# 1010. Pairs of Songs With Total Durations Divisible by 60

## Description

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.

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

## Solutions

• 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;
}
}

• class Solution {
public:
int numPairsDivisibleBy60(vector<int>& time) {
int cnt[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 += 1LL * cnt[0] * (cnt[0] - 1) / 2;
ans += 1LL * cnt[30] * (cnt[30] - 1) / 2;
return ans;
}
};

• 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


• 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
}

• 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;
}