Is Subarray Sum Equals K easy or hard?

Subarray Sum Equals K is classified as a Medium difficulty problem. The brute force solution is simple, but recognizing the prefix sum + hash map optimization requires familiarity with cumulative sum patterns and frequency counting techniques.

Subarray Sum Equals K Python/Java solution

Python and Java implementations both follow the same prefix sum + hash map logic. Maintain a running sum, check map.get(sum - k) to count valid subarrays, and update the map with the current prefix sum frequency. The algorithm runs in O(n) time regardless of language.

How to solve Subarray Sum Equals K in O(n)?

Maintain a running prefix sum while iterating through the array. Store the frequency of every prefix sum in a hash map. For each element, check if prefixSum - k exists in the map; its frequency tells how many subarrays ending at the current index sum to k. Update the map with the current prefix sum after each step.

What is the best approach for Subarray Sum Equals K?

The most efficient approach uses prefix sums with a hash map. As you scan the array, keep a running prefix sum and store how many times each sum appears. If the current sum is curr, any earlier prefix equal to curr - k forms a valid subarray. This method runs in O(n) time with O(n) extra space.

Is Subarray Sum Equals K asked at Google/Amazon/Meta?

Subarray Sum Equals K is a common interview question at companies like Amazon, Google, and Meta because it tests prefix sum reasoning and hash map usage. Variants of this pattern also appear in many array and counting problems during coding interviews.

What data structure is used in Subarray Sum Equals K?

The optimal solution uses a hash map (or dictionary) to store frequencies of prefix sums. Combined with a running prefix sum variable, this allows constant-time lookups to determine whether a previous prefix sum creates a subarray equal to k.

What is the time complexity of Subarray Sum Equals K?

The optimal solution runs in O(n) time using a prefix sum and hash map. Each element is processed once, and hash lookups occur in constant time. A brute force approach that checks all subarrays takes O(n²) time.