Generate All Permutations of Array in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Permutations for Array [1, 2]

What is the output of the following Java code that generates all permutations of the array [1, 2]?

DSA Java

import java.util.*;

public class Permutations {
    public static void permute(int[] arr, int start, List<List<Integer>> result) {
        if (start == arr.length) {
            List<Integer> perm = new ArrayList<>();
            for (int num : arr) perm.add(num);
            result.add(perm);
            return;
        }
        for (int i = start; i < arr.length; i++) {
            int temp = arr[start];
            arr[start] = arr[i];
            arr[i] = temp;
            permute(arr, start + 1, result);
            temp = arr[start];
            arr[start] = arr[i];
            arr[i] = temp;
        }
    }

    public static void main(String[] args) {
        int[] arr = {1, 2};
        List<List<Integer>> result = new ArrayList<>();
        permute(arr, 0, result);
        System.out.println(result);
    }
}

A [[1, 2], [2, 1]]

B [[1, 2]]

C [[2, 1], [1, 2]]

D [[2, 1]]

💡 Hint

Think about how swapping elements and recursion generate all possible orders.

The code swaps elements starting from the current index and recursively generates all permutations. For [1, 2], the permutations are [1, 2] and [2, 1].

🧠 Conceptual

intermediate

Number of Permutations for Array of Length 3

How many permutations are there for an array of length 3?

A 3

B 12

C 9

D 6

💡 Hint

Recall factorial definition: n! = n × (n-1) × ... × 1

The number of permutations of an array of length n is n factorial (n!). For n=3, 3! = 3 × 2 × 1 = 6.

🔧 Debug

advanced

Identify the Error in Permutation Code

What error does the following Java code produce when generating permutations of [1, 2, 3]? public static void permute(int[] arr, int start, List> result) { if (start == arr.length - 1) { List perm = new ArrayList<>(); for (int num : arr) perm.add(num); result.add(perm); return; } for (int i = start; i < arr.length; i++) { int temp = arr[start]; arr[start] = arr[i]; arr[i] = temp; permute(arr, start + 1, result); temp = arr[start]; arr[start] = arr[i]; arr[i] = temp; } }

A Base case off by one; last permutation is missing from result.

B No error; code runs correctly and generates all permutations.

C Missing base case causes infinite recursion and StackOverflowError.

D Swapping logic is incorrect causing ArrayIndexOutOfBoundsException.

💡 Hint

Check the base case condition carefully and how it affects the last recursion step.

No error; the code runs correctly and generates all permutations. The base case `if (start == arr.length - 1)` works equivalently to `if (start == arr.length)` because when `start` reaches the last index, the permutation is complete and no further swaps are needed (the loop at that level would be a no-op).

🚀 Application

advanced

Permutations of Array with Duplicate Elements

Given the array [1, 1, 2], which option shows the correct list of unique permutations generated by a proper algorithm?

A [[1, 1, 2], [2, 1, 1]]

B [[1, 1, 2], [1, 2, 1], [2, 1, 1], [1, 1, 2], [1, 2, 1], [2, 1, 1]]

C [[1, 1, 2], [1, 2, 1], [2, 1, 1]]

D [[1, 2, 1], [2, 1, 1]]

💡 Hint

Duplicates cause repeated permutations; a good algorithm avoids duplicates in the output.

The unique permutations of [1, 1, 2] are three: [1, 1, 2], [1, 2, 1], and [2, 1, 1]. Option C lists these without duplicates.

🧠 Conceptual

expert

Time Complexity of Generating All Permutations

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

A O(n!)

B O(n^2)

C O(2^n)

D O(n)

💡 Hint

Consider how many permutations exist and how much work is done per permutation.

There are n! permutations of an array of length n. Generating each permutation requires at least O(n) work, so total complexity is O(n × n!) which is simplified as O(n!).

Practice

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

easy

A. Sorting the array in ascending order

B. Listing all possible orders of the array elements

C. Finding the largest element in the array

D. Removing duplicate elements from the array

Solution

1. Step 1: Understand permutations

   Permutations are all possible ways to arrange elements in order.

2. Step 2: Apply to array

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

3. Final Answer:

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

4. Quick Check:

   Permutations = all orders [OK]

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

Common Mistakes:

• Confusing permutations with sorting
• Thinking permutations remove duplicates
• Mixing permutations with finding max element

2. Which Java method signature correctly defines a function to generate permutations of an integer array?

easy

A. public void generatePermutations(int[] arr, int start)

B. void generatePermutations(int arr[], int start) public

C. public int generatePermutations(int[] arr, int start)

D. public static generatePermutations(int[] arr, int start)

Solution

1. Step 1: Check method signature syntax

   In Java, method signature starts with access modifier, return type, method name, and parameters.

