<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, arr: List[int], k: int) -> int:
        d = {0: 1}  # Initialize with 0 sum occurring once
        s = 0  # Running sum
        c = 0  # Count of subarrays
        
        for i in range(len(arr)):
            s += arr[i]  # Add current element to the running sum
            
            # Check if there's a prefix sum that makes s - k
            if s - k in d:
                c += d[s - k]  # Add the number of times (s - k) has occurred
            
            # Add the current sum to the dictionary or increment its count
            if s in d:
                d[s] += 1
            else:
                d[s] = 1
        
        return c

19300