How do you solve Find Subarrays With Equal Sum on LeetCode?

Find Subarrays With Equal Sum (LeetCode #2395) 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 a hash set allows for efficient duplicate checking.

What is the brute force solution for Find Subarrays With Equal Sum?

The brute force approach for Find Subarrays With Equal Sum is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible pair of subarrays of length 2 to see if they have the same 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 Find Subarrays With Equal Sum?

Find Subarrays With Equal Sum uses the following concepts: Array, Hash Table. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find Subarrays With Equal Sum?

Always clarify the problem constraints and requirements. Consider edge cases and test your solution with them. Explain your thought process clearly while coding.

What are common mistakes when solving Find Subarrays With Equal Sum?

Not considering the requirement for different starting indices. Overlooking edge cases with negative numbers.

#2395

Find Subarrays With Equal Sum

Easy

ArrayHash TableHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Using a hash set allows us to store the sums of subarrays of length 2 and check for duplicates efficiently. This reduces the time complexity significantly.

⚙️

Algorithm

6 steps

1Step 1: Initialize an empty hash set to store the sums of subarrays.
2Step 2: Iterate through the array from index 0 to n-2.
3Step 3: For each index i, calculate the sum of the subarray nums[i] + nums[i+1].
4Step 4: Check if this sum already exists in the hash set. If it does, return true.
5Step 5: If not, add the sum to the hash set and continue.
6Step 6: If no duplicates are found after the loop, return false.

solution.py11 lines

1# Full working Python code
2nums = [4, 2, 4]
3def hasEqualSumSubarrays(nums):
4    seen = set()
5    for i in range(len(nums) - 1):
6        sum_subarray = nums[i] + nums[i + 1]
7        if sum_subarray in seen:
8            return True
9        seen.add(sum_subarray)
10    return False
11print(hasEqualSumSubarrays(nums))

ℹ

Complexity note: The time complexity is O(n) because we only loop through the array once, calculating sums and checking the hash set in constant time. The space complexity is O(n) due to the storage of sums in the hash set.

1Using a hash set allows for efficient duplicate checking.
2Subarrays must start at different indices, which is crucial.

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