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DSA C++programming~10 mins

Four Sum Problem All Unique Quadruplets in DSA C++ - Interactive Practice

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to sort the input vector before processing.

DSA C++
std::sort(nums.[1](), nums.end());
Drag options to blanks, or click blank then click option'
Arbegin
Brend
Cend
Dbegin
Attempts:
3 left
💡 Hint
Common Mistakes
Using rbegin() or rend() which reverse the order and break the logic.
Forgetting to sort the vector before searching quadruplets.
2fill in blank
medium

Complete the code to skip duplicate elements for the first pointer i.

DSA C++
if (i > 0 && nums[i] == nums[i - 1]) continue; // skip duplicates for [1]
Drag options to blanks, or click blank then click option'
Ai
Bj
Cleft
Dright
Attempts:
3 left
💡 Hint
Common Mistakes
Skipping duplicates for the wrong pointer variable.
Not checking i > 0 before accessing nums[i - 1].
3fill in blank
hard

Fix the error in the two-pointer while loop condition to avoid infinite loops.

DSA C++
while (left [1] right) {
Drag options to blanks, or click blank then click option'
A<
B>
C>=
D<=
Attempts:
3 left
💡 Hint
Common Mistakes
Using '<=' causes pointers to overlap and may cause infinite loops.
Using '>' or '>=' reverses the logic and breaks the loop.
4fill in blank
hard

Fill both blanks to correctly move the left and right pointers after finding a quadruplet.

DSA C++
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+
D-
Attempts:
3 left
💡 Hint
Common Mistakes
Using '-' with left pointer or '+' with right pointer causes wrong movement.
Not skipping duplicates leads to repeated quadruplets.
5fill in blank
hard

Fill all three blanks to complete the quadruplet insertion and pointer updates.

DSA C++
result.push_back({nums[i], nums[j], nums[left], nums[right]});
left[1];
right[2];
while (left < right && nums[left] == nums[left [3]]) left++;
Drag options to blanks, or click blank then click option'
A++
B--
C+1
D-1
Attempts:
3 left
💡 Hint
Common Mistakes
Using -- for left pointer or ++ for right pointer causes wrong pointer movement.
Not skipping duplicates after pointer updates causes repeated quadruplets.

Practice

(1/5)
1. What is the main goal of the Four Sum Problem in data structures and algorithms?
easy
A. Find all unique quadruplets in an array that sum up to a target value
B. Sort the array in ascending order
C. Find the maximum element in the array
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 all unique sets of four numbers in an array that add up to a specific target sum.
  2. Step 2: Identify the goal

    The goal is to find quadruplets (groups of four numbers) that sum exactly to the target, without duplicates.
  3. Final Answer:

    Find all unique quadruplets in an array that sum up to a target value -> Option A
  4. Quick Check:

    Four Sum Problem = Find quadruplets summing to target [OK]
Hint: Look for four numbers adding to target without repeats [OK]
Common Mistakes:
  • Confusing with pair or triplet sum problems
  • Ignoring uniqueness of quadruplets
  • Trying to sort only without finding quadruplets
2. Which of the following C++ code snippets correctly initializes a vector to store quadruplets for the Four Sum Problem?
easy
A. std::vector<int> quadruplets(4);
B. std::vector quadruplets;
C. std::vector<int> quadruplets = {4};
D. std::vector> quadruplets;

Solution

  1. Step 1: Understand the data structure needed

    We need to store multiple quadruplets, each is a vector of 4 integers, so a vector of vector<int> is suitable.
  2. Step 2: Check each option

    std::vector quadruplets; declares a vector of vector<int>, perfect for storing multiple quadruplets. std::vector<int> quadruplets(4); creates a single vector of size 4, not multiple quadruplets. std::vector<int> quadruplets = {4}; initializes a vector with one element 4, incorrect. std::vector> quadruplets; uses pairs, which store only two integers, not four.
  3. Final Answer:

    std::vector quadruplets; -> Option B
  4. Quick Check:

