1046MediumArrayHash TableCounting

Pairs of Tracks Filling Whole Minutes

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Trains the technique from

LeetCode 1010Pairs of Songs With Total Durations Divisible by 60

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 playlist's track lengths, in seconds, read seconds.

Return how many pairs of different positions hold two tracks whose lengths add up to a whole number of minutes, that is to a multiple of sixty seconds. A pair counts once, whichever order it is taken in.

Examples

Example 1

Input: seconds = [30, 30]
Output: 1

The two lengths of thirty add to sixty, exactly one whole minute.

Example 2

Input: seconds = [59, 1, 120]
Output: 1

Fifty-nine and one add to sixty. The track of a hundred and twenty already fills two whole minutes on its own, but a pair needs two tracks, and it leaves neither of the others on a whole minute.

Example 3

Input: seconds = [500, 500, 500, 500]
Output: 0

Five hundred leaves twenty over a whole minute, and twenty and twenty come to forty, never sixty.

Constraints

1 <= seconds.length <= 6 * 10^4
1 <= seconds[i] <= 500

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 num_pairs_divisible_by60(seconds: list[int]) -> int:

Java

public int numPairsDivisibleBy60(int[] seconds)

The intended approach

Time O(n) over the playlist's length.Space O(1), since only sixty remainders exist.

Only each length's remainder against sixty matters. Two tracks fill whole minutes exactly when their remainders add to sixty, or when both remainders are nothing. Walk the playlist once carrying a tally of the remainders already seen. For the track in hand, the earlier tracks that pair with it are those whose remainder is sixty less this one, taken to the remainder itself so that nothing pairs with nothing rather than with sixty. Add that tally to the count, then record the track's own remainder. Counting only against tracks already seen is what stops every pair being counted twice, and it handles two tracks of the same remainder without any special case.

Pythonentry point: num_pairs_divisible_by60

def num_pairs_divisible_by60(seconds):
    tally = [0] * 60
    total = 0
    for length in seconds:
        rest = length % 60
        total += tally[(60 - rest) % 60]
        tally[rest] += 1
    return total

The obvious version

Take every pair of positions with one before the other, add the two lengths, and count the pair when the total divides evenly by sixty. Costs the playlist's length squared rather than one pass with a tally of remainders.

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_num_pairs_divisible_by60(seconds):
    n = len(seconds)
    total = 0
    for i in range(n):
        for j in range(i + 1, n):
            if (seconds[i] + seconds[j]) % 60 == 0:
                total += 1
    return total

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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