Generate All Permutations of Array in DSA C - Build from Scratch

Generate All Permutations of Array

📖 Scenario: You are working on a puzzle game where you need to find all possible orders of a set of unique numbers. This helps the game create different challenges by rearranging the numbers in every possible way.

🎯 Goal: Build a program in C that generates and prints all permutations of a given array of integers.

📋 What You'll Learn

Create an integer array called arr with the exact values 1, 2, 3

Create an integer variable called n that stores the size of arr

Write a function called swap that swaps two integers by their pointers

Write a recursive function called permute that generates all permutations of arr

Print each permutation on a separate line with elements separated by spaces

💡 Why This Matters

🌍 Real World

Generating permutations is useful in games, puzzles, and testing all possible arrangements of items.

💼 Career

Understanding recursion and permutations helps in algorithm design, problem solving, and coding interviews.

1

Create the array and its size

Create an integer array called arr with the exact values 1, 2, 3. Also create an integer variable called n that stores the size of arr.

DSA C

# Create the array arr with values 1, 2, 3
# Create the integer n to store the size of arr

Hint

Use int arr[] = {1, 2, 3}; to create the array and int n = 3; for its size.

2

Write a swap function

Write a function called swap that takes two integer pointers a and b and swaps the values they point to.

DSA C

#include 

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

// Write the swap function below

Hint

Use a temporary variable to hold one value while swapping.

3

Write the recursive permute function

Write a recursive function called permute that takes an integer start. It should generate all permutations of arr by swapping elements starting from index start to n - 1. When start equals n, print the current permutation.

DSA C

#include 

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

void swap(int *a, int *b) {
    int temp = *a;
    *a = *b;
    *b = temp;
}

// Write the permute function below

Hint

Use recursion and swap elements to generate permutations. Remember to swap back after recursive call.

4

Call permute and print all permutations

In the main function, call permute(0) to generate and print all permutations of arr.

DSA C

#include 

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

void swap(int *a, int *b) {
    int temp = *a;
    *a = *b;
    *b = temp;
}

void permute(int start) {
    if (start == n) {
        for (int i = 0; i < n; i++) {
            printf("%d", arr[i]);
            if (i < n - 1) printf(" ");
        }
        printf("\n");
        return;
    }
    for (int i = start; i < n; i++) {
        swap(&arr[start], &arr[i]);
        permute(start + 1);
        swap(&arr[start], &arr[i]);
    }
}

int main() {
    // Call permute with 0 to print all permutations
    
}

Hint

Call permute(0) inside main and return 0.

Practice

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

easy

A. Finding the maximum element in the array

B. Sorting the array in ascending order

C. Listing all possible orders of the array elements

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 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 arrange elements [OK]

Common Mistakes:

• Confusing permutations with sorting
• Thinking permutations remove duplicates
• Mixing permutations with finding max/min

2. Which of the following is the correct base case for a recursive function generating all permutations of an array arr of size n?

easy

A. if (start > n) { print(arr); return; }

B. if (start == n) { print(arr); return; }

C. if (n == 0) { return; }

D. if (start == 0) { print(arr); return; }

Solution

1. Step 1: Understand recursion base case

   The base case occurs when the start index reaches the array size, meaning a full permutation is formed.

2. Step 2: Match base case condition

   Condition start == n means all positions are fixed, so print the permutation and return.

3. Final Answer:

   if (start == n) { print(arr); return; } -> Option B

4. Quick Check:

   Base case when start equals array size [OK]

Hint: Base case: start index equals array length [OK]

Common Mistakes:

• Using start > n instead of start == n
• Printing when start == 0 (start of recursion)
• Missing base case leading to infinite recursion

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

void permute(int *arr, int start, int n) {
    if (start == n) {
        for (int i = 0; i < n; i++) printf("%d ", arr[i]);
        printf("\n");
        return;
    }
    for (int i = start; i < n; i++) {
        int temp = arr[start];
        arr[start] = arr[i];
        arr[i] = temp;
        permute(arr, start + 1, n);
        temp = arr[start];
        arr[start] = arr[i];
        arr[i] = temp;
    }
}

int main() {
    int arr[] = {1, 2};
    permute(arr, 0, 2);
    return 0;
}

medium

A. 1 2 \n 2 1 \n

B. 1 2 \n 1 2 \n

C. 2 1 \n 1 2 \n

D. No output

Solution

1. Step 1: Trace first call permute(arr, 0, 2)

   Start=0, loop i=0 to 1. Swap arr[0] with arr[0], call permute(arr,1,2).

2. Step 2: Trace second call permute(arr, 1, 2)

   Start=1, loop i=1 to 1. Swap arr[1] with arr[1], call permute(arr,2,2). Base case reached, print "1 2".

3. Step 3: Backtrack and swap arr[0] with arr[1]

   Now arr = {2,1}. Call permute(arr,1,2) again, print "2 1".

4. Final Answer:

   1 2 2 1 -> Option A

5. Quick Check:

   All permutations of [1,2] printed [OK]

Hint: Swap and backtrack to generate permutations [OK]

Common Mistakes:

• Not swapping back after recursive call
• Printing incomplete permutations
• Confusing loop indices

4. Identify the error in the following code snippet that generates permutations of an array arr of size n:

void permute(int *arr, int start, int n) {
    if (start == n) {
        for (int i = 0; i < n; i++) printf("%d ", arr[i]);
        printf("\n");
        return;
    }
    for (int i = start; i < n; i++) {
        int temp = arr[start];
        arr[start] = arr[i];
        arr[i] = temp;
        permute(arr, start + 1, n);
        // Missing swap back here
    }
}

medium

A. Missing swap back after recursive call causes incorrect permutations

B. Base case condition should be start > n

C. Loop should run from i = 0 to n

D. Printing inside base case is incorrect

Solution

1. Step 1: Understand swap and backtrack

   After swapping and recursive call, we must swap back to restore original array for next iteration.

2. Step 2: Identify missing swap back

   The comment shows swap back is missing, so array stays changed, causing wrong permutations.

3. Final Answer:

   Missing swap back after recursive call causes incorrect permutations -> Option A

4. Quick Check:

   Swap back needed to restore array [OK]

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

Common Mistakes:

• Forgetting to swap back after recursion
• Changing loop range incorrectly
• Wrong base case condition

5. Given an array arr = {1, 2, 2}, how many unique permutations will a correct algorithm generate that avoids duplicates?

hard

A. 6

B. 2

C. 4

D. 3

Solution

1. Step 1: Calculate total permutations without duplicates

   For 3 elements with two '2's, total permutations = 3! / 2! = 3.

2. Step 2: List unique permutations

   Unique permutations are: [1,2,2], [2,1,2], [2,2,1]. But careful with counting duplicates in generation.

3. Step 3: Consider correct algorithm avoiding duplicates

   Some algorithms generate 4 permutations including one duplicate if not careful, but correct unique count is 3.

4. Final Answer:

   3 -> Option D

5. Quick Check:

   Unique permutations = 3 [OK]

Hint: Use factorial formula with duplicates: n!/k! [OK]

Common Mistakes:

• Counting duplicates as unique permutations
• Ignoring duplicate elements in calculation
• Assuming all permutations are unique