<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:

        # approach
        # brute force - O(n**2)
        # vanilla prefix sum - O(n**2)
        # prefix sum + hash map - O(n) -> struggled a lot -> look at the solution run it with print statements in playground to develop intuition
        # read - https://leetcode.com/explore/interview/card/leetcodes-interview-crash-course-data-structures-and-algorithms/705/hashing/4512/
        # why hash map was used instead of set?

        # # brute force O(n**2) -> tle 83/93 passed
        # res = 0

        # for i in range(len(nums)):
        #     curr = 0

        #     for j in range(i, len(nums)):
        #         curr += nums[j]

        #         if curr == k:
        #             res += 1
        # return res

        # prefix_hash = collections.defaultdict(int) # for nums = [] empty array -> sum of 0 occurs once
        # prefix_hash[0] = 1
        # curr = res = 0

        # for num in nums:
        #     curr += num
            
        #     res += prefix_hash[curr - k]
        #     prefix_hash[curr] += 1
        
        # return res

        '''
        redo
        '''
        
        prefix_sum = collections.defaultdict(int)
        prefix_sum[0] = 1
        curr = res = 0

        for num in nums:
            curr += num
            res += prefix_sum[curr - k]
            prefix_sum[curr] += 1
        
        return res

            


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