Four Sum Problem All Unique Quadruplets in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main goal of the Four Sum problem?

To find all unique sets of four numbers in an array that add up to a given target sum.

beginner

Why do we sort the array before solving the Four Sum problem?

Sorting helps to avoid duplicates easily and allows using two-pointer technique efficiently.

intermediate

In the Four Sum problem, after fixing two numbers, how do we find the other two numbers?

We use two pointers starting from the ends of the remaining array segment and move them inward to find pairs that sum to the remaining target.

intermediate

How do we ensure the quadruplets are unique in the Four Sum problem?

By skipping duplicate values for each fixed number and while moving the two pointers, we avoid adding the same quadruplet multiple times.

intermediate

What is the time complexity of the Four Sum problem solution using sorting and two pointers?

O(n^3), where n is the number of elements in the array, because we fix two numbers and use two pointers for the remaining two.

What is the first step in solving the Four Sum problem efficiently?

A Reverse the array

B Use a hash map

C Pick random quadruplets

D Sort the array

Sorting the array helps in skipping duplicates and using two pointers effectively.

After fixing two numbers, which technique is used to find the remaining two numbers in Four Sum?

A Two pointers

B Binary search

C Recursion

D Brute force

Two pointers move inward from both ends to find pairs that sum to the remaining target.

How do we avoid duplicate quadruplets in the result?

A By checking all quadruplets twice

B By using a set of quadruplets

C By sorting and skipping duplicates during iteration

D By ignoring the target sum

Sorting allows skipping duplicates easily while iterating and moving pointers.

What is the overall time complexity of the Four Sum solution using sorting and two pointers?

A O(n^2)

B O(n^3)

C O(n^4)

D O(n log n)

Fixing two numbers and using two pointers for the rest results in O(n^3) complexity.

Which data structure is NOT typically used in the classic Four Sum solution?

A Hash map

B Sorting

C Two pointers

D Array

The classic solution uses sorting and two pointers, not hash maps.

Explain step-by-step how to find all unique quadruplets that sum to a target in an array.

Think about how sorting and two pointers help reduce complexity and avoid duplicates.

Describe how skipping duplicates works in the Four Sum problem to ensure unique quadruplets.

Duplicates appear consecutively after sorting, so skip repeated values.

Practice

1. What is the main goal of the Four Sum Problem in data structures and algorithms?

easy

A. Sort the array in ascending order

B. Count the number of pairs with a given sum

C. Find the maximum element in the array

D. Find all unique quadruplets in an array that sum up to a target value

Solution

1. Step 1: Understand the problem statement

   The Four Sum Problem asks to find all unique sets of four numbers in an array that add up to a specific target sum.

2. Step 2: Identify the goal

   The goal is not sorting or counting pairs but to find quadruplets that sum to the target without duplicates.

3. Final Answer:

   Find all unique quadruplets in an array that sum up to a target value -> Option D

4. Quick Check:

   Four Sum Problem = Find quadruplets summing to target [OK]

Hint: Look for four numbers adding to target, no duplicates [OK]

Common Mistakes:

• Confusing with two or three sum problems
• Trying to sort only without finding quadruplets
• Counting pairs instead of quadruplets

2. Which of the following Go code snippets correctly initializes a slice to store quadruplets for the Four Sum problem?

easy

A. quadruplets := [][]int{}

B. quadruplets := [4]int{}

C. quadruplets := make([]int, 4)

D. quadruplets := map[int]int{}

Solution

1. Step 1: Understand data structure for quadruplets

   We need a slice of slices of integers to store multiple quadruplets, each quadruplet is a slice of 4 ints.

2. Step 2: Choose correct initialization

   quadruplets := [][]int{} initializes an empty slice of int slices, suitable for storing multiple quadruplets. quadruplets := [4]int{} is a fixed array of 4 ints, not multiple quadruplets. quadruplets := make([]int, 4) is a slice of 4 ints, not a slice of quadruplets. quadruplets := map[int]int{} is a map, not suitable here.

