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

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Permutations for a 3-Element Array

What is the output of the following C code that generates all permutations of the array [1, 2, 3]?

DSA C

#include <stdio.h>

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

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

int main() {
    int arr[] = {1, 2, 3};
    permute(arr, 0, 2);
    return 0;
}

A

1 2 3 
1 3 2 
2 1 3 
2 3 1 
3 2 1 
3 1 2 

B

1 2 3 
2 1 3 
1 3 2 
3 1 2 
2 3 1 
3 2 1 

C

1 2 3 
1 3 2 
2 1 3 
2 3 1 
3 1 2 
3 2 1 

D

1 2 3 
1 3 2 
3 2 1 
2 1 3 
2 3 1 
3 1 2 

💡 Hint

Trace the recursive calls and swaps carefully to see the order of permutations.

The code generates permutations by swapping elements starting from the current index. The order of permutations is lexicographic based on the swapping sequence.

🧠 Conceptual

intermediate

Number of Permutations for an Array

If an array has 5 distinct elements, how many unique permutations can be generated?

A 60

B 25

C 10

D 120

💡 Hint

Recall the factorial function for counting permutations.

The number of permutations of n distinct elements is n! (n factorial). For 5 elements, 5! = 120.

🔧 Debug

advanced

Identify the Bug in Permutation Code

What error will this code produce when generating permutations of [1, 2, 3]?

DSA C

#include <stdio.h>

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

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

int main() {
    int arr[] = {1, 2, 3};
    permute(arr, 0, 2);
    return 0;
}

A Prints incomplete permutations missing the last element

B Segmentation fault due to invalid memory access

C Infinite recursion causing stack overflow

D No output printed

💡 Hint

Check the loop boundary in the base case printing loop.

The base case loop uses 'i < end' instead of 'i <= end', so it misses printing the last element of the array.

📝 Syntax

advanced

Identify Syntax Error in Permutation Code

Which option contains a syntax error in the permutation function?

DSA C

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

A Incorrect printf format specifier

B Missing semicolon after swap(&arr[start], &arr[i])

C Missing braces around for loop

D Incorrect function return type

💡 Hint

Look carefully at the end of each statement inside the for loop.

The line 'swap(&arr[start], &arr[i])' is missing a semicolon, causing a syntax error.

🚀 Application

expert

Count Unique Permutations with Duplicate Elements

Given the array [1, 2, 2, 3], how many unique permutations can be generated?

A 12

B 24

C 6

D 18

💡 Hint

Use the formula for permutations of multiset: n! / (n1! * n2! * ...)

The array has 4 elements with '2' repeated twice. Number of unique permutations = 4! / 2! = 24 / 2 = 12.

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