<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) {

        long long int total = 0;

        // Create an unordered map to store prefix sums and their frequencies.
        unordered_map<long long int, long long int> m;

        // Initialize prefix_sum to 0 to store the running sum.
        long long int prefix_sum = 0;
        m[0] = 1;  // To handle cases where the prefix_sum equals k directly.
        
        for (int i = 0; i < nums.size(); i++) {
            // Add the current element to the prefix sum.
            prefix_sum += nums[i];
            
            // If the difference (prefix_sum - k) is found in the map,
            // it means there is a subarray that sums to k.
            if (m.count(prefix_sum - k)) {
                total += m[prefix_sum - k];  // Increment total by the count of this prefix_sum.
            }
            
            // Increment the count of this prefix_sum in the map.
            m[prefix_sum]++;
        }

        // Return the total count of subarrays with sum equal to k.
        return total;
    }
};

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