Generate All Permutations of Array in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to swap two elements in the array.

DSA C#

void Swap(int[] arr, int i, int j) {
    int temp = arr[i];
    arr[i] = arr[1];
    arr[j] = temp;
}

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

A[j]

B[i]

C[0]

D[j + 1]

Attempts:

3 left

💡 Hint

Common Mistakes

Using arr[i] instead of arr[j] causes no change.

Using an invalid index like j + 1 causes errors.

To swap elements at positions i and j, assign arr[i] = arr[j].

2fill in blank

medium

Complete the code to generate permutations by recursion.

DSA C#

void Permute(int[] arr, int start, int end) {
    if (start == end) {
        PrintArray(arr);
    } else {
        for (int i = start; i <= [1]; i++) {
            Swap(arr, start, i);
            Permute(arr, start + 1, end);
            Swap(arr, start, i);
        }
    }
}

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

Aarr.Length

Bstart

Cend

Dend - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using arr.Length causes index out of range errors.

Using end - 1 misses the last element.

The loop runs from start to end inclusive to cover all permutations.

3fill in blank

hard

Fix the error in the base case condition to correctly print permutations.

DSA C#

void Permute(int[] arr, int start, int end) {
    if (start [1] end) {
        PrintArray(arr);
    } else {
        for (int i = start; i <= end; i++) {
            Swap(arr, start, i);
            Permute(arr, start + 1, end);
            Swap(arr, start, i);
        }
    }
}

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

A==

B<

C>

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' or '>' causes infinite recursion or missed base case.

Using '!=' prints permutations too early.

The base case is when start equals end, meaning one permutation is ready to print.

4fill in blank

hard

Fill both blanks to correctly print the array elements separated by spaces.

DSA C#

void PrintArray(int[] arr) {
    for (int i = 0; i < arr.Length; i++) {
        Console.Write(arr[i][1]);
    }
    Console.WriteLine([2]);
}

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

A + " "

B" "

C", "

D"\n"

Attempts:

3 left

💡 Hint

Common Mistakes

Using ", " adds commas which are not needed.

Using "\n" in WriteLine adds extra new lines.

We add a space after each element and print a space line break after the loop.

5fill in blank

hard

Fill all three blanks to call the permutation function correctly for an array.

DSA C#

int[] arr = {1, 2, 3};
int n = arr[1];
Permute(arr, [2], [3]);

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

A.Length

B0

Carr.Length - 1

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 as start index skips the first element.

Using arr.Length as end index causes out of range error.

We get the array length, start from index 0, and end at last index (length - 1).

Practice

1. What does the process of generating all permutations of an array mean?

easy

A. Removing duplicate elements from the array

B. Sorting the array in ascending order

C. Listing all possible orders of the array elements

D. Finding the largest element in the array

Solution

1. Step 1: Understand permutations

   Permutations are all possible ways to arrange elements in order.

2. Step 2: Apply to array elements

   Generating all permutations means listing every possible order of the array elements.

3. Final Answer:

   Listing all possible orders of the array elements -> Option C

4. Quick Check:

   Permutations = All orders [OK]

Hint: Permutations = all ways to reorder elements [OK]

Common Mistakes:

• Confusing permutations with sorting
• Thinking permutations remove duplicates
• Assuming permutations find max/min

2. Which of the following C# method signatures is correct for a function that generates all permutations of an integer array and returns a list of integer arrays?

easy

A. void GeneratePermutations(int[] nums)

B. List<int> GeneratePermutations(int nums)

C. int GeneratePermutations(List<int> nums)

D. List GeneratePermutations(int[] nums)

Solution

1. Step 1: Identify input and output types

   The input is an integer array, so parameter type is int[]. The output is all permutations, so a list of int arrays is suitable.

2. Step 2: Match method signature

   List GeneratePermutations(int[] nums) returns List and takes int[] as input, matching the requirement.

