<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:
    def subarraySum(self, nums: List[int], k: int) -> int:
        prefix_sums = {0: 1}
        prefix_sum = count = 0

        # for each item, calculate the prefix sum. if the difference
        # between the prefix sum and k exists, update the count
        # otherwise add it to the prefix sums map
        for n in nums:
            prefix_sum += n
            diff = prefix_sum - k
            if diff in prefix_sums:
                count += prefix_sums[diff] 
            prefix_sums[prefix_sum] = prefix_sums.get(prefix_sum, 0) + 1

        return count
19600