Next Permutation of Array in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the first index from the end where the array is not in descending order.

DSA Go

i := len(nums) - 2
for i >= 0 && nums[i] [1] nums[i+1] {
    i--
}

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

A>=

B>

C<

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '>' causes incorrect pivot detection.

Using '>=' includes equal elements which is not correct here.

We look for the first index from the right where nums[i] is greater than nums[i+1]. This helps find the pivot for the next permutation.

2fill in blank

medium

Complete the code to find the index j to swap with i, where nums[j] is just larger than nums[i].

DSA Go

j := len(nums) - 1
for j >= 0 && nums[j] [1] nums[i] {
    j--
}

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

A>=

B<=

C<

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>=' causes wrong swap index.

Using '<=' or '<' in the wrong direction causes infinite loops.

We find the first element from the right that is greater than nums[i] to swap and get the next permutation.

3fill in blank

hard

Fix the error in the code to reverse the subarray after index i correctly.

DSA Go

start, end := i+1, len(nums)-1
for start < end {
    nums[start], nums[end] = nums[end], nums[1]  // swap
    start++
    end--
}

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

A[start]

B[end]

C[i]

D[0]

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping with nums[end] again causes no change.

Using wrong indices causes incorrect reversal.

We swap nums[start] and nums[end] to reverse the subarray. The blank must be nums[start] to complete the swap.

4fill in blank

hard

Fill both blanks to complete the swap and reverse steps for next permutation.

DSA Go

if i >= 0 {
    j := len(nums) - 1
    for nums[j] [1] nums[i] {
        j--
    }
    nums[i], nums[j] = nums[2], nums[j]
}

start, end := i+1, len(nums)-1
for start < end {
    nums[start], nums[end] = nums[end], nums[start]
    start++
    end--
}

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

A>

B<

C[i]

D[j]

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '>' in the loop.

Swapping with wrong index causes wrong permutation.

The first blank needs '>' to find the correct j. The second blank is '[i]' to swap nums[i] with nums[j].

5fill in blank

hard

Fill all three blanks to complete the full next permutation function.

DSA Go

func nextPermutation(nums []int) {
    i := len(nums) - 2
    for i >= 0 && nums[i] [1] nums[i+1] {
        i--
    }
    if i >= 0 {
        j := len(nums) - 1
        for nums[j] [2] nums[i] {
            j--
        }
        nums[i], nums[j] = nums[3], nums[j]
    }
    start, end := i+1, len(nums)-1
    for start < end {
        nums[start], nums[end] = nums[end], nums[start]
        start++
        end--
    }
}

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

A>

B<

C[i]

D[start]

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong comparison operators in loops.

Swapping with wrong indices.

The blanks are: '>' to find pivot i, '<' to find j, and '[i]' to swap nums[i] with nums[j].

Practice

1. What does the next permutation of an array represent?

easy

A. The next lexicographically greater arrangement of the array elements

B. The array sorted in ascending order

C. The array reversed

D. The smallest element moved to the front

Solution

1. Step 1: Understand the concept of permutation order

   Permutations can be ordered lexicographically like words in a dictionary.

2. Step 2: Define next permutation

   The next permutation is the next arrangement that is just larger than the current one in this order.

3. Final Answer:

   The next lexicographically greater arrangement of the array elements -> Option A

4. Quick Check:

   Next permutation = next bigger arrangement [OK]

Hint: Think of next permutation as next dictionary word [OK]

Common Mistakes:

• Confusing next permutation with sorting
• Thinking it always reverses the array
• Assuming it moves smallest element to front

2. Which Go code snippet correctly swaps two elements nums[i] and nums[j] in an array?

easy

A. swap(nums[i], nums[j])

B. temp = nums[i]; nums[i] = nums[j]; nums[j] = temp

C. nums[i], nums[j] = nums[j], nums[i]

D. nums[i] = nums[j]; nums[j] = nums[i]

Solution

1. Step 1: Recall Go multiple assignment syntax

   Go allows swapping two variables in one line using multiple assignment.

2. Step 2: Identify correct swap syntax

   The syntax nums[i], nums[j] = nums[j], nums[i] swaps values correctly without temporary variable.

