974. Subarray Sums Divisible by K
Given an array nums of integers, return the number of (contiguous, non-empty) subarrays that have a sum divisible by k.
Example 1:
Input: nums = [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 <= nums.length <= 30000
- -10000 <= nums[i] <= 10000
- 2 <= k <= 10000
Solution
class Solution {
public int subarraysDivByK(int[] nums, int k) {
Map<Integer, Integer> prefix = new HashMap<>();
prefix.put(0, 1);
int sum = 0;
int count = 0;
for (int i = 0; i < nums.length; i++) {
sum += nums[i];
sum = sum % k;
if (sum < 0) {
sum += k;
}
count += prefix.getOrDefault(sum, 0);
prefix.put(sum, prefix.getOrDefault(sum, 0) + 1);
}
return count;
}
}