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

Time Complexity: Three Sum Problem All Unique Triplets

O(n^2)

Understanding Time Complexity

We want to understand how the time needed to find all unique triplets that sum to zero grows as the input list gets bigger.

How does the number of steps change when we have more numbers to check?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void threeSum(int* nums, int numsSize) {
    qsort(nums, numsSize, sizeof(int), compare);
    for (int i = 0; i < numsSize - 2; i++) {
        if (i > 0 && nums[i] == nums[i-1]) continue;
        int left = i + 1, right = numsSize - 1;
        while (left < right) {
            int sum = nums[i] + nums[left] + nums[right];
            if (sum == 0) {
                // store triplet
                left++; right--;
                while (left < right && nums[left] == nums[left-1]) left++;
                while (left < right && nums[right] == nums[right+1]) right--;
            } else if (sum < 0) {
                left++;
            } else {
                right--;
            }
        }
    }
}
    

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The outer loop runs through each element, and inside it, the two-pointer while loop scans the remaining elements.
• How many times: The outer loop runs about n times, and the inner while loop runs up to n times in total for each outer iteration.

How Execution Grows With Input

As the input size grows, the number of steps grows roughly by the square of the input size.

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 by the square of the input size, so doubling the input makes the work about four times bigger.

Final Time Complexity

Time Complexity: O(n2)

This means the time needed grows roughly with the square of the number of elements in the input.

Common Mistake

[X] Wrong: "Because there are three numbers, the time complexity must be O(n³)"

[OK] Correct: The sorting and two-pointer approach reduces the inner search to linear time, so the total is O(n²), not cubic.

Interview Connect

Understanding this problem's time complexity helps you explain how to handle nested loops efficiently and shows your skill in optimizing search problems.

Self-Check

"What if the input array was already sorted? How would that affect the time complexity?"

Practice

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

easy

A. Find the maximum sum of any three elements

B. Sort the array in ascending order

C. Count the number of pairs that sum to zero

D. Find all unique triplets in an array that sum to 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 whose sum is zero.

2. Step 2: Differentiate from other options

   Options A, B, and D describe different problems like maximum sum, sorting, or pairs, not triplets summing to zero.

3. Final Answer:

   Find all unique triplets in an array that sum to zero -> Option D

4. Quick Check:

   Three Sum Problem = Find triplets summing to zero [OK]

Hint: Remember: Three Sum finds triplets with zero sum [OK]

Common Mistakes:

• Confusing with pair sum problems
• Thinking it finds maximum sums
• Assuming it sorts the array only

2. Which of the following code snippets correctly initializes the array for solving the Three Sum Problem in C?

easy

A. int nums[] = { -1, 0, 1, 2, -1, -4 }; int size = sizeof(nums) / sizeof(nums[0]);

B. int nums = { -1, 0, 1, 2, -1, -4 }; int size = 6;

C. int nums[] = (-1, 0, 1, 2, -1, -4); int size = 6;

D. int nums[] = [-1, 0, 1, 2, -1, -4]; int size = 6;

Solution

1. Step 1: Check array initialization syntax in C

   In C, arrays are initialized with curly braces {}, not parentheses or square brackets.

2. Step 2: Verify size calculation

   int nums[] = { -1, 0, 1, 2, -1, -4 }; int size = sizeof(nums) / sizeof(nums[0]); correctly calculates size using sizeof operator divided by element size.

3. Final Answer:

   int nums[] = { -1, 0, 1, 2, -1, -4 }; int size = sizeof(nums) / sizeof(nums[0]); -> Option A

4. Quick Check:

   C array init with {} and sizeof = int nums[] = { -1, 0, 1, 2, -1, -4 }; int size = sizeof(nums) / sizeof(nums[0]); [OK]

Hint: Use curly braces {} for arrays in C [OK]

Common Mistakes:

• Using parentheses () instead of braces {}
• Using square brackets [] for initialization
• Not calculating size correctly

3. What will be the output of the following C code snippet that attempts to find unique triplets summing to zero?

int nums[] = {-1, 0, 1, 2, -1, -4};
int size = 6;
// Assume function threeSum prints triplets
threeSum(nums, size);
// Output format: each triplet printed as (a, b, c)

medium

A. No triplets found

B. (-4, 2, 2)
(-1, 0, 1)

C. (-1, -1, 2)
(-1, 0, 1)

D. (0, 0, 0)
(-1, 1, 2)

Solution

1. Step 1: Identify triplets that sum to zero

   From the array {-1, 0, 1, 2, -1, -4}, the unique triplets summing to zero are (-1, -1, 2) and (-1, 0, 1).

2. Step 2: Verify output matches these triplets

   (-1, -1, 2) (-1, 0, 1) lists these exact triplets; other options include invalid or non-existent triplets.

3. Final Answer:

   (-1, -1, 2) (-1, 0, 1) -> Option C

4. Quick Check:

   Unique triplets summing zero = (-1, -1, 2) (-1, 0, 1) [OK]

Hint: Check sums of triplets carefully for zero [OK]

Common Mistakes:

• Including duplicates like (-4, 2, 2)
• Assuming zero triplets when some exist
• Mixing up triplet elements

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

void threeSum(int* nums, int size) {
    for (int i = 0; i < size - 2; i++) {
        for (int j = i + 1; j < size - 1; j++) {
            for (int k = j + 1; k < size; k++) {
                if (nums[i] + nums[j] + nums[k] == 0) {
                    printf("(%d, %d, %d)\n", nums[i], nums[j], nums[k]);
                }
            }
        }
    }
}

medium

A. Does not avoid printing duplicate triplets

B. Loops have incorrect ranges causing out-of-bounds access

C. Sum condition should be nums[i] * nums[j] * nums[k] == 0

D. Missing return statement in void function

Solution

1. Step 1: Analyze triplet printing logic

   The code prints any triplet summing to zero but does not check for duplicates, so duplicate triplets may print multiple times.

2. Step 2: Check loop ranges and conditions

   Loop ranges are correct and sum condition is correct; void functions do not require return statements.

3. Final Answer:

   Does not avoid printing duplicate triplets -> Option A

4. Quick Check:

   Duplicate triplets not filtered = Does not avoid printing duplicate triplets [OK]

Hint: Check if duplicates are filtered in nested loops [OK]

Common Mistakes:

• Confusing sum with product condition
• Thinking void needs return
• Assuming loops cause out-of-bounds

5. Given the array int nums[] = {0, 0, 0, 0}; and size 4, what is the correct output of a Three Sum solution that finds all unique triplets summing to zero?

hard

A. No triplets found

B. (0, 0, 0)

C. (0, 0, 0)
(0, 0, 0)

D. (0, 0, 0)
(0, 0, 0)
(0, 0, 0)

Solution

1. Step 1: Identify unique triplets summing to zero

   All elements are zero, so the only unique triplet is (0, 0, 0).

2. Step 2: Avoid duplicates in output

   Even though multiple combinations exist, only one unique triplet should be printed.

3. Final Answer:

   (0, 0, 0) -> Option B

4. Quick Check:

   Unique triplet from zeros = (0, 0, 0) [OK]

Hint: Only print unique triplets, even if duplicates exist [OK]

Common Mistakes:

• Printing duplicate triplets multiple times
• Assuming no triplets because all zeros
• Confusing triplet uniqueness