Four Sum Problem All Unique Quadruplets in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Four Sum Unique Quadruplets Function

What is the output of the following code that finds all unique quadruplets in the array that sum to the target 0?

DSA C

void fourSum(int* nums, int numsSize, int target) {
    // Assume nums is sorted
    for (int i = 0; i < numsSize - 3; i++) {
        if (i > 0 && nums[i] == nums[i-1]) continue;
        for (int j = i + 1; j < numsSize - 2; j++) {
            if (j > i + 1 && nums[j] == nums[j-1]) continue;
            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]);
                    while (left < right && nums[left] == nums[left+1]) left++;
                    while (left < right && nums[right] == nums[right-1]) right--;
                    left++;
                    right--;
                } else if (sum < target) {
                    left++;
                } else {
                    right--;
                }
            }
        }
    }
}

int main() {
    int nums[] = {1, 0, -1, 0, -2, 2};
    int size = sizeof(nums)/sizeof(nums[0]);
    // Sort the array
    for (int i = 0; i < size - 1; i++) {
        for (int j = 0; j < size - i - 1; j++) {
            if (nums[j] > nums[j+1]) {
                int temp = nums[j];
                nums[j] = nums[j+1];
                nums[j+1] = temp;
            }
        }
    }
    fourSum(nums, size, 0);
    return 0;
}

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

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

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

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

💡 Hint

Remember to sort the array first and skip duplicates to get unique quadruplets.

The code sorts the array and uses two nested loops with two pointers to find quadruplets summing to zero. It skips duplicates to ensure unique quadruplets. The correct quadruplets are [-2, -1, 1, 2], [-2, 0, 0, 2], and [-1, 0, 0, 1].

🧠 Conceptual

intermediate

Understanding the Role of Sorting in Four Sum

Why is sorting the array important before finding all unique quadruplets that sum to a target?

A Sorting is not necessary; it only makes the code run slower.

B Sorting helps to easily skip duplicate elements and use two pointers efficiently.

C Sorting changes the target sum to zero automatically.

D Sorting removes negative numbers from the array.

💡 Hint

Think about how duplicates and two pointers work.

Sorting allows the algorithm to skip duplicates easily by comparing adjacent elements and enables the two-pointer technique to find pairs efficiently.

🔧 Debug

advanced

Identify the Bug in Quadruplet Finding Code

What error will occur if the inner while loops that skip duplicates are removed from the four sum code?

DSA C

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]);
        // Duplicate skipping loops removed here
        left++;
        right--;
    } else if (sum < target) {
        left++;
    } else {
        right--;
    }
}

A The code will miss some quadruplets that sum to the target.

B The code will cause a segmentation fault.

C The code will not compile due to syntax errors.

D The code will print duplicate quadruplets multiple times.

💡 Hint

Think about what skipping duplicates prevents.

Without skipping duplicates, the same quadruplet can be printed multiple times because the pointers move without checking for repeated values.

❓ Predict Output

advanced

Output with Negative Target Sum

What is the output of the four sum function when the input array is [-3, -1, 0, 2, 4, 5] and the target is 2?

DSA C

int nums[] = {-3, -1, 0, 2, 4, 5};
int size = sizeof(nums)/sizeof(nums[0]);
// Sort the array
for (int i = 0; i < size - 1; i++) {
    for (int j = 0; j < size - i - 1; j++) {
        if (nums[j] > nums[j+1]) {
            int temp = nums[j];
            nums[j] = nums[j+1];
            nums[j+1] = temp;
        }
    }
}
fourSum(nums, size, 2);

A [[-3, -1, 2, 4]]

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

C [[-3, -1, 2, 4], [-3, 0, 2, 3]]

D [[-3, -1, 0, 6], [-1, 0, 2, 1]]

💡 Hint

Check sums carefully and remember the array is sorted.

Only the quadruplet [-3, -1, 2, 4] sums to 2. Other options include numbers not in the array or sums that don't match.

🧠 Conceptual

expert

Time Complexity of Four Sum Algorithm

What is the time complexity of the standard four sum algorithm that uses sorting and two pointers to find all unique quadruplets?

A O(n^3)

B O(n^4)

C O(n^2 log n)

D O(n^2)

💡 Hint

Consider the nested loops and two-pointer approach.

The algorithm uses two nested loops (O(n^2)) and inside uses two pointers moving linearly (O(n)), resulting in O(n^3) time complexity.

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