1010. song backtracking with total duration divisible by 60

In the song list , The first  ​​i​​​  The duration of a song is  ​​time[i]​​  second .

Return its total duration （ In seconds ） Can be  ​​60​​​  The number of divided song pairs . Formally , We want subscript numbers  ​​i​​​  and  ​​j​​​  Satisfy   ​​i < j​​​  And there are  ​​(time[i] + time[j]) % 60 == 0​​.

Example 1：

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         Input ：time = [30,20,150,100,40]
        
 Output ：3
        
 explain ： The total duration of these three pairs can be  60  to be divisible by ：
        
(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
       
       

        
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Example 2：

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         Input ：time = [60,60,60]
        
 Output ：3
        
 explain ： The total duration of all three pairs is  120, Can be  60  to be divisible by .
       
       

        
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Tips ：

• • 
    
• ​​1 <= time.length <= 6 * 104​​
• ​​1 <= time[i] <= 500​​
  • 
    

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        /**
        
 * @param {number[]} time
        
 * @return {number}
        
 */
        
 //   solution 1  to flash back 
        
// var numPairsDivisibleBy60 = function(time) {
        
//     //2 Permutation 
        
//     let count = 0;
        
//     let  backTrack = (start, path, time) => {
        
//         if (path.length === 2) {
        
//             if ((path[0] + path[1]) % 60 === 0) {
        
//                 count++
        
//             }
        
//             return;
        
//         }
        
//         for(let i = start; i < time.length; i++) {
        
//             path.push(time[i])
        
//             backTrack(i+1, path, time);
        
//             path.pop()
        
//         }
        
//     }
        
//     backTrack(0, [],time)
        
//     return count
        
// };
        

        
var numPairsDivisibleBy60 = function(time) {
        
    let count = 0;
        
    for(let i = 0; i < time.length; i++) {
        
        for(let j = i+1; j<time.length; j++) {
        
            if ((time[i] + time[j]) % 60 === 0) {
        
                count ++
        
            }
        
        }
        
    }
        
    return count
        
}
       
       

        
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