Why Four Sum Problem All Unique Quadruplets in DSA Go?

The Big Idea

What if you could find all unique groups of four numbers adding to a target without checking every possibility by hand?

The Scenario

Imagine you have a list of numbers and want to find all groups of four numbers that add up to a specific target. Doing this by hand means checking every possible group of four, which is like trying to find four friends in a crowd who together have a certain total age.

The Problem

Manually checking every group is slow and tiring because the number of groups grows very fast as the list gets bigger. It's easy to miss some groups or count the same group more than once, making the process error-prone and frustrating.

The Solution

The Four Sum problem solution uses sorting and smart searching to quickly find all unique groups of four numbers that add up to the target. This method skips unnecessary checks and avoids duplicates, making the search fast and reliable.

Before vs After

✗ Before

for i := 0; i < len(nums); i++ {
  for j := i+1; j < len(nums); j++ {
    for k := j+1; k < len(nums); k++ {
      for l := k+1; l < len(nums); l++ {
        if nums[i]+nums[j]+nums[k]+nums[l] == target {
          // add quadruplet
        }
      }
    }
  }
}

✓ After

sort.Ints(nums)
for i := 0; i < len(nums)-3; i++ {
  if i > 0 && nums[i] == nums[i-1] { continue }
  for j := i+1; j < len(nums)-2; j++ {
    if j > i+1 && nums[j] == nums[j-1] { continue }
    left, right := j+1, len(nums)-1
    for left < right {
      sum := nums[i]+nums[j]+nums[left]+nums[right]
      if sum == target {
        // add quadruplet
        left++
        right--
        for left < right && nums[left] == nums[left-1] {
          left++
        }
        for left < right && nums[right] == nums[right+1] {
          right--
        }
      } else if sum < target {
        left++
      } else {
        right--
      }
    }
  }
}

What It Enables

This approach lets you quickly find all unique sets of four numbers that meet your target, even in large lists, without missing or repeating any group.

Real Life Example

Suppose you want to split a group of friends into teams of four where the total height of each team matches a certain number for a fun game. This method helps you find all such teams easily.

Key Takeaways

Manual checking is slow and error-prone for finding quadruplets.

Sorting and two-pointer technique speed up the search and avoid duplicates.

This method works well even for large lists and ensures all unique quadruplets are found.

Practice

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: Identify the goal

   The goal is not sorting or counting pairs but to find quadruplets that sum to the target without duplicates.

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 summing to target [OK]

Hint: Look for four numbers adding to target, no duplicates [OK]

Common Mistakes:

• Confusing with two or three sum problems
• Trying to sort only without finding quadruplets
• Counting pairs instead of quadruplets

2. Which of the following Go code snippets correctly initializes a slice to store quadruplets for the Four Sum problem?

easy

A. quadruplets := [][]int{}

B. quadruplets := [4]int{}

C. quadruplets := make([]int, 4)

D. quadruplets := map[int]int{}

Solution

1. Step 1: Understand data structure for quadruplets

   We need a slice of slices of integers to store multiple quadruplets, each quadruplet is a slice of 4 ints.

2. Step 2: Choose correct initialization

   quadruplets := [][]int{} initializes an empty slice of int slices, suitable for storing multiple quadruplets. quadruplets := [4]int{} is a fixed array of 4 ints, not multiple quadruplets. quadruplets := make([]int, 4) is a slice of 4 ints, not a slice of quadruplets. quadruplets := map[int]int{} is a map, not suitable here.

