<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) {
        vector<int> cands={60,120,180,240,300,360,420,480,540,600,660,720,780,840,900,960};
        map<int,int> mp;
        for(auto it:time){
            mp[it]++;
        }
        long long ans=0;
        
        for(auto it:mp){
            // cout<<it.first<<" "<<it.second<<endl;
            for(auto t:cands){
                int val=t-it.first;
                if(mp.find(val)!=mp.end()){
                    cout<<it.first<<" "<<val<<endl;
                    if(val==it.first){

                        if(it.second>1){
                            
                            long long temp=1;
                            temp*=it.second;
                            temp*=it.second-1;
                            temp/=2;
                            ans+=temp;
                        }
                    }
                    else{
                        ans+=(it.second*mp[val]);
                    }
                    mp[it.first]=0;
                    // mp[val]=0;
                }
            }
        }
        return ans;
    }
};
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