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 \* 10^4
1 <= time[i] <= 500

Solution

Time complexity : O(n)
Space complexity : O(60)

class Solution {
public:
    int numPairsDivisibleBy60(vector<int>& time) {
        int ans = 0;
        vector<int> ht(60, 0);
        for (int t: time) {
            printf("%d ", t);
            t %= 60;
            ans += ht[(60-t)%60];
            ++ht[t];
        }
        return ans;
    }
};

直覺的暴力解法，針對每一個 pair 去判斷是否符合條件，Time complexity 為 O(n^2)，Space complexity 為 O(1)。

此解法是每檢查一個新的元素時，看它的 mod60 值，能否與之前的 mod 值互補為 60。可以的話，則此新的元素可和它們各配成一對。再把它放到對應的容器中。