Question

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

Difficulty:Medium

Category:

Analyze

Solution

class Solution {
public:
  int numPairsDivisibleBy60(vector<int>& time) {
    int ans = 0;
    vector<int> vec(61, 0);
    for (int t : time) {
      t %= 60;
      if (t == 0) ans += vec[0];
      else ans += vec[60-t];
      vec[t]++;
    }
    return ans;
  }
};
By guozetang            Updated: 2020-09-19 13:02:30

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