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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.

  • 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;
    }
}

############

class Solution {
    public int numPairsDivisibleBy60(int[] time) {
        int[] cnt = new int[501];
        int ans = 0;
        for (int t : time) {
            int s = 60;
            for (int i = 0; i < 17; ++i) {
                if (s - t >= 0 && s - t < cnt.length) {
                    ans += cnt[s - t];
                }
                s += 60;
            }
            cnt[t]++;
        }
        return ans;
    }
}

  • // 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;
    }
};

  • class Solution:
    def numPairsDivisibleBy60(self, time: List[int]) -> int:
        cnt = defaultdict(int)
        ans = 0
        for t in time:
            s = 60
            for _ in range(17):
                ans += cnt[s - t]
                s += 60
            cnt[t] += 1
        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


  • func numPairsDivisibleBy60(time []int) (ans int) {
	cnt := [501]int{}
	for _, t := range time {
		s := 60
		for i := 0; i < 17; i++ {
			if s-t >= 0 && s-t < 501 {
				ans += cnt[s-t]
			}
			s += 60
		}
		cnt[t]++
	}
	return
}

  • function numPairsDivisibleBy60(time: number[]): number {
    const cnt: number[] = new Array(60).fill(0);
    for (const t of time) {
        ++cnt[t % 60];
    }
    let ans = 0;
    for (let x = 1; x < 30; ++x) {
        ans += cnt[x] * cnt[60 - x];
    }
    ans += (cnt[0] * (cnt[0] - 1)) / 2;
    ans += (cnt[30] * (cnt[30] - 1)) / 2;
    return ans;
}

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