LeetCode 1013 - Pairs of Songs With Total Durations Divisible by 60

https://leetcode.com/problems/pairs-of-songs-with-total-durations-divisible-by-60/

In a list of songs, the i-th 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 with (time[i] + time[j]) % 60 == 0.

Example 1:

Input: [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: [60,60,60]
Output: 3
Explanation: All three pairs have a total duration of 120, which is divisible by 60.

Note:

1. 1 <= time.length <= 60000
2. 1 <= time[i] <= 500

https://leetcode.com/problems/pairs-of-songs-with-total-durations-divisible-by-60/discuss/256738/JavaC%2B%2BPython-Two-Sum-with-K-60
In Java you can use Math.floorMod(-t, 60) instead of (60 - t % 60) % 60.

Calculate the time % 60 then it will be exactly same as two sum problem.

Java:

    public int numPairsDivisibleBy60(int[] time) {
        int c[]  = new int[60], res = 0;
        for (int t : time) {
            res += c[(60 - t % 60) % 60];
            c[t % 60] += 1;
        }
        return res;
    }

https://leetcode.com/problems/pairs-of-songs-with-total-durations-divisible-by-60/discuss/256726/Java-O(n)-code-w-comment-similar-to-Two-Sum

Let target in Two Sum be 60 and each item in time % 60, the two problems are very similar to each other.

    public int numPairsDivisibleBy60(int[] time) {
        Map<Integer, Integer> count = new HashMap<>();
        int ans = 0;
        for (int t : time) {
            int d = (60 - t % 60) % 60;
            if (count.containsKey(d)) { ans += count.get(d); } // in current HashMap, get the number of songs that if adding t equals to a multiple of 60.
            count.put(t % 60, 1 + count.getOrDefault(t % 60, 0)); // update the number of t % 60.
        }
        return ans;
    }