Three Sum Problem All Unique Triplets in DSA Python - Time & Space Complexity

Time Complexity: Three Sum Problem All Unique Triplets

O(n²)

Understanding Time Complexity

We want to understand how the time needed grows when finding all unique triplets that sum to zero in a list.

How does the number of steps change as the list gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

nums.sort()
result = []
for i in range(len(nums)):
    if i > 0 and nums[i] == nums[i-1]:
        continue
    left, right = i + 1, len(nums) - 1
    while left < right:
        total = nums[i] + nums[left] + nums[right]
        if total < 0:
            left += 1
        elif total > 0:
            right -= 1
        else:
            result.append([nums[i], nums[left], nums[right]])
            while left < right and nums[left] == nums[left + 1]:
                left += 1
            while left < right and nums[right] == nums[right - 1]:
                right -= 1
            left += 1
            right -= 1
    

This code finds all unique triplets in the list that add up to zero using sorting and two pointers.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The outer loop runs through each element once, and inside it, a while loop moves two pointers through the remaining elements.
• How many times: The outer loop runs n times, and the inner while loop moves pointers across the list, together covering the list roughly n times for each outer loop.

How Execution Grows With Input

As the list size grows, the number of steps grows much faster because for each element, we scan the rest with two pointers.

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

Pattern observation: The steps grow roughly with the square of the input size, meaning if the list size doubles, the work increases about four times.

Final Time Complexity

Time Complexity: O(n²)

This means the time needed grows roughly with the square of the list size, so doubling the list makes the work about four times bigger.

Common Mistake

[X] Wrong: "The time complexity is O(n³) because there are three nested loops."

[OK] Correct: The code uses sorting and two pointers inside a single loop, not three nested loops. The inner while loop moves pointers forward without restarting, so total work is less than cubic.

Interview Connect

Understanding this problem's time complexity helps you explain how to handle searching pairs efficiently after sorting, a skill useful in many coding challenges.

Self-Check

"What if we used three nested loops instead of two pointers? How would the time complexity change?"

Practice

1. What is the main goal of the Three Sum Problem in arrays?

easy

A. Find all unique triplets that sum to zero

B. Find the maximum sum of any three elements

C. Sort the array in ascending order

D. Count the number of pairs with sum zero

Solution

1. Step 1: Understand the problem statement

   The Three Sum Problem asks to find all unique sets of three numbers in an array that add up to zero.

2. Step 2: Differentiate from other sum problems

   Unlike finding pairs or maximum sums, this problem focuses on triplets with sum exactly zero and uniqueness.

3. Final Answer:

   Find all unique triplets that sum to zero -> Option A

4. Quick Check:

   Three Sum = unique triplets summing to zero [OK]

Hint: Look for three numbers adding exactly to zero [OK]

Common Mistakes:

• Confusing with pair sum problems
• Looking for maximum sum instead of zero sum
• Ignoring uniqueness of triplets

2. Which of the following Python code snippets correctly initializes an empty list to store unique triplets for the Three Sum Problem?

easy

A. triplets = []

B. triplets = {}

C. triplets = set()

D. triplets = None

Solution

1. Step 1: Identify the correct data structure for storing triplets

   Triplets are stored as lists inside a list, so initializing an empty list is correct.

2. Step 2: Analyze other options

   {} creates a dictionary, set() is for unique single elements, None is not a container.

3. Final Answer:

   triplets = [] -> Option A

4. Quick Check:

   Empty list for triplets = [] [OK]

Hint: Use [] to store multiple triplets as lists [OK]

Common Mistakes:

• Using {} which creates a dictionary
• Using set() which stores unique single elements, not lists
• Assigning None which is not a container

3. What is the output of the following code snippet for the input array nums = [-1, 0, 1, 2, -1, -4]?

def three_sum(nums):
    nums.sort()
    result = []
    for i in range(len(nums)-2):
        if i > 0 and nums[i] == nums[i-1]:
            continue
        left, right = i+1, len(nums)-1
        while left < right:
            total = nums[i] + nums[left] + nums[right]
            if total == 0:
                result.append([nums[i], nums[left], nums[right]])
                while left < right and nums[left] == nums[left+1]:
                    left += 1
                while left < right and nums[right] == nums[right-1]:
                    right -= 1
                left += 1
                right -= 1
            elif total < 0:
                left += 1
            else:
                right -= 1
    return result

print(three_sum(nums))

medium

A. []

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

C. [[0, 0, 0]]

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

Solution

1. Step 1: Sort the input array

   Sorted nums = [-4, -1, -1, 0, 1, 2]

2. Step 2: Find unique triplets summing to zero

   Triplets found are [-1, -1, 2] and [-1, 0, 1]. These are appended to result.

3. Final Answer:

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

4. Quick Check:

   Triplets sum zero = [[-1,-1,2],[-1,0,1]] [OK]

Hint: Sort array and use two pointers to find triplets [OK]

Common Mistakes:

• Not sorting the array before processing
• Including duplicate triplets
• Incorrect pointer movement causing missed triplets

4. Identify the bug in the following code snippet for the Three Sum Problem:

def three_sum(nums):
    nums.sort()
    result = []
    for i in range(len(nums)):
        left, right = i+1, len(nums)-1
        while left < right:
            total = nums[i] + nums[left] + nums[right]
            if total == 0:
                result.append([nums[i], nums[left], nums[right]])
                left += 1
                right -= 1
            elif total < 0:
                left += 1
            else:
                right -= 1
    return result

print(three_sum([-1, 0, 1, 2, -1, -4]))

medium

A. The outer loop should run till len(nums)-2, not len(nums)

B. No check to skip duplicate triplets

C. Missing sorting of the input array

D. Incorrect initialization of left and right pointers

Solution

1. Step 1: Check for duplicate skipping logic

   The code lacks 'if i > 0 and nums[i] == nums[i-1]: continue' and inner while loops to skip duplicate left/right values after finding a triplet. This causes duplicate triplets like [-1, 0, 1] appearing twice.

2. Step 2: Verify other parts

   Sorting is present, pointers initialized correctly, loop range to len(nums) is unnecessary but does not cause errors or duplicates.

3. Final Answer:

   No check to skip duplicate triplets -> Option B

4. Quick Check:

   Missing duplicate skips cause repeated triplets [OK]

Hint: Add duplicate skips for i, left, and right pointers [OK]

Common Mistakes:

• Missing 'if i > 0 and nums[i] == nums[i-1]' check
• No inner while loops to skip duplicate left/right
• Thinking loop range causes index errors (it doesn't)

5. Given the input array nums = [0, 0, 0, 0], what is the correct output of a Three Sum solution that returns all unique triplets summing to zero?

hard

A. [[0, 0, 0], [0, 0, 0]]

B. []

C. [[0, 0, 0]]

D. [[0, 0]]

Solution

1. Step 1: Understand uniqueness requirement

   Even though multiple zeros exist, only one unique triplet [0, 0, 0] should be returned.

2. Step 2: Check triplet formation

   Triplet [0, 0, 0] sums to zero and is valid. Duplicates must be avoided.

3. Final Answer:

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

4. Quick Check:

   Unique triplets = [[0,0,0]] [OK]

Hint: Return only unique triplets, avoid duplicates [OK]

Common Mistakes:

• Returning duplicate triplets multiple times
• Returning pairs instead of triplets
• Returning empty list when triplets exist