[Swift]LeetCode1010. 总持续时间可被 60 整除的歌曲 | Pairs of Songs With Total Durations Divisible by 60

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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

---

在歌曲列表中，第 i 首歌曲的持续时间为 time[i] 秒。

返回其总持续时间（以秒为单位）可被 60 整除的歌曲对的数量。形式上，我们希望索引的数字  i < j 且有 (time[i] + time[j]) % 60 == 0。 

示例 1：

输入：[30,20,150,100,40]
输出：3
解释：这三对的总持续时间可被 60 整数：
(time[0] = 30, time[2] = 150): 总持续时间 180
(time[1] = 20, time[3] = 100): 总持续时间 120
(time[1] = 20, time[4] = 40): 总持续时间 60

示例 2：

输入：[60,60,60]
输出：3
解释：所有三对的总持续时间都是 120，可以被 60 整数。 

提示：

1. 1 <= time.length <= 60000
2. 1 <= time[i] <= 500

---

 212ms

class Solution {
    func numPairsDivisibleBy60(_ time: [Int]) -> Int {
        var res = 0
        var rems: [Int] = Array(repeating: 0, count: 60)
        for t in time {
            rems[t % 60] += 1
        }
        for i in (0..<rems.count - 1) {
            for j in  (i..<rems.count) {
                if (i + j) % 60 == 0 {
                    if i == j {
                       res = res + ((rems[i] * (rems[i] - 1)) / 2)
                    } else if rems[i] != 0 && rems[j] != 0 {
                        res += (rems[i] * rems[j])
                    }
                 }
            }
        }
        return res
    }
}

---

228ms

class Solution {
    func numPairsDivisibleBy60(_ time: [Int]) -> Int {
        if time.count <= 1 { return 0 }
        let count = time.count
        var rs = [Int](repeating: 0, count: count)
        var map = [Int: Int]()
        var pair = 0
        for i in 0..<count {
            let r = time[i] % 60
            if let c = map[r] {
                map[r] = c + 1
            } else {
                map[r] = 1
            }
        }
        for i in 1..<30 {
            guard let c1 = map[i], let c2 = map[60 - i] else { continue }
            pair += c1 * c2
        }
        if let c = map[0], c > 1 {
            pair += (c * (c - 1)) / 2
        }
        if let c = map[30], c > 1 {
            pair += (c * (c - 1)) / 2
        }
        return pair
    }
}

---

Runtime: 232 ms

Memory Usage: 19.4 MB

class Solution {
    func numPairsDivisibleBy60(_ time: [Int]) -> Int {
        var n:Int = time.count
        var cnt:[Int] = [Int](repeating:0,count:60)
        var ans:Int = 0
        for i in 0..<n
        {
            var t:Int = time[i] % 60
            ans += cnt[(60 - t) % 60]
            cnt[time[i] % 60] += 1
        }
        return ans
        
    }
}

---

 

236ms 

class Solution {
    func numPairsDivisibleBy60(_ time: [Int]) -> Int {
    if time.count == 0 {
        return 0
    }

    var tempDic = [Int : Int]()

    for temp in time {
        let v = temp % 60
        let number = tempDic[v] ?? 0
        tempDic[v] = number + 1
    }

    var total = 0
    
    for (key, value) in tempDic {
        
        if key == 30 || key == 0 {
            if value > 1 {
                let temp = value * (value - 1) / 2
                total += temp
            }
        }
        else {
            let v1 = 60 - key
            if let cache = tempDic[v1] {
                total += value * cache
                tempDic[v1] = nil
            }
        }
        tempDic[key] = nil
    }
    return total
   }
}

---

252ms

class Solution {
    func numPairsDivisibleBy60(_ time: [Int]) -> Int {
        var bucket = Array(repeating: 0, count: 60)
        
        for time in time {
            bucket[time % 60] += 1
        }
        
        var result = 0
        for i in 1...29 {
            result += bucket[i] * bucket[60 - i]
        }
        
        if bucket[0] >= 2 {
            result += bucket[0] * (bucket[0] - 1) / 2
        }
        if bucket[30] >= 2 {
            result += bucket[30] * (bucket[30] - 1) / 2
        }
        
        return result
    }
}

---

256ms

class Solution {
    func numPairsDivisibleBy60(_ time: [Int]) -> Int {
        var res = 0
        var val = [Int: Int]()
        
        for t in time {
            let a = t % 60
            let sup = (60 - a) % 60
            res += val[a, default: 0]
            if let v = val[sup] {
                val[sup] = v + 1
            } else {
                val[sup] = 1
            }
        }
        return res
    }
}