    Use vector of vector<int> for quadruplets [OK]
Hint: Use vector of vector<int> to store multiple quadruplets [OK]
Common Mistakes:
  • Using vector<int> for multiple quadruplets
  • Using pairs instead of quadruplets
  • Initializing vector with wrong size or values
3. What will be the output of the following C++ code snippet for input array {1, 0, -1, 0, -2, 2} and target 0?
std::vector<int> nums = {1, 0, -1, 0, -2, 2};
int target = 0;
// Assume fourSum function returns all unique quadruplets
auto result = fourSum(nums, target);
for (auto &quad : result) {
    for (int num : quad) std::cout << num << " ";
    std::cout << std::endl;
}
medium
A. [-2 -1 1 2] [-2 0 0 2] [-1 0 0 1]
B. [-2 -1 0 1] [-1 0 0 2]
C. [-2 -2 1 3] [-1 0 0 1]
D. No quadruplets found

Solution

  1. Step 1: Sort the array and find quadruplets summing to target 0

    Sorted array: [-2, -1, 0, 0, 1, 2]. The unique quadruplets summing to 0 are [-2, -1, 1, 2], [-2, 0, 0, 2], and [-1, 0, 0, 1].
  2. Step 2: Verify output matches quadruplets

    [-2 -1 1 2] [-2 0 0 2] [-1 0 0 1] lists exactly these three quadruplets. Options A, C, and D have incorrect quadruplets or sums.
  3. Final Answer:

    [-2 -1 1 2] [-2 0 0 2] [-1 0 0 1] -> Option A
  4. Quick Check:

    Unique quadruplets summing to 0 = [-2 -1 1 2] [-2 0 0 2] [-1 0 0 1] [OK]
Hint: Sort array first, then find quadruplets summing to target [OK]
Common Mistakes:
  • Not sorting array before searching
  • Including duplicate quadruplets
  • Missing some valid quadruplets
4. Identify the bug in the following C++ code snippet for the Four Sum Problem:
std::vector fourSum(std::vector<int>& nums, int target) {
    std::vector result;
    std::sort(nums.begin(), nums.end());
    for (int i = 0; i < nums.size(); ++i) {
        for (int j = i + 1; j < nums.size(); ++j) {
            int left = j + 1, right = nums.size() - 1;
            while (left < right) {
                int sum = nums[i] + nums[j] + nums[left] + nums[right];
                if (sum == target) {
                    result.push_back({nums[i], nums[j], nums[left], nums[right]});
                    ++left; --right;
                } else if (sum < target) {
                    ++left;
                } else {
                    --right;
                }
            }
        }
    }
    return result;
}
medium
A. Returns result before completing loops
B. Incorrect sorting of the input array
C. Wrong initialization of left and right pointers
D. Does not skip duplicate quadruplets, causing repeated results

Solution

  1. Step 1: Analyze duplicate handling in loops

    The code does not skip duplicates for i, j, left, or right indices, so it can add the same quadruplet multiple times.
  2. Step 2: Check other parts for errors

    Sorting is correct, left and right pointers are initialized properly, and return is after loops. So these are not bugs.
  3. Final Answer:

    Does not skip duplicate quadruplets, causing repeated results -> Option D
  4. Quick Check:

    Missing duplicate checks = Does not skip duplicate quadruplets, causing repeated results [OK]
Hint: Always skip duplicates after finding quadruplets [OK]
Common Mistakes:
  • Not skipping duplicates for i, j, left, right
  • Returning result too early
  • Incorrect pointer initialization
5. Given an array nums = {2, 2, 2, 2, 2} and target 8, how many unique quadruplets does the Four Sum algorithm find, and what are they?
hard
A. 10 quadruplets: including duplicates
B. 5 quadruplets: all combinations of 2s
C. 1 quadruplet: [2, 2, 2, 2]
D. 0 quadruplets: no quadruplets sum to 8

Solution

  1. Step 1: Understand the input and target

    All elements are 2, and target is 8. Sum of four 2s is 8, so quadruplet [2, 2, 2, 2] is valid.
  2. Step 2: Consider uniqueness

    Even though multiple combinations exist, they are duplicates. The algorithm returns only one unique quadruplet.
  3. Final Answer:

    1 quadruplet: [2, 2, 2, 2] -> Option C
  4. Quick Check:

    All 2s sum to 8 once uniquely [OK]
Hint: Duplicates count as one unique quadruplet [OK]
Common Mistakes:
  • Counting duplicate quadruplets multiple times
  • Assuming no quadruplets because of repeated numbers
  • Confusing number of combinations with unique quadruplets
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