0727MediumArrayHash TablePrefix Sum

Spacing Totals Along The Conveyor

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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 conveyor carries parts in slots numbered 0, 1, 2, and so on. codes[i] is the part code sitting in slot i, and two slots hold the same part exactly when their codes are equal.

For each slot i, its spacing total is the sum of |i - j| over every other slot j that holds the same part code. A slot whose code appears in no other slot has a spacing total of 0.

Return an array holding the spacing total of every slot, in slot order.

Examples

Example 1

Input: codes = [4, 9, 4, 4, 6]
Output: [5, 0, 3, 4, 0]

Part 4 sits in slots 0, 2 and 3. Slot 0 is 2 and 3 away from the other two, giving 5; slot 2 is 2 and 1 away, giving 3; slot 3 is 3 and 1 away, giving 4. Parts 9 and 6 appear once each, so their slots score 0.

Example 2

Input: codes = [3, 3, 3, 3]
Output: [6, 4, 4, 6]

Every slot holds the same part. Slot 0 is 1, 2 and 3 away from the others, giving 6, and slot 1 is 1, 1 and 2 away, giving 4.

Example 3

Input: codes = [7, 1, 4, 2, 9]
Output: [0, 0, 0, 0, 0]

All five codes are different, so no slot has a partner and every spacing total is 0.

Constraints

1 <= codes.length <= 10^5
0 <= codes[i] <= 10^9
The largest spacing total these bounds allow is below 5 * 10^9, so a 32-bit integer type is not wide enough to hold one.

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_totals(codes: list[int]) -> list[int]:

Java

public long[] spacingTotals(int[] codes)

The intended approach

Time O(n) for n slots.Space O(n) for the buckets and the returned array.

Slots only ever measure against slots holding the same code, so bucket the slot numbers by code; because the conveyor is read left to right each bucket comes out in increasing order. Within one bucket a slot's total is the distance to the slots before it plus the distance to the slots after it, and both halves follow from running sums: with `p` slots before it summing to `behind`, the near half is `p * s - behind`, and with the remaining slots summing to `ahead`, the far half is `ahead` minus that count times `s`. One pass per bucket fills the answer.

Pythonentry point: spacing_totals

def spacing_totals(codes):
    # Bucket the slot numbers by part code, keeping each bucket in increasing order.
    # Inside one bucket the total for a slot splits into the slots before it and the
    # slots after it: with p earlier slots whose numbers add up to L, the earlier
    # part is p * s - L, and with q later slots adding up to R the later part is
    # R - q * s.  Running L and R along the bucket makes each bucket linear.
    # Time O(n).  Space O(n) for the buckets and the returned array.
    buckets = {}
    for slot, code in enumerate(codes):
        if code in buckets:
            buckets[code].append(slot)
        else:
            buckets[code] = [slot]
    answer = [0] * len(codes)
    for slots in buckets.values():
        size = len(slots)
        if size < 2:
            continue
        behind = 0
        ahead = 0
        for s in slots:
            ahead += s
        for p, s in enumerate(slots):
            ahead -= s
            answer[s] = (p * s - behind) + (ahead - (size - 1 - p) * s)
            behind += s
    return answer

The obvious version

For every slot, walk the whole conveyor and add up the gap to each other slot carrying an equal code.

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_totals(codes):
    n = len(codes)
    out = []
    for i in range(n):
        total = 0
        for j in range(n):
            if j != i and codes[j] == codes[i]:
                total += abs(i - j)
        out.append(total)
    return out

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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