<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:
        '''
        1. Prefix Sum 
        # [0,1,3,6]
        2. DP, store previous sums and count
        3. Insert after counters
        4. 0 equals 1
        '''

        dp = defaultdict(int) # [1,1,1]
        dp[0] = 1
        '''
        0 : 1
        1 : 1
        2 : 1
        '''
        prefixSum = 0 # 2
        result = 0 # 1

        for num in nums:
            prefixSum += num
            result += dp[prefixSum - k]
            dp[prefixSum] += 1
        
        return result


        
20400