Next Permutation of Array in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the next permutation by swapping the rightmost pair where the left is smaller.

DSA Python

for i in range(len(nums) - 2, -1, -1):
    if nums[i] [1] nums[i + 1]:
        break

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

A==

B!=

C<

D>

Attempts:

3 left

💡 Hint

Common Mistakes

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

We look for the first index from the right where the number is less than its next number to find the pivot.

2fill in blank

medium

Complete the code to find the element to swap with the pivot for the next permutation.

DSA Python

for j in range(len(nums) - 1, i, -1):
    if nums[j] [1] nums[i]:
        break

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

A<

B>

C==

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' causes swapping with a smaller element, which is incorrect.

We find the rightmost element greater than the pivot to swap with.

3fill in blank

hard

Fix the error in reversing the suffix after swapping to get the next permutation.

DSA Python

nums[i + 1:] = nums[i + 1:][1]

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

A[::-1]

B.reverse()

C.sort()

D.append()

Attempts:

3 left

💡 Hint

Common Mistakes

Using .reverse() returns None and causes assignment error.

We reverse the suffix using slicing with step -1 to get the smallest order after the pivot.

4fill in blank

hard

Fill both blanks to complete the next permutation function that handles the case when no pivot is found.

DSA Python

if i == -1:
    nums[1] = nums[2]

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

A.reverse()

B[:]

C[::-1]

D.sort()

Attempts:

3 left

💡 Hint

Common Mistakes

Using nums = nums[::-1] creates a new list and does not modify in place.

If no pivot is found, the array is in descending order, so we reverse it to get the smallest permutation.

5fill in blank

hard

Fill all three blanks to complete the next permutation function with pivot swap and suffix reversal.

DSA Python

nums[i], nums[j] = nums[1], nums[2]
nums[i + 1:] = nums[i + 1:][3]

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

A[j]

B[i]

C[::-1]

D.reverse()

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping with wrong indices or using .reverse() incorrectly.

Swap the pivot with the found element, then reverse the suffix to get the next permutation.

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 meaning

   The next permutation is the next arrangement that is just greater 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: Next permutation means next bigger order arrangement [OK]

Common Mistakes:

• Thinking next permutation is just sorting ascending
• Confusing next permutation with reversing the array
• Assuming next permutation always moves smallest element front

2. Which of the following Python code snippets correctly swaps two elements arr[i] and arr[j] in an array?

easy

A. arr[i], arr[j] = arr[j], arr[i]

B. arr[i] = arr[j]; arr[j] = arr[i]

C. swap(arr[i], arr[j])

D. arr[i] = arr[j]

Solution

1. Step 1: Recall Python tuple unpacking for swapping

   Python allows swapping two variables in one line using tuple unpacking.

2. Step 2: Check each option

   arr[i], arr[j] = arr[j], arr[i] correctly swaps using arr[i], arr[j] = arr[j], arr[i]. arr[i] = arr[j]; arr[j] = arr[i] overwrites arr[i] before assigning arr[j]. swap(arr[i], arr[j]) is invalid syntax without a defined function. arr[i] = arr[j] only assigns one element.

3. Final Answer:

   arr[i], arr[j] = arr[j], arr[i] -> Option A

4. Quick Check:

   Python swap uses tuple unpacking [OK]

Hint: Use tuple unpacking to swap in Python [OK]

Common Mistakes:

• Overwriting one element before swapping
• Using undefined swap function
• Assigning only one element instead of both

3. What is the output of the following code snippet?

arr = [1, 3, 2]
# Apply next permutation logic
# Step 1: Find first decreasing element from right (index 0)
# Step 2: Find element just larger than arr[0] from right (index 2)
# Step 3: Swap arr[0] and arr[2]
# Step 4: Reverse elements after index 0

arr[0], arr[2] = arr[2], arr[0]
arr[1:] = arr[1:][::-1]
print(arr)

medium

A. [3, 1, 2]

B. [2, 1, 3]

C. [1, 2, 3]

D. [2, 3, 1]

Solution

1. Step 1: Identify the first decreasing element from right

   From right, arr[1]=3 > arr[2]=2 (not decreasing), arr[0]=1 < arr[1]=3 (decreasing at index 0)

2. Step 2: Find element just larger than arr[0] from right

   From right, arr[2]=2 > arr[0]=1, so index 2 chosen

3. Step 3: Swap arr[0] and arr[2]

   Swap 1 and 2 -> arr becomes [2, 3, 1]

4. Step 4: Reverse elements after index 0

   Reverse arr[1:] -> arr[1:] = [3,1] reversed to [1,3]

5. Final Answer:

   [2, 1, 3] -> Option B

6. Quick Check:

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

Hint: Swap pivot with next bigger, then reverse suffix [OK]

Common Mistakes:

• Not reversing suffix after swap
• Swapping wrong elements
• Confusing increasing and decreasing order

4. The following code attempts to find the next permutation of arr = [3, 2, 1]. What is the error?

def next_permutation(arr):
    i = len(arr) - 2
    while i >= 0 and arr[i] >= arr[i + 1]:
        i -= 1
    if i >= 0:
        j = len(arr) - 1
        while arr[j] <= arr[i]:
            j -= 1
        arr[i], arr[j] = arr[j], arr[i]
    arr[i + 1:] = reversed(arr[i + 1:])

arr = [3, 2, 1]
next_permutation(arr)
print(arr)

medium

A. The while loop condition is incorrect

B. Swapping elements is done in wrong order

C. Using reversed() returns an iterator, not a list, causing wrong assignment

D. The function does not handle empty arrays

Solution

1. Step 1: Analyze the reversal line

   The line arr[i + 1:] = reversed(arr[i + 1:]) assigns an iterator to a list slice, which expects a list.

2. Step 2: Identify correct usage

   To assign reversed elements properly, convert iterator to list: arr[i + 1:] = list(reversed(arr[i + 1:])).

3. Final Answer:

   Using reversed() returns an iterator, not a list, causing wrong assignment -> Option C

4. Quick Check:

   reversed() needs list() for slice assignment [OK]

Hint: Use list(reversed(...)) when assigning slices [OK]

Common Mistakes:

• Assigning reversed() iterator directly to list slice
• Ignoring empty array edge case
• Swapping elements incorrectly

5. Given the array arr = [1, 5, 1], what is the next permutation after applying the standard algorithm? Consider duplicates carefully.

hard

A. [5, 1, 5]

B. [1, 1, 5]

C. [1, 5, 1]

D. [5, 1, 1]

Solution

1. Step 1: Find first decreasing element from right

   From right: arr[1]=5 > arr[2]=1 (not decreasing), arr[0]=1 < arr[1]=5 (decreasing at index 0)

2. Step 2: Find element just larger than arr[0] from right

   From right, arr[2]=1 not > 1, arr[1]=5 > 1, so index 1 chosen

3. Step 3: Swap arr[0] and arr[1]

   Swap 1 and 5 -> arr becomes [5, 1, 1]

4. Step 4: Reverse elements after index 0

   Reverse arr[1:] -> arr[1:] = [1,1] reversed is same [1,1]

5. Final Answer:

   [5, 1, 1] -> Option D

6. Quick Check:

   Duplicates handled, next permutation = [5,1,1] [OK]

Hint: Swap pivot with next bigger, reverse suffix even with duplicates [OK]

Common Mistakes:

• Ignoring duplicates when finding next bigger element
• Not reversing suffix after swap
• Assuming next permutation is sorted ascending always