LeetCode #1010 — MEDIUM

Pairs of Songs With Total Durations Divisible by 60

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given a list of songs where the i^th song has a duration of time[i] seconds.

Return the number of pairs of songs for which their total duration in seconds is divisible by 60. Formally, we want the number of indices i, j such that i < j with (time[i] + time[j]) % 60 == 0.

Example 1:

Input: time = [30,20,150,100,40]
Output: 3
Explanation: Three pairs have a total duration divisible by 60:
(time[0] = 30, time[2] = 150): total duration 180
(time[1] = 20, time[3] = 100): total duration 120
(time[1] = 20, time[4] = 40): total duration 60

Example 2:

Input: time = [60,60,60]
Output: 3
Explanation: All three pairs have a total duration of 120, which is divisible by 60.

Constraints:

1 <= time.length <= 6 * 10⁴
1 <= time[i] <= 500

Step 01

Brute Force Baseline

Problem summary: You are given a list of songs where the ith song has a duration of time[i] seconds. Return the number of pairs of songs for which their total duration in seconds is divisible by 60. Formally, we want the number of indices i, j such that i < j with (time[i] + time[j]) % 60 == 0.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

[30,20,150,100,40]

Example 2

[60,60,60]

Core Insight

What unlocks the optimal approach

We only need to consider each song length modulo 60.
We can count the number of songs having same (length % 60), and store that in an array of size 60.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1010: Pairs of Songs With Total Durations Divisible by 60
class Solution {
    public int numPairsDivisibleBy60(int[] time) {
        int[] cnt = new int[60];
        for (int t : time) {
            ++cnt[t % 60];
        }
        int ans = 0;
        for (int x = 1; x < 30; ++x) {
            ans += cnt[x] * cnt[60 - x];
        }
        ans += (long) cnt[0] * (cnt[0] - 1) / 2;
        ans += (long) cnt[30] * (cnt[30] - 1) / 2;
        return ans;
    }
}

// Accepted solution for LeetCode #1010: Pairs of Songs With Total Durations Divisible by 60
func numPairsDivisibleBy60(time []int) (ans int) {
	cnt := [60]int{}
	for _, t := range time {
		cnt[t%60]++
	}
	for x := 1; x < 30; x++ {
		ans += cnt[x] * cnt[60-x]
	}
	ans += cnt[0] * (cnt[0] - 1) / 2
	ans += cnt[30] * (cnt[30] - 1) / 2
	return
}

# Accepted solution for LeetCode #1010: Pairs of Songs With Total Durations Divisible by 60
class Solution:
    def numPairsDivisibleBy60(self, time: List[int]) -> int:
        cnt = Counter(t % 60 for t in time)
        ans = sum(cnt[x] * cnt[60 - x] for x in range(1, 30))
        ans += cnt[0] * (cnt[0] - 1) // 2
        ans += cnt[30] * (cnt[30] - 1) // 2
        return ans

// Accepted solution for LeetCode #1010: Pairs of Songs With Total Durations Divisible by 60
struct Solution;

impl Solution {
    fn num_pairs_divisible_by60(time: Vec<i32>) -> i32 {
        let mut a: Vec<i32> = vec![0; 60];
        let mut res = 0;
        for x in time {
            let count = a[((600 - x) % 60) as usize];
            if count != 0 {
                res += count;
            }
            a[(x % 60) as usize] += 1;
        }
        res
    }
}

#[test]
fn test() {
    let time = vec![30, 20, 150, 100, 40];
    assert_eq!(Solution::num_pairs_divisible_by60(time), 3);
    let time = vec![60, 60, 60];
    assert_eq!(Solution::num_pairs_divisible_by60(time), 3);
}

// Accepted solution for LeetCode #1010: Pairs of Songs With Total Durations Divisible by 60
function numPairsDivisibleBy60(time: number[]): number {
    const cnt: number[] = new Array(60).fill(0);
    for (const t of time) {
        ++cnt[t % 60];
    }
    let ans = 0;
    for (let x = 1; x < 30; ++x) {
        ans += cnt[x] * cnt[60 - x];
    }
    ans += (cnt[0] * (cnt[0] - 1)) / 2;
    ans += (cnt[30] * (cnt[30] - 1)) / 2;
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.