3. Final Answer:

   quadruplets := [][]int{} -> Option A

4. Quick Check:

   Slice of int slices = [][]int{} [OK]

Hint: Use [][]int{} for list of quadruplets [OK]

Common Mistakes:

• Using fixed arrays instead of slices
• Using map instead of slice
• Initializing slice with wrong dimensions

3. What will be the output of the following Go code snippet that finds unique quadruplets summing to 0?

nums := []int{1, 0, -1, 0, -2, 2}
target := 0
result := fourSum(nums, target)
fmt.Println(result)

// Assume fourSum is correctly implemented

medium

A. [[-2 -1 0 1] [0 0 1 2]]

B. [[-2 -1 1 2] [-2 0 0 2] [-1 0 0 1]]

C. [[1 0 -1 0] [-2 2 0 0]]

D. []

Solution

1. Step 1: Understand input and target

   The input array is [1, 0, -1, 0, -2, 2] and target sum is 0.

2. Step 2: Identify unique quadruplets summing to 0

   Possible unique quadruplets are [-2, -1, 1, 2], [-2, 0, 0, 2], and [-1, 0, 0, 1]. These sum exactly to 0 and are unique.

3. Final Answer:

   [[-2 -1 1 2] [-2 0 0 2] [-1 0 0 1]] -> Option B

4. Quick Check:

   Sum quadruplets = 0 [OK]

Hint: Sort array and find quadruplets summing to target [OK]

Common Mistakes:

• Including duplicate quadruplets
• Incorrect sums in quadruplets
• Missing some valid quadruplets

4. Given the following Go code snippet for the Four Sum problem, what is the main bug causing incorrect results?

func fourSum(nums []int, target int) [][]int {
    sort.Ints(nums)
    var res [][]int
    for i := 0; i < len(nums)-3; i++ {
        for j := i + 1; j < len(nums)-2; j++ {
            left, right := j+1, len(nums)-1
            for left < right {
                sum := nums[i] + nums[j] + nums[left] + nums[right]
                if sum == target {
                    res = append(res, []int{nums[i], nums[j], nums[left], nums[right]})
                    left++
                    right--
                } else if sum < target {
                    left++
                } else {
                    right--
                }
            }
        }
    }
    return res
}

medium

A. No checks to skip duplicate quadruplets causing duplicates in result

B. Incorrect sorting of the input array

C. Wrong initialization of left and right pointers

D. Using append incorrectly causing runtime error

Solution

1. Step 1: Analyze duplicate handling

   The code finds quadruplets but does not skip duplicates after finding a valid quadruplet, so duplicates appear in the result.

2. Step 2: Check other parts

   Sorting is correct, pointers are correctly initialized, and append usage is valid.

3. Final Answer:

   No checks to skip duplicate quadruplets causing duplicates in result -> Option A

4. Quick Check:

   Missing duplicate skip = duplicates in output [OK]

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

Common Mistakes:

• Not skipping duplicates for i, j, left, right
• Incorrect pointer updates after finding quadruplet
• Assuming sorting alone removes duplicates

5. You want to optimize the Four Sum solution to avoid unnecessary computations. Which approach best reduces the search space while ensuring all unique quadruplets are found?

hard

A. Sort the array and use two nested loops with two pointers, skipping duplicates at each step

B. Use four nested loops to check all quadruplets without sorting

C. Sort the array and use a hash map to store sums of pairs, then find complements

D. Use recursion without sorting or duplicate checks

Solution

1. Step 1: Understand optimization goals

   We want to reduce time complexity and avoid duplicates efficiently.

2. Step 2: Evaluate approaches

   Sort the array and use two nested loops with two pointers, skipping duplicates at each step is common but still O(n^3). Use four nested loops to check all quadruplets without sorting is brute force O(n^4). Sort the array and use a hash map to store sums of pairs, then find complements uses a hash map to store pair sums and find complements, reducing complexity and handling duplicates well. Use recursion without sorting or duplicate checks is inefficient without sorting or duplicate checks.

3. Final Answer:

   Sort the array and use a hash map to store sums of pairs, then find complements -> Option C

4. Quick Check:

   Hash map pairs + complements = optimized search [OK]

Hint: Use hash map of pair sums to find quadruplets faster [OK]

Common Mistakes:

• Using brute force four loops
• Ignoring duplicate skipping
• Not sorting before processing