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.