Generate All Permutations of Array in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to swap two elements in the array.

DSA C

void swap(int *a, int *b) {
    int temp = *a;
    *a = *b;
    *b = [1];
}

Drag options to blanks, or click blank then click option'

Aa

Btemp

Cb

D*temp

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning *a to *b instead of using temp.

Using incorrect pointer dereferencing.

The swap function uses a temporary variable to hold the value of *a before assigning *b to *a, then assigns temp to *b to complete the swap.

2fill in blank

medium

Complete the code to print the current permutation of the array.

DSA C

void printArray(int arr[], int size) {
    for (int i = 0; i < [1]; i++) {
        printf("%d ", arr[i]);
    }
    printf("\n");
}

Drag options to blanks, or click blank then click option'

A0

Barr

Ci

Dsize

Attempts:

3 left

💡 Hint

Common Mistakes

Using the array name instead of size in the loop condition.

Looping beyond the array bounds.

The loop should run from 0 to size-1 to print all elements of the array.

3fill in blank

hard

Fix the error in the recursive call to generate permutations.

DSA C

void permute(int arr[], int l, int r) {
    if (l == r) {
        printArray(arr, r + 1);
    } else {
        for (int i = l; i <= r; i++) {
            swap(&arr[l], &arr[i]);
            permute(arr, [1], r);
            swap(&arr[l], &arr[i]);
        }
    }
}

Drag options to blanks, or click blank then click option'

Al + 1

Bi + 1

Cr + 1

Dl

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong index in the recursive call.

Not incrementing the left index causing infinite recursion.

The recursive call should move the left index forward by one to generate permutations of the remaining elements.

4fill in blank

hard

Fill both blanks to complete the for loop and swap calls correctly.

DSA C

for (int i = [1]; i <= [2]; i++) {
    swap(&arr[l], &arr[i]);
    permute(arr, l + 1, r);
    swap(&arr[l], &arr[i]);
}

Drag options to blanks, or click blank then click option'

Al

Br

C0

Dsize

Attempts:

3 left

💡 Hint

Common Mistakes

Starting loop from 0 instead of l.

Looping beyond r causing out-of-bounds errors.

The loop variable i starts at l and goes up to r to swap elements for generating permutations.

5fill in blank

hard

Fill all three blanks to complete the main function setup and call permutation.

DSA C

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

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A{1, 2, 3}

B0

Cn - 1

D{3, 2, 1}

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect array initialization.

Passing wrong indices to permute function.

The array is initialized with {1, 2, 3}, and permute is called with indices 0 and n-1 to cover the full array.

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