<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) {
        unordered_map<long long, int> prefixSumCounts;
        prefixSumCounts[0] = 1; // Base case: prefix sum of 0 occurs once

        long long current_sum = 0;
        int result = 0;

        for (int num : nums) {
            current_sum += num;

            // Check if there is a prefix sum that when removed leaves k
            long long needed_sum = current_sum - (long long)k;
            if (prefixSumCounts.find(needed_sum) != prefixSumCounts.end()) {
                result += prefixSumCounts[needed_sum];
            }

            // Update the map with the current prefix sum
            prefixSumCounts[current_sum]++;
        }

        return result;
    }
};
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