**Problem Statement:**

Given an integer array nums and an integer k, return the number of non-empty subarrays that have a sum divisible by k.

A subarray is a contiguous part of an array.

**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]

**Example 2:**

**Input:** nums = [5], k = 9

**Output:** 0

**Constraints:**

- 1 <= nums.length <= 3 * 104
- -104 <= nums[i] <= 104
- 2 <= k <= 104

**Solution:**

**Algorithm:**

- Create a count array of size k and initialize all elements as 0.
- Initialize sum of elements as 0 and count of subarrays as 0.
- Iterate through the array and for every element arr[i], do following. a) Increment sum by arr[i]. b) If k is non-zero, then update sum as sum = sum % k. c) Increment count of current sum. d) Add count[sub_sum] to the result. e) Increment count[sub_sum] by 1.
- Return result.

**Code:**

```
public class Solution {
public int subarrayDivByK(int[] nums, int k){
int[] count = new int[k];
count[0] = 1;
int sum = 0;
int ans = 0;
for(int i = 0; i < nums.length; i++){
sum += nums[i];
int mod = (sum % k + k) % k;
ans += count[mod];
count[mod]++;
}
return ans;
}
}
```

**Time Complexity:**

O(N)

**Space Complexity:**

O(K)

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