<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 subarraysDivByK(vector<int>& nums, int k) 
    {
        int n = nums.size();
        vector<int> prefixSum(n, 0);
        prefixSum[0] = nums[0] % k;
        if (prefixSum[0] < 0) 
        prefixSum[0] += k; 

        for(int i = 1; i < n; i++)
        {
            prefixSum[i] = (prefixSum[i-1] + nums[i]) % k;
            if (prefixSum[i] < 0) 
                prefixSum[i] += k; 
        }

        unordered_map<int, int> freqMap;
        for(int value : prefixSum)
        {
            freqMap[value]++;
        }

        freqMap[0]++;
        int totalSubarrays = 0;

        for(auto& count : freqMap )
        {
            int freq = count.second;

            if(freq > 1)
            {
                totalSubarrays += (freq*(freq-1))/2;
            }
        }

        return totalSubarrays;

        
    }
};
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