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DSA C#programming~3 mins

Why Four Sum Problem All Unique Quadruplets in DSA C#?

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The Big Idea

What if you could find all groups of four numbers adding to your target without checking every single combination?

The Scenario

Imagine you have a big list of numbers and you want to find four numbers that add up to a special target number. Doing this by checking every possible group of four numbers by hand would take forever!

The Problem

Manually checking all groups means trying many combinations, which is very slow and easy to mess up. It's like trying to find four friends in a huge crowd who together have a certain total height, but you have to check every group one by one.

The Solution

The Four Sum problem uses smart ways to quickly find all unique groups of four numbers that add up to the target. It avoids repeating groups and skips unnecessary checks, making the search much faster and easier.

Before vs After
✗ Before
for (int i = 0; i < nums.Length; i++) {
  for (int j = i + 1; j < nums.Length; j++) {
    for (int k = j + 1; k < nums.Length; k++) {
      for (int l = k + 1; l < nums.Length; l++) {
        if (nums[i] + nums[j] + nums[k] + nums[l] == target) {
          // store quadruplet
        }
      }
    }
  }
}
✓ After
Array.Sort(nums);
for (int i = 0; i < nums.Length - 3; i++) {
  if (i > 0 && nums[i] == nums[i - 1]) continue;
  for (int j = i + 1; j < nums.Length - 2; j++) {
    if (j > i + 1 && nums[j] == nums[j - 1]) continue;
    int left = j + 1, right = nums.Length - 1;
    while (left < right) {
      int sum = nums[i] + nums[j] + nums[left] + nums[right];
      if (sum == target) {
        // store quadruplet
        left++; right--;
      } else if (sum < target) left++;
      else right--;
    }
  }
}
What It Enables

This method lets you quickly find all unique sets of four numbers that add up to a target, even in large lists, without wasting time on repeated or impossible groups.

Real Life Example

Think of a shopping list where you want to pick exactly four items that together cost a certain amount of money. This method helps find all possible combinations without checking every single group manually.

Key Takeaways

Manual checking of all quadruplets is slow and error-prone.

Sorting and using two pointers speeds up finding unique quadruplets.

Efficiently finds all unique groups that sum to the target.

Practice

(1/5)
1. What is the main goal of the Four Sum Problem in data structures and algorithms?
easy
A. Sort the array in ascending order
B. Count the number of pairs with a given sum
C. Find the maximum element in the array
D. Find all unique quadruplets in an array that sum up to a target value

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: Differentiate from other problems

    Unlike sorting or finding maximum, this problem focuses on quadruplets and their sums, not individual elements or pairs.
  3. Final Answer:

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

    Four Sum Problem = Find quadruplets with target sum [OK]
Hint: Look for four numbers adding to target, not sorting or max [OK]
Common Mistakes:
  • Confusing with sorting or max element problems
  • Looking for pairs instead of quadruplets
  • Ignoring uniqueness of quadruplets
2. Which of the following C# code snippets correctly initializes a list to store quadruplets for the Four Sum Problem?
easy
A. var quadruplets = new List<List<int>>();
B. var quadruplets = new List<int>();
C. var quadruplets = new Dictionary<int, int>();
D. var quadruplets = new int[4];

Solution

  1. Step 1: Identify the data structure for storing quadruplets

    Each quadruplet is a list of 4 integers, so a list of lists is needed to store multiple quadruplets.
  2. Step 2: Match with C# syntax

    Using List<List<int>> correctly represents multiple quadruplets, unlike a single list or dictionary.
  3. Final Answer:

    var quadruplets = new List<List<int>>(); -> Option A
  4. Quick Check:

    Use List of List for quadruplets [OK]
Hint: Use List<List<int>> to store multiple quadruplets [OK]
Common Mistakes:
  • Using List<int> for multiple quadruplets
  • Using Dictionary instead of List
  • Using fixed-size array for dynamic quadruplets
3. For the input array nums = [1, 0, -1, 0, -2, 2] and target 0, what does FourSum(nums, 0) return?
Assuming FourSum returns all unique quadruplets that sum to target.
medium
A. []
B. [[1, 0, -1, 0]]
C. [[-2, -1, 1, 2], [-2, 0, 0, 2], [-1, 0, 0, 1]]
D. [[0, 0, 0, 0]]

Solution

  1. Step 1: Sort and find quadruplets summing to 0

    Sorting nums gives [-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 uniqueness and completeness

    All quadruplets are unique and sum exactly to 0, matching the expected output.
  3. Final Answer:

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

    Unique quadruplets sum to 0 = [[-2, -1, 1, 2], [-2, 0, 0, 2], [-1, 0, 0, 1]] [OK]
Hint: Sort array first, then find quadruplets with two pointers [OK]
Common Mistakes:
  • Not sorting array before processing
  • Including duplicate quadruplets
  • Missing some valid quadruplets
4. Identify the bug in this C# snippet for the Four Sum Problem:
for (int i = 0; i < nums.Length - 3; i++) {
    for (int j = i + 1; j < nums.Length - 2; j++) {
        int left = j + 1, right = nums.Length - 1;
        while (left < right) {
            int sum = nums[i] + nums[j] + nums[left] + nums[right];
            if (sum == target) {
                result.Add(new List<int> { 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 duplicates in result
B. Incorrect initialization of left and right pointers
C. Wrong loop ranges for i and j causing index out of range
D. Missing sorting of the input array before processing

Solution

  1. Step 1: Analyze duplicate handling

    The code adds quadruplets without skipping duplicates, so repeated quadruplets can appear in the result.
  2. Step 2: Check other parts

    Left and right pointers are correctly initialized; loop ranges are valid; sorting is assumed done before this snippet.
  3. Final Answer:

    No check to skip duplicate quadruplets causing duplicates in result -> Option A
  4. Quick Check:

    Missing duplicate skip = No check to skip duplicate quadruplets causing duplicates in result [OK]
Hint: Always skip duplicates after finding quadruplets [OK]
Common Mistakes:
  • Not skipping duplicates for i, j, left, right
  • Assuming sorting is done inside the loop
  • Incorrect pointer updates causing infinite loops
5. Given the array nums = [2, 2, 2, 2, 2] and target 8, what is the correct output of the Four Sum Problem algorithm that finds all unique quadruplets?
medium
A. [[2, 2, 2, 2], [2, 2, 2, 2]]
B. [[2, 2, 2, 2]]
C. []
D. [[2, 2, 2]]

Solution

  1. Step 1: Understand the input and target

    All elements are 2, and target is 8, so quadruplets of four 2's sum to 8.
  2. Step 2: Identify unique quadruplets

    Only one unique quadruplet [2, 2, 2, 2] exists despite duplicates in input.
  3. Final Answer:

    [[2, 2, 2, 2]] -> Option B
  4. Quick Check:

    Only one unique quadruplet sums to 8 [OK]
Hint: Duplicates form one unique quadruplet if sum matches target [OK]
Common Mistakes:
  • Counting duplicate quadruplets multiple times
  • Returning empty list when quadruplet exists
  • Returning triplets instead of quadruplets
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