1010

1010. Pairs of Songs With Total Durations Divisible by 60 (Medium)

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.

 

Example 1:

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

 

Constraints:

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

Companies:
Amazon, Citadel, ServiceNow, Bloomberg, Cisco, VMware, Mathworks, Salesforce, Twilio

Related Topics:
Array, Hash Table, Counting

Solution 1. Prefix State Map

// OJ: https://leetcode.com/problems/pairs-of-songs-with-total-durations-divisible-by-60/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int numPairsDivisibleBy60(vector<int>& A) {
        int cnt[60] = {}, ans = 0;
        for (int n : A) {
            n %= 60;
            ans += cnt[(60 - n) % 60];
            cnt[n]++;
        }
        return ans;
    }
};