Four Sum Problem All Unique Quadruplets in DSA C++ - Time & Space Complexity

Time Complexity: Four Sum Problem All Unique Quadruplets

O(n³)

Understanding Time Complexity

We want to understand how the time needed grows when finding all groups of four numbers that add up to a target.

How does the work increase as the list of numbers gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

vector<vector<int>> fourSum(vector<int>& nums, int target) {
    sort(nums.begin(), nums.end());
    vector<vector<int>> result;
    int n = nums.size();
    for (int i = 0; i < n - 3; i++) {
        for (int j = i + 1; j < n - 2; j++) {
            int left = j + 1, right = n - 1;
            while (left < right) {
                int sum = nums[i] + nums[j] + nums[left] + nums[right];
                if (sum == target) {
                    result.push_back({nums[i], nums[j], nums[left], nums[right]});
                    left++; right--;
                } else if (sum < target) {
                    left++;
                } else {
                    right--;
                }
            }
        }
    }
    return result;
}
    

This code finds all sets of four numbers in the list that add up to the target value.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Two nested loops plus a two-pointer scan inside the innermost loop.
• How many times: Outer loop runs about n times, second loop about n times, and the two-pointer scan runs up to n times per inner loop.

How Execution Grows With Input

As the list size grows, the number of operations grows roughly as the cube of the list size.

Input Size (n)	Approx. Operations
10	About 1,000
100	About 1,000,000
1000	About 1,000,000,000

Pattern observation: The work grows very fast, roughly like n³.

Final Time Complexity

Time Complexity: O(n³)

This means the time needed grows roughly with the cube of the input size, so doubling the input makes the work about eight times more.

Common Mistake

[X] Wrong: "The time complexity is O(n⁴) because there are four numbers to pick."

[OK] Correct: The innermost two-pointer scan is linear, not a full loop, so it reduces one level of nesting, making it O(n³) instead of O(n⁴).

Interview Connect

Understanding this time complexity helps you explain how nested loops and two-pointer techniques combine, a common pattern in coding challenges.

Self-Check

"What if we changed the problem to find unique triplets instead of quadruplets? How would the time complexity change?"

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 with a given sum

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 to find quadruplets (groups of four numbers) that sum exactly to the target, without duplicates.

3. Final Answer:

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

4. Quick Check:

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

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

Common Mistakes:

• Confusing with pair or triplet sum problems
• Ignoring uniqueness of quadruplets
• Trying to sort only without finding quadruplets

2. Which of the following C++ code snippets correctly initializes a vector to store quadruplets for the Four Sum Problem?

easy

A. std::vector<int> quadruplets(4);

B. std::vector quadruplets;

C. std::vector<int> quadruplets = {4};

D. std::vector> quadruplets;

Solution

1. Step 1: Understand the data structure needed

   We need to store multiple quadruplets, each is a vector of 4 integers, so a vector of vector<int> is suitable.

2. Step 2: Check each option

   std::vector quadruplets; declares a vector of vector<int>, perfect for storing multiple quadruplets. std::vector<int> quadruplets(4); creates a single vector of size 4, not multiple quadruplets. std::vector<int> quadruplets = {4}; initializes a vector with one element 4, incorrect. std::vector> quadruplets; uses pairs, which store only two integers, not four.

3. Final Answer:

   std::vector quadruplets; -> Option B

4. Quick Check:

   Use vector of vector<int> for quadruplets [OK]

Hint: Use vector of vector<int> to store multiple quadruplets [OK]

Common Mistakes:

• Using vector<int> for multiple quadruplets
• Using pairs instead of quadruplets
• Initializing vector with wrong size or values

3. What will be the output of the following C++ code snippet for input array {1, 0, -1, 0, -2, 2} and target 0?

std::vector<int> nums = {1, 0, -1, 0, -2, 2};
int target = 0;
// Assume fourSum function returns all unique quadruplets
auto result = fourSum(nums, target);
for (auto &quad : result) {
    for (int num : quad) std::cout << num << " ";
    std::cout << std::endl;
}

medium

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

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

C. [-2 -2 1 3]
[-1 0 0 1]

D. No quadruplets found

Solution

1. Step 1: Sort the array and find quadruplets summing to target 0

   Sorted array: [-2, -1, 0, 0, 1, 2]. The unique quadruplets summing to 0 are [-2, -1, 1, 2], [-2, 0, 0, 2], and [-1, 0, 0, 1].

2. Step 2: Verify output matches quadruplets

   [-2 -1 1 2] [-2 0 0 2] [-1 0 0 1] lists exactly these three quadruplets. Options A, C, and D have incorrect quadruplets or sums.

3. Final Answer:

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

4. Quick Check:

   Unique quadruplets summing to 0 = [-2 -1 1 2] [-2 0 0 2] [-1 0 0 1] [OK]

Hint: Sort array first, then find quadruplets summing to target [OK]

Common Mistakes:

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

4. Identify the bug in the following C++ code snippet for the Four Sum Problem:

std::vector fourSum(std::vector<int>& nums, int target) {
    std::vector result;
    std::sort(nums.begin(), nums.end());
    for (int i = 0; i < nums.size(); ++i) {
        for (int j = i + 1; j < nums.size(); ++j) {
            int left = j + 1, right = nums.size() - 1;
            while (left < right) {
                int sum = nums[i] + nums[j] + nums[left] + nums[right];
                if (sum == target) {
                    result.push_back({nums[i], nums[j], nums[left], nums[right]});
                    ++left; --right;
                } else if (sum < target) {
                    ++left;
                } else {
                    --right;
                }
            }
        }
    }
    return result;
}

medium

A. Returns result before completing loops

B. Incorrect sorting of the input array

C. Wrong initialization of left and right pointers

D. Does not skip duplicate quadruplets, causing repeated results

Solution

1. Step 1: Analyze duplicate handling in loops

   The code does not skip duplicates for i, j, left, or right indices, so it can add the same quadruplet multiple times.

2. Step 2: Check other parts for errors

   Sorting is correct, left and right pointers are initialized properly, and return is after loops. So these are not bugs.

3. Final Answer:

   Does not skip duplicate quadruplets, causing repeated results -> Option D

4. Quick Check:

   Missing duplicate checks = Does not skip duplicate quadruplets, causing repeated results [OK]

Hint: Always skip duplicates after finding quadruplets [OK]

Common Mistakes:

• Not skipping duplicates for i, j, left, right
• Returning result too early
• Incorrect pointer initialization

5. Given an array nums = {2, 2, 2, 2, 2} and target 8, how many unique quadruplets does the Four Sum algorithm find, and what are they?

hard

A. 10 quadruplets: including duplicates

B. 5 quadruplets: all combinations of 2s

C. 1 quadruplet: [2, 2, 2, 2]

D. 0 quadruplets: no quadruplets sum to 8

Solution

1. Step 1: Understand the input and target

   All elements are 2, and target is 8. Sum of four 2s is 8, so quadruplet [2, 2, 2, 2] is valid.

2. Step 2: Consider uniqueness

   Even though multiple combinations exist, they are duplicates. The algorithm returns only one unique quadruplet.

3. Final Answer:

   1 quadruplet: [2, 2, 2, 2] -> Option C

4. Quick Check:

   All 2s sum to 8 once uniquely [OK]

Hint: Duplicates count as one unique quadruplet [OK]

Common Mistakes:

• Counting duplicate quadruplets multiple times
• Assuming no quadruplets because of repeated numbers
• Confusing number of combinations with unique quadruplets