3. Final Answer:

   quadruplets := [][]int{} -> Option A

4. Quick Check:

   Slice of int slices = [][]int{} [OK]

Hint: Use [][]int{} for list of quadruplets [OK]

Common Mistakes:

• Using fixed arrays instead of slices
• Using map instead of slice
• Initializing slice with wrong dimensions

3. What will be the output of the following Go code snippet that finds unique quadruplets summing to 0?

nums := []int{1, 0, -1, 0, -2, 2}
target := 0
result := fourSum(nums, target)
fmt.Println(result)

// Assume fourSum is correctly implemented

medium

A. [[-2 -1 0 1] [0 0 1 2]]

B. [[-2 -1 1 2] [-2 0 0 2] [-1 0 0 1]]

C. [[1 0 -1 0] [-2 2 0 0]]

D. []

Solution

1. Step 1: Understand input and target

   The input array is [1, 0, -1, 0, -2, 2] and target sum is 0.

2. Step 2: Identify unique quadruplets summing to 0

   Possible unique quadruplets are [-2, -1, 1, 2], [-2, 0, 0, 2], and [-1, 0, 0, 1]. These sum exactly to 0 and are unique.

3. Final Answer:

   [[-2 -1 1 2] [-2 0 0 2] [-1 0 0 1]] -> Option B

4. Quick Check:

   Sum quadruplets = 0 [OK]

Hint: Sort array and find quadruplets summing to target [OK]

Common Mistakes:

• Including duplicate quadruplets
• Incorrect sums in quadruplets
• Missing some valid quadruplets

4. Given the following Go code snippet for the Four Sum problem, what is the main bug causing incorrect results?

func fourSum(nums []int, target int) [][]int {
    sort.Ints(nums)
    var res [][]int
    for i := 0; i < len(nums)-3; i++ {
        for j := i + 1; j < len(nums)-2; j++ {
            left, right := j+1, len(nums)-1
            for left < right {
                sum := nums[i] + nums[j] + nums[left] + nums[right]
                if sum == target {
                    res = append(res, []int{nums[i], nums[j], nums[left], nums[right]})
                    left++
                    right--
                } else if sum < target {
                    left++
                } else {
                    right--
                }
            }
        }
    }
    return res
}

medium

A. No checks to skip duplicate quadruplets causing duplicates in result

B. Incorrect sorting of the input array

C. Wrong initialization of left and right pointers

D. Using append incorrectly causing runtime error

Solution

1. Step 1: Analyze duplicate handling

   The code finds quadruplets but does not skip duplicates after finding a valid quadruplet, so duplicates appear in the result.

2. Step 2: Check other parts

   Sorting is correct, pointers are correctly initialized, and append usage is valid.

3. Final Answer:

   No checks to skip duplicate quadruplets causing duplicates in result -> Option A

4. Quick Check:

   Missing duplicate skip = duplicates in output [OK]

Hint: Skip duplicates after finding quadruplets to avoid repeats [OK]

Common Mistakes:

• Not skipping duplicates for i, j, left, right
• Incorrect pointer updates after finding quadruplet
• Assuming sorting alone removes duplicates

5. You want to optimize the Four Sum solution to avoid unnecessary computations. Which approach best reduces the search space while ensuring all unique quadruplets are found?

hard

A. Sort the array and use two nested loops with two pointers, skipping duplicates at each step

B. Use four nested loops to check all quadruplets without sorting

C. Sort the array and use a hash map to store sums of pairs, then find complements

D. Use recursion without sorting or duplicate checks

Solution

1. Step 1: Understand optimization goals

   We want to reduce time complexity and avoid duplicates efficiently.

2. Step 2: Evaluate approaches

   Sort the array and use two nested loops with two pointers, skipping duplicates at each step is common but still O(n^3). Use four nested loops to check all quadruplets without sorting is brute force O(n^4). Sort the array and use a hash map to store sums of pairs, then find complements uses a hash map to store pair sums and find complements, reducing complexity and handling duplicates well. Use recursion without sorting or duplicate checks is inefficient without sorting or duplicate checks.

3. Final Answer:

   Sort the array and use a hash map to store sums of pairs, then find complements -> Option C

4. Quick Check:

   Hash map pairs + complements = optimized search [OK]

Hint: Use hash map of pair sums to find quadruplets faster [OK]

Common Mistakes:

• Using brute force four loops
• Ignoring duplicate skipping
• Not sorting before processing