Three Sum Problem All Unique Triplets in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort the array before processing.

DSA C

qsort(nums, numsSize, sizeof(int), [1]);

Drag options to blanks, or click blank then click option'

Acmpfunc

BcompareInts

CcompareFunc

Dcompare

Attempts:

3 left

💡 Hint

Common Mistakes

Using a wrong function name for the comparator.

Not passing the comparator function pointer.

The qsort function requires a comparison function pointer. 'cmpfunc' is the standard name used here.

2fill in blank

medium

Complete the code to skip duplicate elements in the outer loop.

DSA C

for (int i = 0; i < numsSize - 2; i++) {
    if (i > 0 && nums[i] == nums[1]) continue;

Drag options to blanks, or click blank then click option'

Ai + 1

Bi - 1

Ci

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with the next element instead of previous.

Not checking i > 0 before accessing i-1.

To skip duplicates, compare current element with the previous one at index i-1.

3fill in blank

hard

Fix the error in the while loop condition to move the left pointer correctly.

DSA C

while (left < right && nums[left] == nums[1]) left++;

Drag options to blanks, or click blank then click option'

Aright

Bleft - 1

Cleft

Dleft + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with left - 1 which is behind the pointer.

Comparing with left itself causing infinite loop.

To skip duplicates, compare nums[left] with nums[left + 1] and then move left forward.

4fill in blank

hard

Fill both blanks to correctly move the right pointer and skip duplicates.

DSA C

while (left < right && nums[right] == nums[1]) [2];

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Aright - 1

Bright + 1

Cright--

Dright++

Attempts:

3 left

💡 Hint

Common Mistakes

Using right + 1 which goes out of bounds.

Incrementing right pointer instead of decrementing.

Compare nums[right] with nums[right - 1] and decrement right pointer to skip duplicates.

5fill in blank

hard

Fill all three blanks to correctly add the triplet to the result array.

DSA C

result[*returnSize] = malloc(sizeof(int) * 3);
result[*returnSize][0] = nums[[1]];
result[*returnSize][1] = nums[[2]];
result[*returnSize][2] = nums[[3]];
(*returnSize)++;

Drag options to blanks, or click blank then click option'

Ai

Bleft

Cright

Dk

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong indices like k which is undefined here.

Mixing up left and right indices.

The triplet consists of nums[i], nums[left], and nums[right].

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