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

Time Complexity: Four Sum Problem All Unique Quadruplets

O(n³)

Understanding Time Complexity

We want to understand how the time needed to find all sets of four numbers that add up to a target grows as the input list gets bigger.

How does the number of steps change when the list size increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

nums.sort()
quadruplets = []
n = len(nums)
for i in range(n - 3):
    for j in range(i + 1, n - 2):
        left, right = j + 1, n - 1
        while left < right:
            total = nums[i] + nums[j] + nums[left] + nums[right]
            if total == target:
                quadruplets.append([nums[i], nums[j], nums[left], nums[right]])
                left += 1
                right -= 1
            elif total < target:
                left += 1
            else:
                right -= 1

This code finds all groups of four numbers in a list that add up to a given target value.

Identify Repeating Operations

• Primary operation: Nested loops and a two-pointer scan inside the innermost loop.
• How many times: The outer two loops run roughly n² times, and for each pair, the two-pointer scan runs up to n times.

How Execution Grows With Input

As the list size grows, the number of steps grows much faster because of the nested loops and the two-pointer scan inside.

Input Size (n)	Approx. Operations
10	About 10³ = 1,000 steps
100	About 100³ = 1,000,000 steps
1000	About 1000³ = 1,000,000,000 steps

Pattern observation: The steps grow roughly by the cube of the input size, so doubling the list size makes the work about eight times bigger.

Final Time Complexity

Time Complexity: O(n³)

This means the time needed grows roughly with the cube of the input size, so bigger lists take much more time.

Common Mistake

[X] Wrong: "Because we use two pointers inside, the time complexity is O(n²)."

[OK] Correct: The two-pointer scan runs inside two nested loops, so the total work multiplies, making it O(n³), not just O(n²).

Interview Connect

Understanding this time complexity helps you explain how nested loops and scanning affect performance, a key skill in solving real problems efficiently.

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 that sum to a target

Solution

1. Step 1: Understand the problem statement

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

2. Step 2: Identify the goal

   The goal is not sorting or counting pairs but to find quadruplets with a specific sum 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 = unique quadruplets summing to target [OK]

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

Common Mistakes:

• Confusing Four Sum with Two Sum or Three Sum problems
• Thinking the problem is about sorting or counting pairs
• Ignoring the uniqueness of quadruplets

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

easy

A. quadruplets = ()

B. quadruplets = []

C. quadruplets = {}

D. quadruplets = set()

Solution

1. Step 1: Identify the correct data structure for storing multiple quadruplets

   Quadruplets are multiple groups of four numbers, so a list is suitable to store them.

2. Step 2: Check the initialization syntax

   Using [] initializes an empty list, which is correct. Curly braces {} create a dictionary, parentheses () create a tuple, and set() creates a set which is unordered and not ideal here.

3. Final Answer:

   quadruplets = [] -> Option B

4. Quick Check:

   Empty list for quadruplets = [] [OK]

Hint: Use [] to store multiple quadruplets in a list [OK]

Common Mistakes:

• Using {} which creates a dictionary, not a list
• Using () which creates an immutable tuple
• Using set() which is unordered and may lose duplicates

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

nums = [1, 0, -1, 0, -2, 2]
target = 0
# After sorting
nums.sort()
quadruplets = []
for i in range(len(nums)-3):
    for j in range(i+1, len(nums)-2):
        left, right = j+1, len(nums)-1
        while left < right:
            total = nums[i] + nums[j] + nums[left] + nums[right]
            if total == target:
                quadruplets.append([nums[i], nums[j], nums[left], nums[right]])
                left += 1
                right -= 1
            elif total < target:
                left += 1
            else:
                right -= 1
print(quadruplets)

medium

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

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

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

D. []

Solution

1. Step 1: Sort the array and understand the loops

   Sorted nums = [-2, -1, 0, 0, 1, 2]. The nested loops and two-pointer approach check all quadruplets summing to target 0.

2. Step 2: Identify quadruplets that sum to zero

   Valid quadruplets are [-2, -1, 1, 2], [-2, 0, 0, 2], and [-1, 0, 0, 1]. These are appended to quadruplets list.

3. Final Answer:

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

4. Quick Check:

   Sum quadruplets = 0 matches output [OK]

Hint: Sort array, use two pointers inside nested loops [OK]

Common Mistakes:

• Not sorting the array before processing
• Missing some quadruplets due to pointer errors
• Appending duplicates without checks

4. Identify the bug in the following code snippet for the Four Sum problem and select the correct fix.

nums = [2, 2, 2, 2, 2]
target = 8
nums.sort()
quadruplets = []
for i in range(len(nums)-3):
    for j in range(i+1, len(nums)-2):
        left, right = j+1, len(nums)-1
        while left < right:
            total = nums[i] + nums[j] + nums[left] + nums[right]
            if total == target:
                quadruplets.append([nums[i], nums[j], nums[left], nums[right]])
                left += 1
                right -= 1
            elif total < target:
                left += 1
            else:
                right -= 1
print(quadruplets)

medium

A. Add checks to skip duplicates for i, j, left, and right indices

B. Change quadruplets to a set to avoid duplicates

C. Remove sorting to improve performance

D. Replace while loop with a for loop

Solution

1. Step 1: Identify the problem with duplicates

   The code appends quadruplets but does not skip duplicates, so it will add the same quadruplet multiple times.

2. Step 2: Fix by skipping duplicates

   Adding checks to skip duplicate values for i, j, left, and right indices after finding a quadruplet prevents repeated quadruplets.

3. Final Answer:

   Add checks to skip duplicates for i, j, left, and right indices -> Option A

4. Quick Check:

   Skipping duplicates avoids repeated quadruplets [OK]

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

Common Mistakes:

• Not skipping duplicates causing repeated quadruplets
• Using set without sorting can cause unordered results
• Removing sorting breaks two-pointer logic

5. Given the array nums = [1, 0, -1, 0, -2, 2, 2] and target 2, which of the following lists contains all unique quadruplets that sum to the target?

hard

A. [[-2, -1, 2, 3], [0, 0, 1, 1]]

B. [[-2, 0, 1, 3], [-1, 0, 0, 3]]

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

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

Solution

1. Step 1: Sort the array and find quadruplets summing to 2

   Sorted nums = [-2, -1, 0, 0, 1, 2, 2]. We look for quadruplets where sum equals 2.

2. Step 2: Identify valid quadruplets

   Valid quadruplets are [-2, 0, 2, 2] (sum = 2) and [-1, 0, 1, 2] (sum = 2). Other options contain invalid or impossible sums.

3. Final Answer:

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

4. Quick Check:

   Sum quadruplets = 2 matches output [OK]

Hint: Sort and check quadruplets summing exactly to target [OK]

Common Mistakes:

• Including quadruplets with wrong sums
• Ignoring duplicates or invalid numbers
• Not sorting before searching quadruplets