<p>You are given a list of songs where the <code>i<sup>th</sup></code> song has a duration of <code>time[i]</code> seconds.</p>

<p>Return <em>the number of pairs of songs for which their total duration in seconds is divisible by</em> <code>60</code>. Formally, we want the number of indices <code>i</code>, <code>j</code> such that <code>i &lt; j</code> with <code>(time[i] + time[j]) % 60 == 0</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> time = [30,20,150,100,40]
<strong>Output:</strong> 3
<strong>Explanation:</strong> 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
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> time = [60,60,60]
<strong>Output:</strong> 3
<strong>Explanation:</strong> All three pairs have a total duration of 120, which is divisible by 60.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= time.length &lt;= 6 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= time[i] &lt;= 500</code></li>
</ul>

class Solution {
public:
    int numPairsDivisibleBy60(vector<int>& time) {
        map<int, int> mp;

        for(int t: time) {
            mp[t%60]++;
        }

        unordered_map<int, int> vis;

        long long ans=0;
        for(auto it: mp) {
            int rem = it.first;
            long long count = it.second;

            // cout<<"rem:"<<rem<<endl;

            if(vis[60-rem] == 1) continue;

            vis[rem]=1;


            long long count2 = 0;
            if(mp[60-rem]!=0)
                count2 = mp[60-rem];

            // cout<<"rem:"<<rem<<endl;
            if(rem==30 || rem==0) {
                ans= ans + (count*(count-1))/2;
            } else {
                ans = ans + (count * count2);
            }

        }

        return ans;
        
    }
};
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