Four Sum Problem All Unique Quadruplets 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'

AcompareInts

Bcmp

CcompareFunc

Dcompare

Attempts:

3 left

💡 Hint

Common Mistakes

Using a wrong comparator function name.

Passing NULL instead of a comparator.

The qsort function requires a comparison function pointer. 'compareInts' is the correct comparator for integers.

2fill in blank

medium

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

DSA C

for (int i = 0; i < numsSize - 3; i++) {
    if (i > 0 && nums[i] == nums[[1]]) continue;

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

Ai + 1

Bi

Ci - 1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with i+1 which is the next element.

Comparing with i itself causing no skip.

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 avoid infinite loops.

DSA C

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

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Aright

Bleft + 1

Cleft - 1

Dright - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with left+1 which may cause out-of-bound errors.

Comparing with left-1 which is invalid when left is 0.

We compare nums[left] with nums[right] to skip duplicates after finding a quadruplet.

4fill in blank

hard

Fill both blanks to correctly update pointers after finding a quadruplet.

DSA C

quadruplets[quadrupletCount][0] = nums[i];
quadruplets[quadrupletCount][1] = nums[j];
quadruplets[quadrupletCount][2] = nums[left];
quadruplets[quadrupletCount][3] = nums[right];
quadrupletCount++;
while (left < right && nums[left] == nums[left [1]]) left++;
while (left < right && nums[right] == nums[right [2]]) right--;

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

A+1

B-1

C+2

D-2

Attempts:

3 left

💡 Hint

Common Mistakes

Using -1 for left pointer which moves it backward incorrectly.

Using +1 for right pointer which moves it forward incorrectly.

After adding a quadruplet, move left pointer forward by 1 and right pointer backward by 1 to skip duplicates.

5fill in blank

hard

Fill all three blanks to complete the condition checking if the sum matches the target.

DSA C

int sum = nums[i] + nums[j] + nums[left] + nums[right];
if (sum [1] target) {
    left++;
} else if (sum [2] target) {
    right--;
} else if (sum == target) {
    // add quadruplet
    [3];
}

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

A<

B>

Csum = target

D/* add quadruplet code here */

Attempts:

3 left

💡 Hint

Common Mistakes

Using assignment '=' instead of comparison operators.

Not moving pointers correctly based on sum comparison.

If sum is less than target, move left pointer forward; if greater, move right pointer backward; if equal, add quadruplet.

Practice

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

easy

A. Find the maximum element in the array

B. Sort the array in ascending order

C. Find all unique quadruplets in an array that sum up to a target value

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 four numbers in an array whose sum equals a target value.

2. Step 2: Identify the expected output

   We want all unique sets of four numbers (quadruplets) that add up to the target.

3. Final Answer:

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

4. Quick Check:

   Four Sum = unique quadruplets with target sum [OK]

Hint: Look for four numbers adding to target, not just pairs [OK]

Common Mistakes:

• Confusing with two or three sum problems
• Ignoring uniqueness of quadruplets
• Trying to find maximum or minimum instead of sum

2. Which of the following code snippets correctly initializes the array and target for the Four Sum problem in C?

easy

A. int nums[] = [1, 0, -1, 0, -2, 2]; int target = 0;

B. int nums = {1, 0, -1, 0, -2, 2}; int target = 0;

C. int nums[] = (1, 0, -1, 0, -2, 2); int target = 0;

D. int nums[] = {1, 0, -1, 0, -2, 2}; int target = 0;

Solution

1. Step 1: Check array initialization syntax in C

   Arrays in C are initialized with curly braces {} and square brackets [] for declaration.

2. Step 2: Validate each option

   int nums[] = {1, 0, -1, 0, -2, 2}; int target = 0; uses correct syntax: int nums[] = { ... }; int nums = {1, 0, -1, 0, -2, 2}; int target = 0; misses brackets, C uses parentheses, D uses square brackets incorrectly.

