Formatted question description: https://leetcode.ca/all/1010.html

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

You are given a list of songs where 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 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:
Goldman Sachs

Related Topics:
Array

Solution 1.

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

Java

• Java
• C++
• Python

• class Solution {
      public int numPairsDivisibleBy60(int[] time) {
          int length = time.length;
          int[] remainder = new int[length];
          for (int i = 0; i < length; i++)
              remainder[i] = time[i] % 60;
          Arrays.sort(remainder);
          int beginIndex = 0;
          while (beginIndex < length && remainder[beginIndex] == 0)
              beginIndex++;
          int pairs = beginIndex * (beginIndex - 1) / 2;
          int low = beginIndex, high = length - 1;
          while (low < high) {
              int sum = remainder[low] + remainder[high];
              if (sum < 60)
                  low++;
              else if (sum > 60)
                  high--;
              else {
                  if (remainder[low] == 30) {
                      int count = high - low + 1;
                      pairs += count * (count - 1) / 2;
                      break;
                  } else {
                      int lowEnd = low, highEnd = high;
                      int remainderLow = remainder[low], remainderHigh = remainder[high];
                      while (lowEnd < high && remainder[lowEnd] == remainderLow)
                          lowEnd++;
                      while (highEnd > low && remainder[highEnd] == remainderHigh)
                          highEnd--;
                      pairs += (lowEnd - low) * (high - highEnd);
                      low = lowEnd;
                      high = highEnd;
                  }
              }
          }
          return pairs;
      }
  }
  

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

• # 1010. Pairs of Songs With Total Durations Divisible by 60
  # https://leetcode.com/problems/pairs-of-songs-with-total-durations-divisible-by-60/
  
  class Solution:
      def numPairsDivisibleBy60(self, time: List[int]) -> int:
          s = res = 0
          mp = Counter()
          
          for x in time:
              target = (60 - x) % 60
              
              res += mp[target]
              
              mp[x % 60] += 1
          
          return res
  
  

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