974. Subarray Sums Divisible by K

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

類似於Subarray Sum equal to K的題目，可以用preSum的概念套用在這，假設[0, i]的和取mod K的結果和[0, j]的和取mod K的結果一樣，那代表[i + 1, j]的和可以被K整除。所以用一個unordered_map來記錄mod對應到的出現次數。

// Similar to PreSum Method with unordered_map
int subarraysDivByK(vector<int>& A, int K) { // time: O(n); space: O(K)
    unordered_map<int, int> preMod; // mod -> count
    preMod[0] = 1;
    int res = 0, sum = 0;
    for (int num : A) {
        sum = (sum + num) % K;
        if (sum < 0) sum += K;
        res += preMod[sum]++;
    }
    return res;
}

由於mod只可能有K種可能，所以可把unordered_map換成array來代替。

// Similar to PreSum Method with array
int subarraysDivByK(vector<int>& A, int K) { // time: O(n); space: O(K)
    vector<int> preMod(K, 0); // mod -> count
    preMod[0] = 1;
    int res = 0, sum = 0;
    for (int num : A) {
        sum = (sum + num) % K;
        if (sum < 0) sum += K;
        res += preMod[sum]++;
    }
    return res;
}