<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) {
        vector<int> pre(nums.size(), 0);
        pre[0] = nums[0];
        for(int i=1;i<nums.size();i++){
            pre[i] = nums[i] + pre[i-1];
        }
        unordered_map<int, int> m;
        int count = 0;
        for(int i=0;i<nums.size();i++){
            if(pre[i] == k){
                count++;
            }
            
            int rem = pre[i] - k;
            if(m.find(rem) != m.end()){
                count += m[rem];
            }
            m[pre[i]]++;
        }
        return count;
    }
};
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