Four Sum Problem All Unique Quadruplets in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort the input slice before processing.

DSA Go

sort.[1](nums)

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

AReverse

BInts

CStrings

DFloat64s

Attempts:

3 left

💡 Hint

Common Mistakes

Using sort.Reverse which requires a different type.

Using sort.Float64s on an int slice.

Using sort.Strings on an int slice.

We use sort.Ints to sort a slice of integers in Go.

2fill in blank

medium

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

DSA Go

if i > 0 && nums[i] == nums[[1]] {
    continue
}

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

Alen(nums)-1

Bi+1

C0

Di-1

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with nums[i+1] which may cause index out of range.

Comparing with nums[0] which is not the previous element.

Comparing with nums[len(nums)-1] which is unrelated.

We compare nums[i] with nums[i-1] to skip duplicates at index i.

3fill in blank

hard

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

DSA Go

for left < [1] {
    // loop body
}

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

Aright

Bleft

Clen(nums)

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Using left < left which never advances.

Using left < len(nums) which may cause wrong logic.

Using left < i which is unrelated.

The loop continues while left is less than right to avoid overlap.

4fill in blank

hard

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

DSA Go

left++
for left < right && nums[left] [1] nums[left-1] {
    left[2]
}

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

A==

B++

C!=

D--

Attempts:

3 left

💡 Hint

Common Mistakes

Using != instead of == causes skipping wrong elements.

Decrementing left pointer causes infinite loop.

We move left pointer forward skipping duplicates by checking equality and incrementing.

5fill in blank

hard

Fill all three blanks to append the quadruplet and move both pointers skipping duplicates.

DSA Go

result = append(result, []int[1], [2], [3], nums[right]})

left++
right--

for left < right && nums[left] == nums[left-1] {
    left++
}
for left < right && nums[right] == nums[right+1] {
    right--
}

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

Anums[i]

Bnums[j]

Cnums[left]

Dnums[right]

Attempts:

3 left

💡 Hint

Common Mistakes

Using nums[right] twice in the quadruplet.

Not including nums[j] or nums[i].

Appending wrong indices causing incorrect quadruplets.

We append the quadruplet with nums[i], nums[j], nums[left], nums[right] and then move pointers skipping duplicates.

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