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1010. Pairs of Songs with Total Durations Divisible by 60

MediumOpen on LeetCodeProblem statement

Problem Statement

1010. Pairs of Songs With Total Durations Divisible by 60

Medium

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:

C++

Source file
class Solution {
public:
    int numPairsDivisibleBy60(vector<int>& time) {
        array <int, 60> v;
        for ( int i=0; i<60; i++){
            v[i] = 0;
        }
        int res = 0;
        for ( int i=0; i<time.size(); i++){
            v[time[i]%60] += 1;
        }
        for ( int i=1; i<=29; i++){
            if (!(v[i]))
                continue;
            res += v[i] * v[60-i];
            cout << res;
        }
        res += ((v[0]*(v[0]-1))/2 + (v[30]*(v[30]-1))/2);
        cout << v[30];
        return res;
    }
};