What is the brute force approach and complexity for 4Sum?

Four loops; store each matching group sorted in a set with time complexity O(N⁴ · log M) and space complexity O(M).

What is the optimal approach and complexity for 4Sum?

Sort; fix a and b (skipping repeats); two pointers for the last pair with optimal time complexity O(N³) and space complexity O(1).

How do you identify the pattern for 4Sum?

Look for key signals in the prompt: "four values that add up to target" indicates Fix two, then Two Sum II on the rest → k-sum with k = 4.; "unique quadruplets" indicates Sort and skip equal neighbours at every fixed level and after every hit.; "values up to 10^9" indicates The sum of four overflows `int` → use `long long`..

4Sum

Medium·LeetCode #18 Pro OnlyPattern Transfer

LeetCode

Problem Statement

You are given an integer array nums and a number target.

Find every group of four values, taken from four different positions, whose sum is exactly target.

Each group may appear only once: two groups with the same four values count as the same, even if they came from different positions. The order of the groups, and of the values inside them, does not matter.

Input

An integer array nums and an integer target.

Output

A list of all unique quadruplets (lists of 4 values) that add up to target.

Example 1

Input: nums = [1, 0, -1, 0, -2, 2], target = 0

Output: [[-2, -1, 1, 2], [-2, 0, 0, 2], [-1, 0, 0, 1]]

These are the only three different groups of values that add up to 0. Some can be built from different positions (there are two 0s), but each is listed once.

Example 2

Input: nums = [2, 2, 2, 2, 2], target = 8

Output: [[2, 2, 2, 2]]

Many position-groups give 2 + 2 + 2 + 2, but it is one answer.

Constraints

• 1 ≤ nums.length ≤ 200
• −10^9 ≤ nums[i] ≤ 10^9
• −10^9 ≤ target ≤ 10^9
• A sum of four values can reach ±4·10^9, which does not fit in an `int`.

Pattern Recognition Clues

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

“four values that add up to target”

→ Fix two, then Two Sum II on the rest → k-sum with k = 4.

“unique quadruplets”

→ Sort and skip equal neighbours at every fixed level and after every hit.

“values up to 10^9”

→ The sum of four overflows `int` → use `long long`.

Approach Complexity Ladder

Brute Force — every quadruplet + a setO(N⁴ · log M)O(M)

Four loops; store each matching group sorted in a set

Better — fix three, hash set for the fourthO(N³ · log M)O(N + M)

Fix a, b; for each c, look up the needed fourth value in a set

Optimal — sort, fix two, two pointers, skip duplicatesO(N³)O(1)

Sort; fix a and b (skipping repeats); two pointers for the last pair

Brute Force Intuition

Try every a < b < c < d. When the four values add up to target, sort them and put them in a set, which drops repeats.

Recognition rule for this sub-pattern:

Sort, fix one element (or more), and solve the rest as a two-pointer pair problem. Skip duplicates carefully.

Remember this · Two Pointers

Trigger

Sorted array, pairs/triplets with a target, comparing from both ends, compacting an array in place.

Think

If I compare the elements at two positions, can I safely discard one side?

Key question

Which pointer moves, and why can the skipped elements never be the answer?

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