[Swift]LeetCode974. 和可被 K 整除的子数组 | Subarray Sums Divisible by K

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Given an array A of integers, return the number of (contiguous, non-empty) subarrays that have a sum divisible by K.

Example 1:

Input: A = [4,5,0,-2,-3,1], K = 5
Output: 7
Explanation: There are 7 subarrays with a sum divisible by K = 5:
[4, 5, 0, -2, -3, 1], [5], [5, 0], [5, 0, -2, -3], [0], [0, -2, -3], [-2, -3] 

Note:

1. 1 <= A.length <= 30000
2. -10000 <= A[i] <= 10000
3. 2 <= K <= 10000

---

给定一个整数数组 A，返回其中元素之和可被 K 整除的（连续、非空）子数组的数目。 

示例：

输入：A = [4,5,0,-2,-3,1], K = 5
输出：7
解释：
有 7 个子数组满足其元素之和可被 K = 5 整除：
[4, 5, 0, -2, -3, 1], [5], [5, 0], [5, 0, -2, -3], [0], [0, -2, -3], [-2, -3] 

提示：

1. 1 <= A.length <= 30000
2. -10000 <= A[i] <= 10000
3. 2 <= K <= 10000

---

 368ms 

class Solution {
    func subarraysDivByK(_ A: [Int], _ K: Int) -> Int {
        var n = A.count
        var f:[Int] = [Int](repeating:0,count:K)
        f[0] = 1
        var x:Int = 0 
        for v in A
        {
            x += v
            x %= K
            if x < 0
            {
                x += K
            }
            f[x] += 1
        }
        var ret:Int = 0
        for i in  0..<K
        {
            ret += f[i]*(f[i]-1)/2
        }
        return ret
    }
}

---

372ms

class Solution {
    func subarraysDivByK(_ A: [Int], _ K: Int) -> Int {
     
        var res = 0
        var prefixArr = Array(repeating:0, count:K)
        var prefix = 0
        prefixArr[0] = 1
        
        for num in A {
            prefix = (prefix + num % K + K) % K
            res += prefixArr[prefix]
            prefixArr[prefix] += 1
        }
        
        return res
    }
}

---

392ms

class Solution {
    
    func subarraysDivByK(_ values: [Int], _ k: Int) -> Int {
        var reminderCount = [Int: Int]() 
        var sum = 0

        for i in 0..<values.count {
            sum += values[i]
            let reminder = (sum % k + k) % k
            reminderCount[reminder, default: 0] += 1
        }

        var subArraysCount = 0
        subArraysCount += reminderCount[0] ?? 0

        for key in reminderCount.keys {
            if let count = reminderCount[key], count > 0 {
                subArraysCount += (count * (count - 1))/2
            }
        }

        return subArraysCount
    }    
}

---

400ms

class Solution {
    func subarraysDivByK(_ A: [Int], _ K: Int) -> Int {
        let N = A.count
        var P = [Int](repeating: 0, count: N+1)
        for i in 0..<N {
            P[i+1] = P[i] + A[i]
        }

        var count = [Int](repeating: 0, count: K)
        for x in P {
            count[(x%K + K)%K] = count[(x%K + K)%K] + 1
        }

        var ans = 0;
        for v in count {
            ans += v*(v-1)/2
        }
        return ans
    }
}

---

404ms

class Solution {
    func subarraysDivByK(_ A: [Int], _ K: Int) -> Int {
        var mod = [Int:Int]()
        var result = 0
        var sum = 0
        mod[0] = 1
        for num in A{
            sum = (sum + num)%K
            if sum < 0{
                sum = sum + K
            }
            
            if let value = mod[sum]{
                result += value
                mod[sum] = value + 1
            }else{
                mod[sum] = 1
            }
        }
        
        return result
    }
}

---

408ms

class Solution {
    func subarraysDivByK(_ A: [Int], _ K: Int) -> Int {
        var sums: [Int] = Array(repeating: 0, count: K)
        var sum: Int = 0
        for (index, a) in A.enumerated() {
            sum += a
            sum %= K
            if sum < 0 { sum += K }
            sums[sum] += 1
        }
        var res = sums[0]
        for s in sums {
            if s > 1 {
                res += s*(s-1)/2
            }
        }
        
        return res
    }
}

---

452ms

class Solution {
    func subarraysDivByK(_ A: [Int], _ K: Int) -> Int {
        var counts = [0:1]
        
        var current = 0
        var result = 0
        for a in A {
            current = (current + a) % K
            while current < 0 {
                current += K
            }
            result += counts[current, default:0]
            counts[current] = counts[current, default:0] + 1
        }
        return result
    }
}