Published: Oct 26, 2022

## Introduction

The problem gives us an array and asks about sum which is k. This signals the prefix sum and hash table. It is a kind of 2-sum problem. That’s why the hash table works. The keys of the hash table are accumulated sums. While going over the given array, if accumulated sum minus k exists in the table, the sum equals to k is found.

## Problem Description

Given an array of integers

`nums`

and an integer`k`

, return the total number of subarrays whose sum equals to`k`

.A subarray is a contiguous non-empty sequence of elements within an array.

Constraints:

`1 <= nums.length <= 2 * 10**4`

`-1000 <= nums[i] <= 1000`

`-10**7 <= k <= 10**7`

## Examples

```
Example 1
Input: nums = [1,1,1], k = 2
Output: 2
```

```
Example 2
Input: nums = [1,2,3], k = 3
Output: 2
```

## Analysis

The solution here starts from initializing the accumulated value and hash table. The value of the hash table is a number of times the accumulated value appeared. So, the initial value is sum 0 and count 1. While checking the array value one by one, find accumulated value minus k. Like 2-sum problem, the accumulated sum minus k exists in the hash table, add up the count. Then, count up at the current accumulated sum.

## Solution

```
class SubarraySumEqualsK:
def subarraySum(self, nums: List[int], k: int) -> int:
acc, memo = 0, {0: 1}
count = 0
for v in nums:
acc += v
diff = acc - k
if diff in memo:
count += memo[diff]
if acc in memo:
memo[acc] += 1
else:
memo[acc] = 1
return count
```

## Complexities

- Time:
`O(n)`

- Space:
`O(n)`