<p>Given an array of integers <code>nums</code> and an integer <code>k</code>, return <em>the total number of subarrays whose sum equals to</em> <code>k</code>.</p>

<p>A subarray is a contiguous <strong>non-empty</strong> sequence of elements within an array.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> nums = [1,1,1], k = 2
<strong>Output:</strong> 2
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> nums = [1,2,3], k = 3
<strong>Output:</strong> 2
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>-1000 &lt;= nums[i] &lt;= 1000</code></li>
	<li><code>-10<sup>7</sup> &lt;= k &lt;= 10<sup>7</sup></code></li>
</ul>

class Solution {
public:
    int subarraySum(vector<int>& nums, int k) 
    {
        // use prefix sum to calculate sum(i,j)= prefsum(j-1)- prefsum(i);
        
        vector<long> prefsum;

        int sum=0;
        prefsum.push_back(0);
        for(int i=0; i<nums.size(); i++)
        {
            sum+= nums[i];
            prefsum.push_back(sum);
        }

        // for(auto h: prefsum) cout<<h<<" ";
        // cout<<endl;

        int count=0;
        for(int j=0; j<prefsum.size(); j++)
        {
            for(int i=0; i<j; i++)
            {
                if(prefsum[j]- prefsum[i]==k)count++;
            }
        }
        return count;
        
    }
};
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