<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) {
     // [1,1,1]
    //  if(nums.size()==1){
    //     return 1;
    //  }
     map<int,int>mp;
     vector<int>pref(nums.size());
     pref[0]=nums[0];
     int total=0;
      mp[0]=1;
       if (mp.find(pref[0] - k) != mp.end()) {
            total += mp[pref[0] - k]; // Add the count of subarrays that sum to k
        }

     mp[pref[0]]++;
    

     for(int i=1;i<nums.size();i++){
         pref[i]=pref[i-1]+nums[i];
        if(mp.find(pref[i]-k)!=mp.end()){
            total +=mp[pref[i]-k];
        }
               mp[pref[i]]++;
        }

return total;
     
    }   
};
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