# A solution to LeetCode Problem 560. Subarray Sum Equals K in JavaScript

### Problem Statement:

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.

Example 1:

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

Example 2:

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

Constraints:

`1 <= nums.length <= 2 * 104`

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

`-107 <= k <= 107`

### Solution:

**use a brute force method**, where we check every possible subarray and see if its sum is equal to k. However, this approach has a

**time complexity of O(n^2),**which is not efficient for large arrays.

**prefix sum array**, where we

**store the sum of the elements from the start of the array up to the current index**. We can then use this prefix sum array to find the sum of any subarray in constant time.

`function subarraySum(nums, k) { let count = 0; let sum = 0; const map = new Map(); map.set(0, 1); for (let i = 0; i < nums.length; i++) { sum += nums[i]; if (map.has(sum - k)) { count += map.get(sum - k); } map.set(sum, (map.get(sum) || 0) + 1); } return count; }`

In the above code, we use a map to store the frequency of each prefix sum. We also initialize the map with a key of 0 and a value of 1, as there is always at least one subarray with a sum of 0 (the empty subarray).

As we iterate through the array, we add the current element to the sum and check if there is a previous element such that the difference between the two elements is equal to k. If we find such an element, we increment the count by the frequency of that element in the map. We also update the frequency of the current sum in the map.

#### This solution has a time complexity of O(n) and a space complexity of O(n), making it more efficient than the brute force method.

I hope this blog post was helpful in understanding how to solve the LeetCode Problem 560: Subarray Sum Equals K in JavaScript. If you have any questions or comments, feel free to leave them below.