0453EasyArrayHash TableMathCounting

Matched Spoke Pairs

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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 wheelbuilder keeps loose spokes in a rack. lengths[i] is the length in millimetres of the spoke in slot i.

Two spokes make a matched pair when they sit in different slots and read the same length. A pair is named by its two slots, so the pair from slots 2 and 5 is the same pair as the one from slots 5 and 2, and a spoke never pairs with itself.

Return the number of matched pairs in the rack. Pairs are counted over all slots at once, so a length held in four slots contributes every pair those four slots make.

Examples

Example 1

Input: lengths = [64, 17, 64, 64, 29]
Output: 3

Length 64 sits in slots 0, 2 and 3, which make the pairs (0,2), (0,3) and (2,3). The spokes at 17 and 29 have no partner.

Example 2

Input: lengths = [46, 82, 91]
Output: 0

The three spokes all read different lengths, so the rack holds no matched pair.

Example 3

Input: lengths = [7, 7, 7, 7, 7, 7]
Output: 15

All six slots read 7, and six slots make fifteen distinct pairs.

Constraints

1 <= lengths.length <= 100
1 <= lengths[i] <= 100

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 matched_spoke_pairs(lengths: list[int]) -> int:

Java

public int matchedSpokePairs(int[] lengths)

The intended approach

Time O(n) for n slots.Space O(d) for d distinct lengths in the rack.

Sweep the rack once keeping a tally of how many spokes of each length have already gone by. The spoke now in hand pairs with exactly those, so add the tally to the running answer before bumping it. Every pair is credited once, at the moment its later slot is reached.

Pythonentry point: matched_spoke_pairs

def matched_spoke_pairs(lengths):
    # Walk the rack once. Every spoke of the same length already passed forms
    # one fresh pair with the spoke now in hand.
    # O(n) time, O(d) space for d distinct lengths.
    seen = {}
    pairs = 0
    for L in lengths:
        pairs += seen.get(L, 0)
        seen[L] = seen.get(L, 0) + 1
    return pairs

The obvious version

Test every slot against every later slot and count the equal ones. O(n^2).

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_matched_spoke_pairs(lengths):
    total = 0
    n = len(lengths)
    for i in range(n):
        for j in range(i + 1, n):
            if lengths[i] == lengths[j]:
                total += 1
    return total

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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