<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>& arr, int k) {
        
        // count  = 0
        // 1 1 1 1 1 1 1
        // 1 2 3
        // sorted atleast
        int n=arr.size();
        vector<int> prefix(n,0);
        int count =0;

        prefix[0]=arr[0];
        for(int i=1; i< arr.size(); i++){
            prefix[i]=prefix[i-1]+arr[i];
        }

        
        unordered_map<int , int > mp;
        mp[0]=1;
        for(int j=0; j<arr.size();j++){
            //     1 3 6      3
            //      i
            // 0 2  2 1                                               
            //                     
            if(mp.count(prefix[j]-k)){
                count=count+ mp[prefix[j]-k];
                
            // mp[prefix[j]-k]++;
            }
            mp[prefix[j]]++;
            
        }
        return count;
    }
};
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