<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>

/*
// target = sum - k
m[{1, 1}, {2, 1}]

I know that contiguous subarray sums can be calculated with difference of sums.

i=0, sum = 1, target =-1, answer = 0;
i=1, sum = 2, target = 0, answer = 1;
i=2, sum = 3, target = 1 answer = 0;
*/


class Solution {
public:
    int subarraySum(vector<int>& nums, int k) {
        int n = nums.size(), answer = 0, sum = 0;
        std::unordered_map<int, int> m;
        m.reserve(n*3);
        m[0] = 1;
        for(int i = 0; i < n; i++) {
            sum += nums[i];
            answer += m[sum - k];
            m[sum]++;
        }
        return answer;
    }
};
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