0591MediumArrayHash TablePrefix Sum

Shortest Run to Pull From the Load

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Trains the technique from

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 freight car is loaded with pallets in a single row, and nums[i] is the weight of the pallet in position i. The yard will only release the car when the total weight on board is a multiple of p.

You may pull out one unbroken run of pallets, and only one. You may also pull out nothing at all. You may not pull out every pallet: at least one pallet has to stay on board.

Return the length of the shortest run you can pull out so that the weight left on board is a multiple of p. If no choice of run works, return -1. Pulling out nothing counts as a run of length 0.

Examples

Example 1

Input: nums = [9, 2, 6, 4, 5], p = 11
Output: 1

The load weighs 26, which is not a multiple of 11. Pulling out the single pallet in position 3, weight 4, leaves 22 on board, and 22 is 11 twice over.

Example 2

Input: nums = [5, 5, 5], p = 7
Output: -1

The load weighs 15. The runs available are of weight 5, 5, 5, 10 and 10, and none of them leaves a multiple of 7 behind. Pulling out all three pallets would leave 0, which is a multiple of 7, but the car may not be emptied.

Example 3

Input: nums = [1, 2, 3], p = 1
Output: 0

Every whole number is a multiple of 1, so the load already qualifies and nothing has to come off.

Constraints

1 <= nums.length <= 10^5
1 <= nums[i] <= 10^9
1 <= p <= 10^9

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_subarray(nums: list[int], p: int) -> int:

Java

public int minSubarray(int[] nums, int p)

The intended approach

Time O(n) — one pass with constant-time dictionary work per slot.Space O(min(n, p)) for the remainder dictionary.

Let `t` be the remainder of the whole load when divided by `p`. If `t` is zero nothing has to come off. Otherwise the run pulled out must itself leave remainder `t`, and the weight of the run from slot `a + 1` to slot `b` is the running total at `b` minus the running total at `a`. So walk the row keeping the running total modulo `p`, and at each slot look for the most recent earlier slot whose running remainder is the current remainder minus `t`, taken modulo `p`. A dictionary from remainder to latest slot makes that lookup constant time, and the closest such slot gives the shortest run ending here. A best length equal to the whole row would mean removing everything, which is not allowed, so that case reports -1.

Pythonentry point: min_subarray

def min_subarray(nums, p):
    # O(n) time, O(min(n, p)) space: the remaining weight is divisible by p when
    # the removed stretch leaves the same remainder as the whole load does. One
    # pass over running remainders, with a dictionary holding the latest slot at
    # which each remainder was seen, finds the closest matching earlier slot.
    n = len(nums)
    total = 0
    for weight in nums:
        total += weight
    target = total % p
    if target == 0:
        return 0
    best = n
    latest = {0: -1}
    running = 0
    for i in range(n):
        running = (running + nums[i]) % p
        wanted = (running - target) % p
        if wanted in latest:
            span = i - latest[wanted]
            if span < best:
                best = span
        latest[running] = i
    return best if best < n else -1

The obvious version

Try lengths in increasing order, from 1 up to one less than the number of pallets. For each length slide a window across the row, keeping its weight by adding the entering pallet and dropping the leaving one, and return that length the moment the remaining weight divides evenly. Report -1 if no length works.

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_subarray(nums, p):
    n = len(nums)
    total = sum(nums)
    if total % p == 0:
        return 0
    best = -1
    for length in range(1, n):
        window = sum(nums[0:length])
        for start in range(0, n - length + 1):
            if start > 0:
                window = window - nums[start - 1] + nums[start + length - 1]
            if (total - window) % p == 0:
                return length
    return best

How this problem was checked

Reference solution executed against 11 test cases — 11 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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