In a list of songs, 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 with (time[i] + time[j]) % 60 == 0.
Example 1:
Input: [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: [60,60,60]
Output: 3
Explanation: All three pairs have a total duration of 120, which is divisible by 60.
Note:
1 <= time.length <= 60000
1 <= time[i] <= 500
解题思路
最简单的思路是使用两个for循环，判断加起来是否是60的倍数，然后果断超时了。
改进后的思路是：使用一个数组来存储余数(t % 60)的个数，然后每个元素可以组成的pair数目就是数组中位置为(60 - t % 60) % 60的元素的大小。
要满足条件：(a + b) % 60 == 0，那么对于大多数例子是：
a % 60 + b % 60 == 60，同时也需要注意一个特殊的例子：a % 60 == 0 && b % 60 == 0;
实现代码
// Runtime: 3 ms, faster than 84.18% of Java online submissions for Pairs of Songs With Total Durations Divisible by 60.
// Memory Usage: 42.4 MB, less than 100.00% of Java online submissions for Pairs of Songs With Total Durations Divisible by 60.
class Solution {
public int numPairsDivisibleBy60(int[] time) {
if (time.length < 2) {
return 0;
}
int[] cnt = new int[60];
int result = 0;
for (int t : time) {
result += cnt[(60 - t % 60) % 60];
cnt[t % 60] += 1;
}
return result;
}
}
