] == 1:
            if prev < 0:
                reach = i
            else:
                reach = (i - prev) // 2
            if reach > best:
                best = reach
            prev = i
    tail = n - 1 - prev
    if tail > best:
        best = tail
    return best

The obvious version

For every free locker, measure the distance to each assigned locker and keep the smallest, then report the largest such value. Quadratic in the number of lockers.

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_max_dist_to_closest(lockers):
    taken = [i for i, v in enumerate(lockers) if v == 1]
    best = 0
    for i, v in enumerate(lockers):
        if v == 1:
            continue
        near = min(abs(i - j) for j in taken)
        if near > best:
            best = near
    return best

How this problem was checked

Reference solution executed against 10 test cases — 10 passed
4 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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