1010. Pairs of Songs With Total Durations Divisible by 60

Description

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

Solutions

Solution 1

Python Code
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class Solution:
    def numPairsDivisibleBy60(self, time: List[int]) -> int:
        cnt = Counter(t % 60 for t in time)
        ans = sum(cnt[x] * cnt[60 - x] for x in range(1, 30))
        ans += cnt[0] * (cnt[0] - 1) // 2
        ans += cnt[30] * (cnt[30] - 1) // 2
        return ans

Java Code
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class Solution {
    public int numPairsDivisibleBy60(int[] time) {
        int[] cnt = new int[60];
        for (int t : time) {
            ++cnt[t % 60];
        }
        int ans = 0;
        for (int x = 1; x < 30; ++x) {
            ans += cnt[x] * cnt[60 - x];
        }
        ans += (long) cnt[0] * (cnt[0] - 1) / 2;
        ans += (long) cnt[30] * (cnt[30] - 1) / 2;
        return ans;
    }
}

C++ Code
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class Solution {
public:
    int numPairsDivisibleBy60(vector<int>& time) {
        int cnt[60]{};
        for (int& t : time) {
            ++cnt[t % 60];
        }
        int ans = 0;
        for (int x = 1; x < 30; ++x) {
            ans += cnt[x] * cnt[60 - x];
        }
        ans += 1LL * cnt[0] * (cnt[0] - 1) / 2;
        ans += 1LL * cnt[30] * (cnt[30] - 1) / 2;
        return ans;
    }
};

Go Code
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func numPairsDivisibleBy60(time []int) (ans int) {
	cnt := [60]int{}
	for _, t := range time {
		cnt[t%60]++
	}
	for 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
}

TypeScript Code
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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;
}

Solution 2

Python Code
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class Solution:
    def numPairsDivisibleBy60(self, time: List[int]) -> int:
        cnt = Counter()
        ans = 0
        for x in time:
            x %= 60
            y = (60 - x) % 60
            ans += cnt[y]
            cnt[x] += 1
        return ans

Java Code
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class Solution {
    public int numPairsDivisibleBy60(int[] time) {
        int[] cnt = new int[60];
        int ans = 0;
        for (int x : time) {
            x %= 60;
            int y = (60 - x) % 60;
            ans += cnt[y];
            ++cnt[x];
        }
        return ans;
    }
}

C++ Code
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class Solution {
public:
    int numPairsDivisibleBy60(vector<int>& time) {
        int cnt[60]{};
        int ans = 0;
        for (int x : time) {
            x %= 60;
            int y = (60 - x) % 60;
            ans += cnt[y];
            ++cnt[x];
        }
        return ans;
    }
};

Go Code
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func numPairsDivisibleBy60(time []int) (ans int) {
	cnt := [60]int{}
	for _, x := range time {
		x %= 60
		y := (60 - x) % 60
		ans += cnt[y]
		cnt[x]++
	}
	return
}

TypeScript Code
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function numPairsDivisibleBy60(time: number[]): number {
    const cnt: number[] = new Array(60).fill(0);
    let ans: number = 0;
    for (let x of time) {
        x %= 60;
        const y = (60 - x) % 60;
        ans += cnt[y];
        ++cnt[x];
    }
    return ans;
}