<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>

// Method 1 -> Brute Force

// class Solution {
// public:
//     int subarraySum(vector<int>& nums, int k) {
        
//         int count = 0;

//         int n = nums.size();

//         for(int i = 0; i < n; i++){
            
//             int sum = 0;

//             for(int j = i; j < n; j++){

//                 sum += nums[j];

//                 if(sum == k) count++;
//             }
//         }

//         return count;
//     }
// };

// sliding window solution will not work because there are negative numbers in the subarray also.

// Method 2 -> Prefix sum

class Solution {
public:
    int subarraySum(vector<int>& nums, int k) {

        unordered_map<int,int> mp;
        mp[0] = 1;

        int sum = 0;
        int count = 0;

        for(int i = 0; i < nums.size(); i++){
            
            sum += nums[i];

            if(mp.find(sum-k) != mp.end()){
                count += mp[sum-k];
            }
            // adding it after increasing count is important to handle the case of k == 0. think?
            mp[sum]++;
        }

        return count;
    }
};

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