How do you solve Left and Right Sum Differences on LeetCode?

Left and Right Sum Differences (LeetCode #2574) 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: Understanding the relationship between left and right sums is crucial for optimizing the solution.

What is the brute force solution for Left and Right Sum Differences?

The brute force approach for Left and Right Sum Differences is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we calculate the left and right sums for each index by iterating through the array multiple times. 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 Left and Right Sum Differences?

Left and Right Sum Differences uses the following concepts: Array, Prefix Sum. The recommended patterns to study are: Prefix Sum, Array.

What are the interview tips for Left and Right Sum Differences?

Always consider the brute force approach first to understand the problem deeply. Look for patterns that can help reduce time complexity, such as maintaining running totals. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Left and Right Sum Differences?

Forgetting to handle edge cases, such as arrays of size 1. Not correctly updating leftSum or rightSum during iterations.

#2574

Left and Right Sum Differences

Easy

ArrayPrefix SumPrefix SumArray

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)

The optimal solution uses a single pass to calculate the total sum of the array and then derives left and right sums dynamically, which significantly reduces the number of operations.

⚙️

Algorithm

6 steps

1Step 1: Calculate the total sum of the array 'nums'.
2Step 2: Initialize leftSum to 0 and create an empty array 'answer'.
3Step 3: Iterate through each index 'i' in 'nums'. For each index, update rightSum as totalSum - leftSum - nums[i].
4Step 4: Calculate the absolute difference |leftSum - rightSum| and store it in 'answer[i]'.
5Step 5: Update leftSum by adding nums[i] for the next iteration.
6Step 6: Return the 'answer' array.

solution.py14 lines

1# Full working Python code
2
3def left_right_sum_difference(nums):
4    totalSum = sum(nums)
5    leftSum = 0
6    answer = []
7    for num in nums:
8        rightSum = totalSum - leftSum - num
9        answer.append(abs(leftSum - rightSum))
10        leftSum += num
11    return answer
12
13# Example usage
14print(left_right_sum_difference([10, 4, 8, 3]))

ℹ

Complexity note: The time complexity is O(n) because we traverse the array a constant number of times, and the space complexity is O(n) due to the output array.

1Understanding the relationship between left and right sums is crucial for optimizing the solution.
2Using a single pass to maintain running totals can greatly improve efficiency.

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