3. Final Answer:

   int nums[] = {1, 0, -1, 0, -2, 2}; int target = 0; -> Option D

4. Quick Check:

   Correct C array syntax = int nums[] = {1, 0, -1, 0, -2, 2}; int target = 0; [OK]

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

Common Mistakes:

• Using parentheses () instead of braces {}
• Missing square brackets [] in declaration
• Using square brackets [] for initialization

3. Given the following C code snippet, what will be the printed output for the quadruplets found?

int nums[] = {1, 0, -1, 0, -2, 2};
int target = 0;
// Assume function fourSum returns all unique quadruplets
// and prints them as quadruplets separated by spaces
// Output format: a b c d\n for each quadruplet
fourSum(nums, 6, target);

medium

A. 1 0 -1 0
-2 2 0 0
-1 1 0 0

B. -2 -1 1 2
-2 0 0 2
-1 0 0 1

C. -2 -1 0 1
-2 0 0 2
-1 0 1 2

D. No quadruplets found

Solution

1. Step 1: Sort the array

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

2. Step 2: Find all unique quadruplets summing to 0

   Quadruplets are [-2, -1, 1, 2], [-2, 0, 0, 2], and [-1, 0, 0, 1].

3. Final Answer:

   -2 -1 1 2 -2 0 0 2 -1 0 0 1 -> Option B

4. Quick Check:

   Sum of each quadruplet = 0 [OK]

Hint: Sort array first, then use two pointers for pairs [OK]

Common Mistakes:

• Not sorting array before processing
• Including duplicate quadruplets
• Incorrect sum calculation

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

void fourSum(int* nums, int numsSize, int target) {
    for (int i = 0; i < numsSize - 3; i++) {
        for (int j = i + 1; j < numsSize - 2; j++) {
            int left = j + 1, right = numsSize - 1;
            while (left < right) {
                int sum = nums[i] + nums[j] + nums[left] + nums[right];
                if (sum == target) {
                    printf("%d %d %d %d\n", nums[i], nums[j], nums[left], nums[right]);
                    left++;
                    right--;
                } else if (sum < target) {
                    left++;
                } else {
                    right--;
                }
            }
        }
    }
}

medium

A. No check to skip duplicate quadruplets causing repeated outputs

B. Incorrect initialization of left and right pointers

C. Wrong loop ranges causing out-of-bounds access

D. Missing return statement in void function

Solution

1. Step 1: Analyze duplicate handling

   The code prints quadruplets but does not skip duplicates, so repeated quadruplets appear.

2. Step 2: Check other parts

   Left and right pointers are correctly initialized; loop ranges are valid; return statement is not needed in void function.

3. Final Answer:

   No check to skip duplicate quadruplets causing repeated outputs -> Option A

4. Quick Check:

   Duplicates not skipped = No check to skip duplicate quadruplets causing repeated outputs [OK]

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

Common Mistakes:

• Not skipping duplicates for i, j, left, right
• Incorrect pointer updates causing infinite loops
• Accessing array out of bounds

5. You are given an array nums = {2, 2, 2, 2, 2} and target 8. How many unique quadruplets sum to the target, and what are they?

hard

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

B. 5 quadruplets: all combinations of 2's

C. 0 quadruplets: no unique quadruplets

D. 4 quadruplets: different indices but same values

Solution

1. Step 1: Understand array and target

   All elements are 2, target is 8, so quadruplet sum must be 2+2+2+2=8.

2. Step 2: Count unique quadruplets

   Even though multiple indices exist, the quadruplet values are the same, so only one unique quadruplet exists: [2, 2, 2, 2].

3. Final Answer:

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

4. Quick Check:

   Sum 2+2+2+2=8 unique quadruplet [OK]

Hint: Same values quadruplet counts as one unique quadruplet [OK]

Common Mistakes:

• Counting quadruplets by indices, not unique values
• Assuming multiple quadruplets due to duplicates
• Ignoring uniqueness condition