Pairs of Songs

/*
    Problem Name: Pairs of Songs With Total Durations Divisible by 60.
    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.
*/

#include<bits/stdc++.h>
using namespace std;

int numPairsDivisibleBy60(vector<int>& time) 
{

    unordered_map<int,int>m;
    for(int i=0;i<time.size();++i)
    {
        time[i]%=60;
        m[time[i]]++;
    }

    int cnt=0;
    for(auto it:m)
    {
        if(it.first==0 || it.first==30)
            cnt+=((it.second-1)*(it.second))/2;
        else if(it.first<30 && m.count(60-it.first))
            cnt+=(m[it.first]*m[60-it.first]);
    }
    return cnt;
}

int main()
{
    vector<int>time;
    int n;
    cin>>n;
    for(int i=0;i<n;++i)
    {
        int data;
        cin>>data;
        time.push_back(data);
    }
    cout<<numPairsDivisibleBy60(time);
    return 0;
}