2. Step 2: Validate options

   public void generatePermutations(int[] arr, int start) correctly uses 'public void' and parameter types. void generatePermutations(int arr[], int start) public has wrong order. public int generatePermutations(int[] arr, int start) returns int but should be void. public static generatePermutations(int[] arr, int start) misses return type.

3. Final Answer:

   public void generatePermutations(int[] arr, int start) -> Option A

4. Quick Check:

   Correct Java method syntax = public void generatePermutations(int[] arr, int start) [OK]

Hint: Java method = access + return type + name + params [OK]

Common Mistakes:

• Missing return type in method signature
• Incorrect order of access modifier and return type
• Wrong return type for void methods

3. What is the output of the following Java code snippet?

public class PermTest {
  public static void permute(int[] arr, int start) {
    if (start == arr.length - 1) {
      for (int num : arr) System.out.print(num + " ");
      System.out.println();
      return;
    }
    for (int i = start; i < arr.length; i++) {
      int temp = arr[start];
      arr[start] = arr[i];
      arr[i] = temp;
      permute(arr, start + 1);
      temp = arr[start];
      arr[start] = arr[i];
      arr[i] = temp;
    }
  }
  public static void main(String[] args) {
    int[] arr = {1, 2};
    permute(arr, 0);
  }
}

medium

A. 1 2 \n2 1 \n

B. 1 2 \n1 2 \n

C. 2 1 \n1 2 \n

D. No output

Solution

1. Step 1: Trace recursive calls for arr = {1, 2}

   Start with start=0, swap arr[0] with arr[0] (no change), recurse with start=1, print 1 2. Then swap back, swap arr[0] with arr[1], arr becomes {2,1}, recurse start=1, print 2 1.

2. Step 2: Confirm output lines

   Two lines printed: '1 2' and '2 1', each followed by newline.

3. Final Answer:

   1 2 \n2 1 \n -> Option A

4. Quick Check:

   All permutations of [1,2] = 1 2 and 2 1 [OK]

Hint: Swap and recurse to generate permutations [OK]

Common Mistakes:

• Forgetting to swap back after recursion
• Printing before reaching last index
• Confusing order of swaps

4. Identify the error in this Java method for generating permutations:

public void permute(int[] arr, int start) {
  if (start == arr.length) {
    System.out.println(Arrays.toString(arr));
    return;
  }
  for (int i = start; i < arr.length; i++) {
    int temp = arr[start];
    arr[start] = arr[i];
    arr[i] = temp;
    permute(arr, start + 1);
  }
}

medium

A. Using Arrays.toString incorrectly

B. Incorrect base case condition

C. Missing swap back after recursive call

D. Wrong loop range in for loop

Solution

1. Step 1: Analyze swapping logic

   After swapping arr[start] and arr[i], the method calls recursion but does not swap back, so array stays changed for next iterations.

2. Step 2: Identify missing swap back

   To restore original array before next iteration, swap back is needed after recursive call.

3. Final Answer:

   Missing swap back after recursive call -> Option C

4. Quick Check:

   Swap back needed to restore array [OK]

Hint: Always swap back after recursion to restore array [OK]

Common Mistakes:

• Not swapping back after recursion
• Incorrect base case index
• Misusing Arrays.toString for printing

5. You want to generate all permutations of an array with possible duplicate elements, but avoid printing duplicate permutations. Which approach is best?

hard

A. Use a HashSet to store and print only unique permutations after generating all

B. Only swap elements if they are different values

C. Generate all permutations without any checks and print all

D. Sort the array first and skip swapping duplicate elements during recursion

Solution

1. Step 1: Understand duplicates in permutations

   Duplicates cause repeated permutations if not handled. Sorting helps group duplicates together.

2. Step 2: Skip swapping duplicates during recursion

   By sorting and skipping swaps of same values at the same recursion level, duplicates are avoided efficiently.

3. Final Answer:

   Sort the array first and skip swapping duplicate elements during recursion -> Option D

4. Quick Check:

   Sort + skip duplicates = unique permutations [OK]

Hint: Sort array and skip duplicate swaps to avoid repeats [OK]

Common Mistakes:

• Generating all then filtering duplicates (inefficient)
• Not sorting before recursion
• Only swapping if values differ misses some permutations