Generate All Permutations of Array in DSA C# - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a permutation in the context of an array?

A permutation is a way of arranging all the elements of an array in every possible order without missing any combination.

beginner

What is the base case in a recursive function that generates all permutations of an array?

The base case occurs when the starting index reaches the length of the array, meaning all positions are fixed and one complete permutation is formed.

intermediate

Why do we swap elements during permutation generation?

Swapping elements helps to fix one element at the current position and then recursively generate permutations for the remaining positions, ensuring all unique orders are explored.

intermediate

What is the time complexity of generating all permutations of an array of size n?

The time complexity is O(n!), because there are n factorial possible permutations to generate and print.

advanced

How can you avoid duplicate permutations if the array contains repeated elements?

You can avoid duplicates by sorting the array first and skipping swaps that would produce the same element in the current position, or by using a hash set to track used elements at each recursion level.

What does the swap operation do in permutation generation?

A Duplicates the array

B Fixes an element at the current position to explore permutations

C Sorts the array elements

D Deletes an element from the array

Swapping fixes one element at the current position so the function can recursively generate permutations for the rest.

When does the recursive function stop generating permutations?

A When the array is empty

B After one permutation is generated

C When the starting index equals the array length

D When the array is sorted

The recursion stops when the starting index reaches the array length, meaning a full permutation is formed.

How many permutations does an array of 3 elements have?

A 6

B 3

C 9

D 12

An array of 3 elements has 3! = 6 permutations.

Which of these is NOT a step in generating permutations recursively?

A Backtracking by swapping back

B Recursing on the next index

C Swapping elements

D Sorting the array at each step

Sorting the array at each step is not required; swapping, recursion, and backtracking are the main steps.

What is the main challenge when the array has duplicate elements?

A Generating duplicate permutations

B Sorting the array

C Swapping elements

D Printing permutations

Duplicate elements can cause the algorithm to generate the same permutation multiple times.

Explain how recursion and swapping work together to generate all permutations of an array.

Think about how you fix one element and then permute the rest.

Describe how you would modify the permutation algorithm to avoid duplicates when the array contains repeated elements.

Consider how to prevent repeating the same element in the same position.

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