<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) {
        int n = nums.size();
        vector<int> left(n, 0);

        for (int i=1; i<n; i++) {
            left[i] = left[i-1]+nums[i-1];
        }
        vector<int> right(n, 0);

        for (int i=n-2; i>-1; i--) {
            right[i] = right[i+1]+nums[i+1];
        }

        int total = nums[0]+right[0];
        int res = 0;

        for (int start=0; start<n; start++) {
            for (int end=start; end<n; end++) {
                if (total - left[start]-right[end] == k)
                    res++;
            }
        }

        return res;
    }
};
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