3. Final Answer:

   List GeneratePermutations(int[] nums) -> Option D

4. Quick Check:

   Input int[], output List of int[] [OK]

Hint: Input array, output list of arrays for permutations [OK]

Common Mistakes:

• Using void return type when output is needed
• Wrong parameter types like List<int> instead of int[]
• Returning int instead of list of arrays

3. What is the output of the following C# code snippet?

var result = GeneratePermutations(new int[] {1, 2});
foreach(var perm in result) {
    Console.WriteLine(string.Join(",", perm));
}

List GeneratePermutations(int[] nums) {
    var list = new List();
    void Backtrack(int start) {
        if (start == nums.Length) {
            list.Add((int[])nums.Clone());
            return;
        }
        for (int i = start; i < nums.Length; i++) {
            int temp = nums[start];
            nums[start] = nums[i];
            nums[i] = temp;
            Backtrack(start + 1);
            temp = nums[start];
            nums[start] = nums[i];
            nums[i] = temp;
        }
    }
    Backtrack(0);
    return list;
}

medium

A. 1,2
2,1

B. 1,2
1,2

C. 2,1
1,2

D. 2,1
2,1

Solution

1. Step 1: Understand the backtracking process

   The code swaps elements starting from index 0 and generates permutations by recursive calls.

2. Step 2: Trace permutations generated

   For input [1,2], permutations are [1,2] and [2,1]. The code prints these in order.

3. Final Answer:

   1,2 2,1 -> Option A

4. Quick Check:

   Permutations of [1,2] = 1,2 and 2,1 [OK]

Hint: Trace swaps and recursive calls carefully [OK]

Common Mistakes:

• Ignoring swap back to original after recursion
• Printing duplicates due to missing cloning
• Confusing order of permutations

4. The following code attempts to generate all permutations of an array but produces incorrect results. What is the main bug?

void GeneratePermutations(int[] nums, int start, List result) {
    if (start == nums.Length) {
        result.Add(nums);
        return;
    }
    for (int i = start; i < nums.Length; i++) {
        int temp = nums[start];
        nums[start] = nums[i];
        nums[i] = temp;
        GeneratePermutations(nums, start + 1, result);
        temp = nums[start];
        nums[start] = nums[i];
        nums[i] = temp;
    }
}

medium

A. The array is added directly without cloning, causing all entries to be the same

B. The swap back after recursion is missing

C. The base case condition is incorrect

D. The loop should start from 0 instead of start

Solution

1. Step 1: Analyze how results are stored

   The code adds the original array reference to the result list without cloning.

2. Step 2: Understand effect of adding reference

   Since the array is modified later, all entries in result point to the same array, causing incorrect results.

3. Final Answer:

   The array is added directly without cloning, causing all entries to be the same -> Option A

4. Quick Check:

   Must clone array before adding to result [OK]

Hint: Always clone array before adding to results [OK]

Common Mistakes:

• Forgetting to clone array before adding
• Missing swap back causing wrong permutations
• Incorrect base case stopping too early

5. You want to generate all unique permutations of an integer array that may contain duplicates, such as [1, 1, 2]. Which approach correctly handles duplicates to avoid repeated permutations?

hard

A. Only swap elements if they are different values

B. Sort the array first, then skip swapping duplicate elements during recursion

C. Generate all permutations without any checks and then remove duplicates at the end

D. Use a HashSet to store all permutations after generating them

Solution

1. Step 1: Sort the array to group duplicates

   Sorting helps detect duplicates easily during recursion.

2. Step 2: Skip swapping duplicate elements

   During recursion, skip swapping if the element is the same as a previously swapped one at the same recursion level to avoid repeated permutations.

3. Final Answer:

   Sort the array first, then skip swapping duplicate elements during recursion -> Option B

4. Quick Check:

   Sort + skip duplicates during recursion = unique permutations [OK]

Hint: Sort array and skip duplicates while swapping [OK]

Common Mistakes:

• Generating all then removing duplicates (inefficient)
• Not sorting array before recursion
• Only swapping if values differ ignores position importance