<p>Given an integer array <code>nums</code> and an integer <code>k</code>, return <em>the number of non-empty <strong>subarrays</strong> that have a sum divisible by </em><code>k</code>.</p>

<p>A <strong>subarray</strong> is a <strong>contiguous</strong> part of an array.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [4,5,0,-2,-3,1], k = 5
<strong>Output:</strong> 7
<strong>Explanation:</strong> 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]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [5], k = 9
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
	<li><code>2 &lt;= k &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    int positiveMod(int a, int k) {
        auto res = a%k;
        return res < 0 ? res + k : res;
    }

    int subarraysDivByK(vector<int>& nums, int k) {
        // sum array a = sum array b - sum array c
        // mod (som array a, k) = mod(sum array b - sum array c, k)
        // mod (sum array a, k) = (mod(sum array b, k) - mod(sum array c, k))mod k
        // mod (sum array a, k) == 0 ==> mod(sum arrayb, k) == mod(sum array c, k)
        map<int, int> modCount;
        int sum = 0;
        int count = 0;
        for(int num : nums) {
            sum += num;
            auto mod = positiveMod(sum, k);
            count += modCount[mod];
            ++modCount[mod];
            if(mod == 0) {
                ++count;
            }            
        }

        return count;
    }
};
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