How do you solve Subarray Sum Equals K on LeetCode?

Subarray Sum Equals K (LeetCode #560) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using cumulative sums allows us to efficiently track the sum of subarrays.

What is the brute force solution for Subarray Sum Equals K?

The brute force approach for Subarray Sum Equals K is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible subarray and calculating its sum. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Subarray Sum Equals K?

Subarray Sum Equals K uses the following concepts: Array, Hash Table, Prefix Sum. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Subarray Sum Equals K?

Always consider edge cases, such as empty arrays or arrays with all negative numbers. Explain your thought process clearly while coding; it helps the interviewer understand your approach. Practice writing clean and efficient code, as readability is important in interviews.

What are common mistakes when solving Subarray Sum Equals K?

Not considering negative numbers in the array, which can affect the cumulative sum. Failing to initialize the HashMap correctly to account for zero-sum subarrays.

#560

Subarray Sum Equals K

Medium

ArrayHash TablePrefix SumHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal solution uses a HashMap to store the cumulative sums and their frequencies. This allows us to efficiently find how many times a subarray sums to k by checking if the difference between the current cumulative sum and k exists in the HashMap.

⚙️

Algorithm

5 steps

1Step 1: Initialize a HashMap to store cumulative sums and a variable to keep track of the current sum.
2Step 2: Initialize a counter to zero to count valid subarrays.
3Step 3: Iterate through the array, updating the current sum at each step.
4Step 4: Check if (current_sum - k) exists in the HashMap. If it does, add its frequency to the counter.
5Step 5: Update the HashMap with the current sum, incrementing its frequency.

solution.py10 lines

1def subarraySum(nums, k):
2    count = 0
3    current_sum = 0
4    sum_map = {0: 1}
5    for num in nums:
6        current_sum += num
7        if (current_sum - k) in sum_map:
8            count += sum_map[current_sum - k]
9        sum_map[current_sum] = sum_map.get(current_sum, 0) + 1
10    return count

ℹ

Complexity note: This complexity arises because we traverse the array once and use a HashMap to store cumulative sums, which allows for constant time lookups.

1Using cumulative sums allows us to efficiently track the sum of subarrays.
2HashMaps can significantly reduce the time complexity by enabling quick lookups.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.