Generate All Permutations of Array in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to swap two elements in the array.

DSA Go

func swap(arr []int, i, j int) {
    temp := arr[i]
    arr[i] = arr[[1]]
    arr[j] = temp
}

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

Aj

Bi

C0

Dlen(arr)

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong index like 0 or len(arr) causes runtime errors.

Swapping arr[i] with itself does nothing.

We swap arr[i] with arr[j], so the blank must be 'j'.

2fill in blank

medium

Complete the code to generate permutations by recursion.

DSA Go

func permute(arr []int, start int) {
    if start == len(arr) {
        fmt.Println(arr)
        return
    }
    for i := start; i < len(arr); i++ {
        swap(arr, start, [1])
        permute(arr, start+1)
        swap(arr, start, i)
    }
}

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

Astart

Bi

Clen(arr)

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping with start instead of i causes no change.

Using len(arr) as index causes out-of-range error.

We swap arr[start] with arr[i] to generate permutations.

3fill in blank

hard

Fix the error in the base case condition to stop recursion correctly.

DSA Go

func permute(arr []int, start int) {
    if start [1] len(arr) {
        fmt.Println(arr)
        return
    }
    for i := start; i < len(arr); i++ {
        swap(arr, start, i)
        permute(arr, start+1)
        swap(arr, start, i)
    }
}

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

A<=

B<

C>

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' causes the base case to never trigger.

Using '>' or '<=' causes incorrect stopping condition.

The base case is when start equals length of array, meaning all positions fixed.

4fill in blank

hard

Fill both blanks to create a function that returns all permutations as a slice of slices.

DSA Go

func permuteAll(arr []int, start int, result *[][]int) {
    if start == len(arr) {
        temp := make([]int, len(arr))
        copy(temp, arr)
        *result = append(*result, temp)
        return
    }
    for i := start; i [1] len(arr); i++ {
        swap(arr, start, i)
        permuteAll(arr, start+1, result)
        swap(arr, start, [2])
    }
}

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

A<

B<=

Ci

Dstart

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' causes index out of range error.

Swapping back with start instead of i breaks permutation correctness.

The loop runs while i < len(arr), and swap back uses i to restore original order.

5fill in blank

hard

Fill all three blanks to create a map of permutations with their string keys.

DSA Go

func permuteMap(arr []int, start int, result map[string][]int) {
    if start == len(arr) {
        key := fmt.Sprint(arr)
        result[[1]] = [2]
        return
    }
    for i := start; i < len(arr); i++ {
        swap(arr, start, i)
        permuteMap(arr, start+1, result)
        swap(arr, start, i)
    }
}

// Usage:
// result := make(map[string][]int)
// permuteMap(arr, 0, result)

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

Akey

Barr

Cappend([]int(nil), arr...)

Darr[:]

Attempts:

3 left

💡 Hint

Common Mistakes

Using arr directly as value causes all map entries to point to same slice.

Using arr[:] is a slice but still references original array.

We use 'key' as map key, and a copy of arr as value to avoid mutation; key is used again as map key.

Practice

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

easy

A. Finding the largest 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: Compare options

   Only Listing all possible orders of the array elements describes listing all possible orders of 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 element

2. Which Go code snippet correctly starts a recursive function to generate permutations of an integer slice arr?

easy

A. func permute(arr int[], start int) {}

B. func permute(arr []int, start int) {}

C. func permute(arr []int, start string) {}

D. func permute(arr []int) {}

Solution

1. Step 1: Check Go function syntax

   Go uses []int for slices and int for integers; parameters must match expected types.

2. Step 2: Validate parameters for permutation

   We need the array slice and an integer index to track recursion start.

3. Final Answer:

   func permute(arr []int, start int) {} -> Option B

4. Quick Check:

   Correct Go slice and int params [OK]

Hint: Use []int for slices and int for indices [OK]

Common Mistakes:

• Using incorrect slice syntax like int[]
• Using wrong parameter types like string for index
• Omitting necessary parameters for recursion

3. What will be the output of this Go code snippet that generates permutations of []int{1, 2} and prints them?

func permute(arr []int, start int) {
  if start == len(arr) {
    fmt.Println(arr)
    return
  }
  for i := start; i < len(arr); i++ {
    arr[start], arr[i] = arr[i], arr[start]
    permute(arr, start+1)
    arr[start], arr[i] = arr[i], arr[start]
  }
}

func main() {
  permute([]int{1, 2}, 0)
}

medium

A. Compilation error

B. [2 1]
[1 2]

C. [1 2]

D. [1 2]
[2 1]

Solution

1. Step 1: Trace recursion for arr = [1, 2]

   Start=0: swap arr[0] with arr[0] -> [1,2], recurse start=1; then swap arr[0] with arr[1] -> [2,1], recurse start=1.

2. Step 2: Print when start == len(arr)

   At start=2, print current arr. So prints [1 2] then [2 1].

3. Final Answer:

   [1 2] [2 1] -> Option D

4. Quick Check:

   All permutations printed in order [OK]

Hint: Swap and backtrack to print all permutations [OK]

Common Mistakes:

• Not swapping back after recursion
• Printing before full permutation formed
• Confusing order of printed permutations

4. Identify the bug in this Go function intended to generate all permutations of arr:

func permute(arr []int, start int) {
  if start == len(arr) {
    fmt.Println(arr)
    return
  }
  for i := 0; i < len(arr); i++ {
    arr[start], arr[i] = arr[i], arr[start]
    permute(arr, start+1)
    arr[start], arr[i] = arr[i], arr[start]
  }
}

medium

A. Loop should start from 'start' instead of 0

B. Missing base case for recursion

C. Swapping elements is incorrect

D. Function should return a slice, not print

Solution

1. Step 1: Analyze loop range

   The loop runs from i=0 to len(arr), but it should start from 'start' to avoid duplicate swaps.

2. Step 2: Understand impact of loop start

   Starting from 0 causes repeated permutations and incorrect swapping order.

3. Final Answer:

   Loop should start from 'start' instead of 0 -> Option A

4. Quick Check:

   Loop start index = start [OK]

Hint: Loop from start index to avoid duplicates [OK]

Common Mistakes:

• Looping from zero causing repeated permutations
• Not swapping back after recursion
• Ignoring base case correctness

5. Given an array []int{1, 2, 3}, how many unique permutations will be generated by a correct permutation function, and why?

hard

A. 6, because 3! (3 factorial) = 6 permutations

B. 3, because there are 3 elements

C. 9, because 3 * 3 = 9 combinations

D. 1, because the array is fixed

Solution

1. Step 1: Calculate permutations count for 3 elements

   Number of permutations = factorial of number of elements = 3! = 3*2*1 = 6.

2. Step 2: Understand why other options are incorrect

   3 is just count of elements, 9 is product not permutations, 1 is incorrect as array can reorder.

3. Final Answer:

   6, because 3! (3 factorial) = 6 permutations -> Option A

4. Quick Check:

   Permutations of 3 elements = 6 [OK]

Hint: Permutations count = factorial of array length [OK]

Common Mistakes:

• Confusing permutations count with array length
• Using multiplication instead of factorial
• Thinking only one arrangement exists