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

For each index, loop over the left side and the right side with time complexity O(N²) and space complexity O(1).

What is the optimal approach and complexity for Left and Right Sum Differences?

One running left sum; right = total − left − nums[i] with optimal time complexity O(N) and space complexity O(1).

How do you identify the pattern for Left and Right Sum Differences?

Look for key signals in the prompt: "sum of the numbers before i" indicates Prefix sum ending just before i — a running total from the left.; "sum of the numbers after i" indicates Suffix sum starting just after i — or total minus the rest.; "for each index" indicates Every index needs both sides → left pass + right pass (or total + running left)..

Left and Right Sum Differences

Easy·LeetCode #2574 Pro OnlyBasic Reinforcement

LeetCode

Problem Statement

You are given an integer array nums of length n.

For each index i, let left be the sum of the numbers before i, and right be the sum of the numbers after i. nums[i] itself is in neither. An empty side has sum 0.

Return an array answer of length n where answer[i] = |left − right|.

Input

An integer array nums of length n.

Output

An array of n integers: answer[i] is the absolute difference between the sum before i and the sum after i.

Example 1

Input: nums = [10, 4, 8, 3]

Output: [15, 1, 11, 22]

Left sums are [0, 10, 14, 22] and right sums are [15, 11, 3, 0]. Their differences are [15, 1, 11, 22].

Example 2

Input: nums = [1]

Output: [0]

Both sides of the only index are empty, so |0 − 0| = 0.

Constraints

• 1 ≤ n ≤ 1000
• 1 ≤ nums[i] ≤ 10^5

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“sum of the numbers before i”

→ Prefix sum ending just before i — a running total from the left.

“sum of the numbers after i”

→ Suffix sum starting just after i — or total minus the rest.

“for each index”

→ Every index needs both sides → left pass + right pass (or total + running left).

Approach Complexity Ladder

Brute Force — sum both sides for every indexO(N²)O(1)

For each index, loop over the left side and the right side

Better — left-sum array + right-sum arrayO(N)O(N)

Build all left sums in one pass and all right sums in another, then subtract

Optimal — total minus running leftO(N)O(1)

One running left sum; right = total − left − nums[i]

Brute Force Intuition

For each i, add up nums[0 … i−1] and nums[i+1 … n−1] with two loops and store the absolute difference.

Recognition rule for this sub-pattern:

Each index needs information from its left and its right. Build a left pass and a right pass, then combine.

Remember this · Prefix Sum

Trigger

Range sums, "subarray sum equals K" (especially with negatives), equal counts, divisible by K.

Think

If I knew the total up to every index, would the answer become a difference of two totals?

Key question

Which earlier prefix value am I looking for when I stand at index i?

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