1116MediumArrayStringPrefix Sum

Gathering the Crates Into Every Bay

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Trains the technique from

LeetCode 1769Minimum Number of Operations to Move All Balls to Each Box

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 corridor has bays in a row, given by bays, where 1 marks a bay holding a crate and 0 marks an empty bay.

One step shifts a crate to a bay next door. A bay may hold any number of crates at once.

Return a list whose entry i is the fewest steps needed to gather every crate into bay i, each entry worked out on its own as though the corridor had never been touched.

Examples

Example 1

Input: bays = "1001"
Output: [3, 3, 3, 3]

Two crates stand three bays apart. Gathering them into any bay of the corridor always comes to three steps, since every step towards one is a step away from the other.

Example 2

Input: bays = "111"
Output: [3, 2, 3]

For the middle bay the two outer crates move one step each and the crate already there moves none. For either end bay the crates move one and two steps.

Example 3

Input: bays = "10"
Output: [0, 1]

The crate already sits in the first bay, so nothing moves, and reaching the second bay takes one step.

Constraints

1 <= bays.length <= 2000
every character of bays is 0 or 1

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 min_operations(bays: str) -> list[int]:

Java

public int[] minOperations(String bays)

The intended approach

Time O(n) over the bays.Space O(n) for the answer.

The answer for one bay is the sum of the distances from every crate to it, so working each bay out on its own costs the corridor length squared. Splitting the sum into the crates on the left and the crates on the right makes both halves fall out in a single sweep each. Sweep left to right holding two running numbers: how many crates lie at or behind the current bay, and the total steps those crates need to reach it. Moving one bay along costs every crate behind exactly one more step, so the total grows by the crate count, and then the current bay's own crate joins the count. That gives each bay its left-hand share. Sweeping right to left with the same two numbers gives the right-hand share, and adding the two sweeps position by position finishes the job. A crate sitting in the bay itself adds nothing to either share, which the ordering inside the sweep takes care of.

Pythonentry point: min_operations

def min_operations(bays):
    n = len(bays)
    answer = [0] * n
    crates = 0
    steps = 0
    for i in range(n):
        answer[i] += steps
        if bays[i] == "1":
            crates += 1
        steps += crates
    crates = 0
    steps = 0
    for i in range(n - 1, -1, -1):
        answer[i] += steps
        if bays[i] == "1":
            crates += 1
        steps += crates
    return answer

The obvious version

List the bays holding crates, then for each bay in turn add up the gap to every one of them. Costs the corridor length times the crate count.

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_min_operations(bays):
    n = len(bays)
    spots = []
    for j in range(n):
        if bays[j] == "1":
            spots.append(j)
    out = []
    for i in range(n):
        total = 0
        for j in spots:
            total += j - i if j > i else i - j
        out.append(total)
    return out

How this problem was checked

Reference solution executed against 15 test cases — 15 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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