1010. Pairs of Songs With Total Durations Divisible by 60

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

## 题目 #

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`

## 解题思路 #

• 简单题。先将数组每个元素对 60 取余，将它们都转换到 [0,59] 之间。然后在数组中找两两元素之和等于 60 的数对。可以在 0-30 之内对半查找符合条件的数对。对 0 和 30 单独计算。因为多个 0 相加，余数还为 0 。2 个 30 相加之和为 60。

## 代码 #

``````func numPairsDivisibleBy60(time []int) int {
counts := make([]int, 60)
for _, v := range time {
v %= 60
counts[v]++
}
res := 0
for i := 1; i < len(counts)/2; i++ {
res += counts[i] * counts[60-i]
}
res += (counts[0] * (counts[0] - 1)) / 2
res += (counts[30] * (counts[30] - 1)) / 2
return res
}
``````

Apr 8, 2023