3. Final Answer:

   nums[i], nums[j] = nums[j], nums[i] -> Option C

4. Quick Check:

   Go swap uses multiple assignment [OK]

Hint: Use Go's multiple assignment for swapping [OK]

Common Mistakes:

• Using undefined swap function
• Forgetting temporary variable in manual swap
• Assigning both to same value causing loss

3. What is the output of the following Go code that finds the next permutation of nums := []int{1, 3, 2}?

func nextPermutation(nums []int) {
    i := len(nums) - 2
    for i >= 0 && nums[i] >= nums[i+1] {
        i--
    }
    if i >= 0 {
        j := len(nums) - 1
        for nums[j] <= nums[i] {
            j--
        }
        nums[i], nums[j] = nums[j], nums[i]
    }
    left, right := i+1, len(nums)-1
    for left < right {
        nums[left], nums[right] = nums[right], nums[left]
        left++
        right--
    }
}

func main() {
    nums := []int{1, 3, 2}
    nextPermutation(nums)
    fmt.Println(nums)
}

medium

A. [1, 2, 3]

B. [2, 1, 3]

C. [1, 3, 2]

D. [3, 1, 2]

Solution

1. Step 1: Find first decreasing element from right

   Starting from index 1 (value 3), check if nums[i] >= nums[i+1]. At i=1, 3 >= 2 true, i=0, 1 >= 3 false, so i=0.

2. Step 2: Find element just larger than nums[i]

   From right, find j where nums[j] > nums[i] (nums[0]=1). j=2 (value 2) > 1, swap nums[0] and nums[2]. Array becomes [2,3,1]. Reverse suffix from i+1=1 to end: reverse [3,1] to [1,3]. Final array: [2,1,3].

3. Final Answer:

   [2, 1, 3] -> Option B

4. Quick Check:

   Next permutation of [1,3,2] = [2,1,3] [OK]

Hint: Swap and reverse suffix after first decreasing index [OK]

Common Mistakes:

• Not reversing suffix after swap
• Swapping wrong elements
• Stopping at first decreasing element without swap

4. Identify the bug in this Go code snippet for next permutation:

func nextPermutation(nums []int) {
    i := len(nums) - 1
    for i > 0 && nums[i] <= nums[i-1] {
        i--
    }
    if i > 0 {
        j := len(nums) - 1
        for nums[j] <= nums[i-1] {
            j--
        }
        nums[i-1], nums[j] = nums[j], nums[i-1]
    }
    left, right := i, len(nums)-1
    for left < right {
        nums[left], nums[right] = nums[right], nums[left]
        left++
        right--
    }
}

medium

A. The initial index i should start at len(nums) - 2, not len(nums) - 1

B. The swap should be between nums[i] and nums[j], not nums[i-1] and nums[j]

C. The reversal should start from i-1, not i

D. The loop conditions should use < instead of >

Solution

1. Step 1: Check starting index for finding first decreasing element

   The algorithm must start checking from second last element (len(nums)-2) to compare with next element.

2. Step 2: Identify impact of starting at len(nums)-1

   Starting at len(nums)-1 compares last element with out-of-range index, causing incorrect behavior or skipping needed checks.

3. Final Answer:

   The initial index i should start at len(nums) - 2, not len(nums) - 1 -> Option A

4. Quick Check:

   Start i at len(nums)-2 for correct next permutation [OK]

Hint: Start i at second last index to find decreasing pair [OK]

Common Mistakes:

• Starting i at last index causing index errors
• Swapping wrong elements
• Incorrect reversal range

5. Given an array nums := []int{3, 2, 1}, what will be the next permutation after applying the standard next permutation algorithm? Explain why.

hard

A. [3, 2, 1] because it is already the highest permutation

B. [3, 1, 2] because swapping first and last elements gives next permutation

C. [2, 3, 1] because reversing the array gives next permutation

D. [1, 2, 3] because the array is in descending order, so next permutation is the lowest ascending order

Solution

1. Step 1: Check if array is in descending order

   Array [3, 2, 1] is strictly descending, meaning it is the highest permutation possible.

2. Step 2: Apply next permutation rule for highest permutation

   When highest permutation is reached, next permutation is the lowest ascending order, which is [1, 2, 3].

3. Final Answer:

   [1, 2, 3] because the array is in descending order, so next permutation is the lowest ascending order -> Option D

4. Quick Check:

   Descending array next permutation = ascending array [OK]

Hint: If descending, next permutation is ascending sorted array [OK]

Common Mistakes:

• Thinking next permutation is same as current if descending
• Swapping wrong elements instead of reversing whole array
• Not resetting to lowest permutation after highest