<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) {
        // brute force, double for loop adding up all possible subarrays (n^2)
        // optimized: lots of recalculation. maintain a prefix sum array. to get subarrays, find if difference of sums at certain indices = k. use 2 pointers Turns into O(n) time, O(N) space
        int count = 0;
        vector<int> prefix(nums.size());
        prefix[0] = nums[0];
        for(int i = 1; i < nums.size(); i++)
        {
            prefix[i] = prefix[i - 1] + nums[i];
        }
        unordered_map<int, int> freq;
        for(int i = 0; i < prefix.size(); i++)
        {
            if(prefix[i] == k) // from 0 to index
                count++;
            if(freq[prefix[i] - k] > 0) // found a subarray sum
                count += freq[prefix[i] - k];
            freq[prefix[i]]++;
        }
        return count;
    }
};

// 1, 3, 6
// 3, 1,2
51903