Subarray Sums Divisible by K

update Feb 15 2019, 15:30

LeetCode

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

Basic Idea:

既然是subarray sum的问题，仍然要从前缀和的角度考虑。题目所求的是sum为k的倍数的subarray的个数，brute force是 O(N^2)，所以我们的目标是 O(N) 时间。

我们发现对于当前preSum，如果有 currPreSum - prevPreSum == c * K, 则之间的部分sum就是K的倍数。进一步，如果有 currPreSum % K == prevPreSum % K, 这和前面的条件等价。

于是我们就有了一个解法，用一个长度为K的count数组记录之前出现的preSum % K的值的个数。每次计算出新的preSum之后，count数组中对应相同余数出现过的个数就是到当前num结尾的符合条件subarray的个数。然后更新count数组，加一。

Java Code:

注意Java中mod会产生负数，需要特别处理。

class Solution {
    public int subarraysDivByK(int[] A, int K) {
        int[] count = new int[K];
        count[0] = 1;
        int preSum = 0, ret = 0;
        for (int num : A) {
            preSum += num;
            int remainder = preSum % K;
            if (remainder < 0) remainder += K;
            ret += count[remainder];
            count[remainder]++;
        }
        return ret;
    }
}