<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();
        int count = 0;
        vector<int> prefixSum(n + 1, 0); // Initialize prefix sum array

        // Calculate prefix sum
        for (int i = 0; i < n; ++i) {
            prefixSum[i + 1] = prefixSum[i] + nums[i];
        }

        // Create a frequency array for remainders
        vector<int> remainderFreq(k, 0);
        remainderFreq[0] = 1; // Initialize with 1 to account for subarrays starting at index 0

        // Calculate subarray sums and update the remainder frequency
        for (int i = 1; i <= n; ++i) {
            int remainder = (prefixSum[i] % k + k) % k; // Handle negative sums
            count += remainderFreq[remainder];
            remainderFreq[remainder]++;
        }

        return count;
        // int n=nums.size();
        // int count=0,sum;
        // vector<int> Sub;
        // for(int i=0;i<n;i++){
        //     for(int j=i+1;j<n;j++){
        //          sum=0;
        //         for(int k=i;k<=j;k++){
        //             sum+=Sub[k];
        //         }
        //         if(sum % k==0 && sum>=0)
        //         count++;
        //     }
        // }
        // return count;
    }
};
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