Generate All Permutations of Array in DSA Java - Time & Space Complexity

Time Complexity: Generate All Permutations of Array

O(n * n!)

Understanding Time Complexity

When we generate all permutations of an array, we want to know how the work grows as the array gets bigger.

We ask: How many steps does it take to list every possible order of the array?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public void permute(int[] nums, int start) {
    if (start == nums.length) {
        // print or store the permutation
        return;
    }
    for (int i = start; i < nums.length; i++) {
        swap(nums, start, i);
        permute(nums, start + 1);
        swap(nums, start, i); // backtrack
    }
}
    

This code generates all possible orders (permutations) of the array by swapping elements and exploring each choice recursively.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls combined with a loop that swaps elements.
• How many times: The recursion explores every possible order, which is the factorial of the array size (n!).

How Execution Grows With Input

As the array size grows, the number of permutations grows very fast because each new element can be placed in many positions.

Input Size (n)	Approx. Operations
3	6 (3 x 2 x 1)
5	120 (5 x 4 x 3 x 2 x 1)
7	5,040 (7 x 6 x 5 x 4 x 3 x 2 x 1)

Pattern observation: The work grows factorially, which means it increases much faster than doubling or squaring.

Final Time Complexity

Time Complexity: O(n * n!)

This means the time grows very fast because we generate every possible order, and each order takes linear time to process.

Common Mistake

[X] Wrong: "Generating permutations is just like looping through the array once, so it should be O(n)."

[OK] Correct: Each element can be in many positions, so the number of permutations multiplies quickly, not just adds up.

Interview Connect

Understanding this time complexity helps you explain why some problems with many possible answers take longer to solve, showing your grasp of algorithm limits.

Self-Check

"What if we only wanted permutations of length k from the array instead of all n elements? How would the time complexity change?"

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