Next Permutation of Array in DSA Go - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output after applying next permutation?

Given the array [1, 3, 2], what is the array after applying the next permutation algorithm once?

DSA Go

arr := []int{1, 3, 2}
// Apply next permutation algorithm here
// Print arr after modification
fmt.Println(arr)

A [1, 2, 3]

B [2, 1, 3]

C [3, 1, 2]

D [1, 3, 2]

💡 Hint

Look for the first pair from the right where the left number is smaller than the right number.

The next permutation of [1, 3, 2] is [2, 1, 3]. So option B is correct.

❓ Predict Output

intermediate

What is the output after next permutation on sorted descending array?

Given the array [3, 2, 1], what is the array after applying the next permutation algorithm once?

DSA Go

arr := []int{3, 2, 1}
// Apply next permutation algorithm here
// Print arr after modification
fmt.Println(arr)

A [2, 3, 1]

B [3, 1, 2]

C [3, 2, 1]

D [1, 2, 3]

💡 Hint

If the array is in descending order, the next permutation is the ascending order.

The array [3, 2, 1] is the highest permutation. The next permutation resets it to the lowest permutation [1, 2, 3].

🔧 Debug

advanced

Identify the error in this next permutation code snippet

What error will this Go code produce when trying to find the next permutation of [3, 2, 1]? func nextPermutation(nums []int) { i := len(nums) - 2 for i >= 0 && nums[i] >= nums[i+1] { i-- } j := len(nums) - 1 for i >= 0 && nums[j] <= nums[i] { j-- } if i >= 0 { nums[i], nums[j] = nums[j], nums[i] } reverse(nums[i+1:]) } func reverse(nums []int) { for i, j := 0, len(nums)-1; i < j; i, j = i+1, j-1 { nums[i], nums[j] = nums[j], nums[i] } }

DSA Go

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

func reverse(nums []int) {
  for i, j := 0, len(nums)-1; i < j; i, j = i+1, j-1 {
    nums[i], nums[j] = nums[j], nums[i]
  }
}

A Runtime panic: index out of range

B Logical error: does not reverse the suffix

C Compilation error: missing return statement

D No error, outputs [1, 2, 3]

💡 Hint

Check what happens when i becomes -1 and is used as an index.

If the array is already the highest permutation, i becomes -1. Then nums[i] causes an index out of range panic.

🧠 Conceptual

advanced

What is the time complexity of the next permutation algorithm?

Consider the standard next permutation algorithm that finds the next lexicographical permutation of an array of length n. What is its time complexity?

A O(n)

B O(n^2)

C O(log n)

D O(1)

💡 Hint

Think about how many times the array is scanned and reversed.

The algorithm scans the array from the end once to find the pivot, then scans again to find the swap element, and finally reverses a suffix. All these steps are linear in n.

🚀 Application

expert

How many unique permutations can be generated from [1, 2, 2] using next permutation?

Starting from the sorted array [1, 2, 2], how many unique permutations will the next permutation algorithm generate before returning to the original array?

A 5

B 6

C 3

D 4

💡 Hint

Count all unique permutations of [1, 2, 2].

The unique permutations of [1, 2, 2] are: [1, 2, 2], [2, 1, 2], and [2, 2, 1]. So there are 3 unique permutations.

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