Four Sum Problem All Unique Quadruplets in DSA Java - 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 pointers to find pairs efficiently.

intermediate

In the Four Sum problem, after fixing two numbers, what technique is used to find the remaining two numbers?

Use two pointers starting from the ends of the remaining array segment to find pairs that sum to the remaining target.

intermediate

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

Skip duplicate numbers while iterating through the array and after finding a valid quadruplet, move pointers past duplicates.

intermediate

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

O(n^3), because we use two nested loops and a two-pointer scan inside.

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

A Reverse the array

B Sort the array

C Use a hash map

D Split the array into halves

Sorting the array helps to handle duplicates and use two pointers effectively.

After fixing two numbers, which method is used to find the other two numbers in Four Sum?

A Binary search

B Recursion

C Two pointers

D Brute force

Two pointers scan the remaining array segment to find pairs that sum to the target.

How do you avoid duplicate quadruplets in the Four Sum problem?

A Skip duplicates while iterating and after finding quadruplets

B Ignore duplicates only at the end

C Use a set to store all quadruplets

D Sort quadruplets before adding

Skipping duplicates during iteration and pointer movement prevents repeated quadruplets.

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

A O(n^3)

B O(n^4)

C O(n^2)

D O(n log n)

Two nested loops (O(n^2)) and a two-pointer scan (O(n)) combine to O(n^3).

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

A Array

B Sorting

C Two pointers

D HashSet

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

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

Think about fixing two numbers and then searching for pairs.

Describe how skipping duplicates works in the Four Sum problem and why it is important.

Duplicates appear when consecutive numbers are the same.

Practice

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

easy

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

B. Sort the array in ascending order

C. Find the maximum element in the array

D. Count the number of pairs that sum to a target

Solution

1. Step 1: Understand the problem statement

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

2. Step 2: Identify the expected output

   We want all unique quadruplets, meaning no repeated sets, that sum exactly to the target.

3. Final Answer:

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

4. Quick Check:

   Four Sum = unique quadruplets summing to target [OK]

Hint: Look for four numbers adding to target without duplicates [OK]

Common Mistakes:

• Confusing Four Sum with Two Sum or Three Sum problems
• Ignoring uniqueness of quadruplets
• Trying to find maximum or minimum instead of sum

2. Which of the following Java code snippets correctly initializes an empty list to store quadruplets for the Four Sum problem?

easy

A. List quadruplets = new List<>();

B. List quadruplets = new ArrayList<>();

C. ArrayList<int> quadruplets = new ArrayList<>();

D. List<List> quadruplets = new ArrayList();

Solution

1. Step 1: Check Java syntax for list initialization

   Correct generic syntax for a list of lists of integers is List quadruplets = new ArrayList<>();

2. Step 2: Identify errors in other options

   List quadruplets = new List<>(); uses raw List without generics and wrong constructor; ArrayList<int> quadruplets = new ArrayList<>(); uses primitive int which is invalid; List<List> quadruplets = new ArrayList(); mixes generics incorrectly.

3. Final Answer:

   List quadruplets = new ArrayList<>(); -> Option B

4. Quick Check:

   Correct generic list initialization = List quadruplets = new ArrayList<>(); [OK]

Hint: Use List with ArrayList constructor [OK]

Common Mistakes:

• Using raw types without generics
• Trying to use primitive types in generics
• Incorrect constructor calls for List

3. What will be the output of the following Java code snippet for the Four Sum problem?

int[] nums = {1, 0, -1, 0, -2, 2};
int target = 0;
List result = fourSum(nums, target);
System.out.println(result);

Assuming fourSum correctly finds all unique quadruplets summing to target.

medium

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

B. [[1, 0, -1, 0]]

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

D. []

Solution

1. Step 1: Sort and find quadruplets summing to zero

   Sorted array: [-2, -1, 0, 0, 1, 2]. Quadruplets summing to 0 are [-2,-1,1,2], [-2,0,0,2], and [-1,0,0,1].

2. Step 2: Verify uniqueness and completeness

   All quadruplets are unique and sum exactly to target 0. No other quadruplets exist.

3. Final Answer:

   [[-2, -1, 1, 2], [-2, 0, 0, 2], [-1, 0, 0, 1]] -> Option C

4. Quick Check:

   Unique quadruplets sum to 0 = [[-2, -1, 1, 2], [-2, 0, 0, 2], [-1, 0, 0, 1]] [OK]

Hint: Sort array first, then find quadruplets with two pointers [OK]

Common Mistakes:

• Not sorting array before processing
• Including duplicate quadruplets
• Missing some valid quadruplets

4. Identify the bug in this Java snippet for the Four Sum problem:

for (int i = 0; i < nums.length - 3; i++) {
  for (int j = i + 1; j < nums.length - 2; j++) {
    int left = j + 1, right = nums.length - 1;
    while (left < right) {
      int sum = nums[i] + nums[j] + nums[left] + nums[right];
      if (sum == target) {
        result.add(Arrays.asList(nums[i], nums[j], nums[left], nums[right]));
        left++;
        right--;
      } else if (sum < target) {
        left++;
      } else {
        right--;
      }
    }
  }
}

medium

A. Sum calculation uses wrong indices

B. Incorrect initialization of left and right pointers

C. Wrong loop conditions for i and j

D. Missing checks to skip duplicate elements for i, j, left, and right

Solution

1. Step 1: Analyze loop and pointer usage

   Loops and pointers are correctly set to find quadruplets, and sum calculation is correct.

2. Step 2: Identify missing duplicate skipping

   Code lacks checks to skip duplicate values for i, j, left, and right after finding a quadruplet, which can cause duplicate quadruplets in result.

3. Final Answer:

   Missing checks to skip duplicate elements for i, j, left, and right -> Option D

4. Quick Check:

   Duplicate skipping missing = Missing checks to skip duplicate elements for i, j, left, and right [OK]

Hint: Always skip duplicates after adding quadruplet to avoid repeats [OK]

Common Mistakes:

• Not skipping duplicates causing repeated quadruplets
• Changing pointers incorrectly inside loops
• Incorrect loop boundaries causing index errors

5. Given the array nums = {2, 2, 2, 2, 2} and target 8, what is the correct output of the Four Sum problem solution that finds all unique quadruplets?

hard

A. [[2, 2, 2, 2]]

B. [[2, 2, 2, 2], [2, 2, 2, 2]]

C. []

D. [[2, 2, 2]]

Solution

1. Step 1: Check if quadruplets sum to target

   All elements are 2, so quadruplet [2, 2, 2, 2] sums to 8.

2. Step 2: Ensure uniqueness of quadruplets

   Even though elements repeat, only one unique quadruplet exists. Duplicates must be avoided.

3. Final Answer:

   [[2, 2, 2, 2]] -> Option A

4. Quick Check:

   Unique quadruplet summing to 8 = [[2, 2, 2, 2]] [OK]

Hint: Same numbers can form one unique quadruplet if sum matches target [OK]

Common Mistakes:

• Returning duplicate quadruplets multiple times
• Returning empty list when quadruplet exists
• Confusing triple with quadruplet