Subarray Sums Divisible by K LT974

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

Idea 1.  prefix sums + HashMap + modular rules, note: count[0] = 1, (X + X%K)%K for negative values

Time complexity: O(n)

Space complexity: O(K)

 1 class Solution {
 2     public int subarraysDivByK(int[] A, int K) {
 3        int[] count = new int[K];
 4        count[0] = 1;
 5        int prefixSum = 0;
 6         
 7        int result = 0;
 8        for(int a: A) {
 9            prefixSum += a;
10            prefixSum = (prefixSum % K + K)%K;
11            result += count[prefixSum];
12            ++count[prefixSum];
13        }
14         
15         return result;
16     }
17 }

Idea 1.a count pairs

 1 class Solution {
 2     public int subarraysDivByK(int[] A, int K) {
 3        int[] count = new int[K];
 4        count[0] = 1;
 5        int prefixSum = 0;
 6         
 7        for(int a: A) {
 8            prefixSum += a;
 9            prefixSum = (prefixSum % K + K)%K;
10            ++count[prefixSum];
11        }
12         
13        int result = 0;
14        for(int val: count) {
15            result += val*(val-1)/2;
16        }
17         
18        return result;
19     }
20 }