Subarray Sum Equals K

1. Define the problem

Given an array of integers nums and an integer k, return the total number of contiguous subarrays whose sum equals k . Keep a running sum and a hashmap of how many times each running sum value has occurred. At every index, sum - k already seen tells you exactly how many earlier stops complete a subarray ending here that sums to k.

Constraints

1 ≤ nums.length ≤ 2 × 10⁴
-1000 ≤ nums_i ≤ 1000
-10⁷ ≤ k ≤ 10⁷

Example

Inputnums = [1, 1, 1], k = 2

Output2

Explanation Two subarrays sum to 2: nums[0..1] = [1, 1] and nums[1..2] = [1, 1].

2. Know the words first

In plain terms

Contiguous subarray: A run of consecutive elements taken from the array without skipping any, like nums[2..4].

3. Visualize the solution

Check sum - k against a hashmap of running sums seen so far

Statusinit

sum=1. Need sum-k=-1, not seen. count stays 0. Record sum=1.

What happens in this step

seenCount = {0: 1}   sum = 0   count = 0

nums[0]=1: sum = 0 + 1 = 1
  need sum-k = 1-2 = -1 → seenCount has no key -1 → add 0
  count stays 0
  seenCount[1] = (0 ?? 0) + 1 → seenCount = {0:1, 1:1}

Step 1 of 3

Steps to visualize

Start a hashmap with {0: 1} — the empty prefix has been seen once.
Walk the array, adding each value onto a running sum.
At each index, add seenCount[sum - k] onto the answer, if that key exists.
Increment seenCount_sum to record this running sum for future indices.

4. Walk through the code

Walk through the code

Same walkthrough, now with the code. Press Next to move one step and watch which lines run.

Check sum - k against a hashmap of running sums seen so far

Statusinit

sum=1. Need sum-k=-1, not seen. count stays 0. Record sum=1.

What happens in this step

seenCount = {0: 1}   sum = 0   count = 0

nums[0]=1: sum = 0 + 1 = 1
  need sum-k = 1-2 = -1 → seenCount has no key -1 → add 0
  count stays 0
  seenCount[1] = (0 ?? 0) + 1 → seenCount = {0:1, 1:1}

Step 1 of 3

5. Solution

Solution

solution.tsTypeScript

function subarraySum(nums, k) {
  const seenCount = new Map([[0, 1]]);
  let sum = 0;
  let count = 0;

  for (const value of nums) {
    sum += value;
    count += seenCount.get(sum - k) ?? 0;
    seenCount.set(sum, (seenCount.get(sum) ?? 0) + 1);
  }

  return count;
}

Time: O(n)
Space: O(n)

6. Test cases

Test cases

Input	Expected	Covers
`nums = [1, 1, 1], k = 2`	`2`	example from the docstring
`nums = [1, 2, 3], k = 3`	`2`	a single-element match and a two-element match
`nums = [1, -1, 0], k = 0`	`3`	negative values producing multiple zero-sum subarrays
`nums = [1, 2, 3], k = 7`	`0`	target impossible to reach, no subarray matches
`nums = [5], k = 5`	`1`	smallest valid input, one element equal to k
`nums = [-1, -1, 1], k = 0`	`1`	all-negative values with exactly one zero-sum subarray

Subarray Sum Equals K