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 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:
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

  • 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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