Next Permutation of Array in DSA Rust - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the purpose of the Next Permutation algorithm?

It finds the next lexicographically greater permutation of an array of numbers. If no such permutation exists, it rearranges the array to the lowest possible order (sorted ascending).

intermediate

In the Next Permutation algorithm, what is the significance of finding the 'pivot' index?

The pivot is the first index from the right where the current element is less than the next element. It marks where the next permutation can be generated by swapping and rearranging.

intermediate

Why do we reverse the suffix after swapping in the Next Permutation algorithm?

Because the suffix after the pivot is in descending order, reversing it makes it the smallest ascending order, ensuring the next permutation is just one step greater.

beginner

Show the Rust code snippet to reverse a slice of an array in place.

arr[i..j].reverse(); // This reverses the elements from index i up to but not including j in place.

beginner

What happens if the entire array is in descending order when applying Next Permutation?

It means the array is the highest permutation. The algorithm reverses the entire array to the lowest permutation (ascending order).

What is the first step in finding the next permutation of an array?

A Find the pivot where arr[i] < arr[i+1] from right to left

B Reverse the entire array

C Swap the first and last elements

D Sort the array in ascending order

The algorithm starts by finding the pivot index from the right where the element is less than its next element.

After finding the pivot, what is the next step in the Next Permutation algorithm?

A Find the rightmost successor to the pivot and swap them

B Reverse the entire array

C Sort the array in descending order

D Do nothing

You find the rightmost element greater than the pivot and swap them to get the next bigger permutation.

Why do we reverse the suffix after swapping in the Next Permutation algorithm?

A To remove duplicates

B To sort the entire array

C To get the largest lexicographical order for the suffix

D To get the smallest lexicographical order for the suffix

Reversing the suffix ensures it is in ascending order, making the permutation just one step bigger.

What does the Next Permutation algorithm return if the array is already the highest permutation?

A An error

B The same array

C The lowest permutation (sorted ascending)

D The highest permutation again

If no next permutation exists, the algorithm resets the array to the lowest permutation by reversing it.

In Rust, which method reverses a slice of an array in place?

A .clone()

B .reverse()

C .swap()

D .sort()

The .reverse() method reverses elements of a slice in place.

Explain the steps of the Next Permutation algorithm and why each step is necessary.

Think about how to get the next bigger arrangement of numbers.

Write a Rust function that modifies an array to its next permutation in place.

Use slice methods like .reverse() and simple loops.

Practice

1. What does the next_permutation algorithm do to an array of numbers?

easy

A. It rearranges the array to the next lexicographically greater permutation.

B. It sorts the array in ascending order.

C. It reverses the array.

D. It finds the maximum element in the array.

Solution

1. Step 1: Understand the purpose of next_permutation

   The algorithm finds the next arrangement of elements that is just larger than the current one in dictionary order.

2. Step 2: Compare with other options

   Sorting or reversing changes order but does not guarantee the next lexicographical permutation. Finding max element is unrelated.

3. Final Answer:

   It rearranges the array to the next lexicographically greater permutation. -> Option A

4. Quick Check:

   Next permutation = next lexicographically greater order [OK]

Hint: Next permutation means next bigger order arrangement [OK]

Common Mistakes:

• Confusing next permutation with sorting
• Thinking it reverses the array always
• Assuming it finds max element

2. Which of the following Rust code snippets correctly swaps two elements in a vector nums at indices i and j?

easy

A. nums.swap(i, j) -> ();

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

C. nums.swap(i, j);

D. nums.swap_elements(i, j);

Solution

1. Step 1: Recall Rust vector swap method

   Rust's standard Vec<T> has a method swap(i, j) to exchange elements at indices i and j.

2. Step 2: Check other options for correctness

   nums[i] = nums[j]; nums[j] = nums[i]; incorrectly assigns values without temporary storage, causing wrong swap. nums.swap_elements(i, j); uses a non-existent method. nums.swap(i, j) -> (); has invalid syntax.

3. Final Answer:

   nums.swap(i, j); -> Option C

4. Quick Check:

   Correct swap method = nums.swap(i, j) [OK]

Hint: Use nums.swap(i, j) to swap elements in Rust vector [OK]

Common Mistakes:

• Assigning without temp variable causing overwrite
• Using non-existent methods
• Incorrect syntax with arrows or extra symbols

3. What is the output of the following Rust code that finds the next permutation of nums = vec![1, 3, 2]?

let mut nums = vec![1, 3, 2];
if next_permutation(&mut nums) {
    println!("{:?}", nums);
} else {
    println!("No next permutation");
}

fn next_permutation(nums: &mut Vec<i32>) -> bool {
    let n = nums.len();
    if n < 2 { return false; }
    let mut i = n - 2;
    while i != usize::MAX && nums[i] >= nums[i + 1] {
        if i == 0 { break; }
        i -= 1;
    }
    if i == usize::MAX || nums[i] >= nums[i + 1] {
        nums.reverse();
        return false;
    }
    let mut j = n - 1;
    while nums[j] <= nums[i] {
        j -= 1;
    }
    nums.swap(i, j);
    nums[i + 1..].reverse();
    true
}

medium

A. [2, 1, 3]

B. [1, 2, 3]

C. [1, 3, 2]

D. No next permutation

Solution

1. Step 1: Identify pivot index i

   Starting from right, find i where nums[i] < nums[i+1]. For [1,3,2], i=0 because 1 < 3.

2. Step 2: Find j to swap with i

   From right, find j where nums[j] > nums[i]. Here j=2 (value 2 > 1).

3. Step 3: Swap and reverse suffix

   Swap nums[0] and nums[2]: [2,3,1]. Then reverse suffix after i: reverse [3,1] to [1,3]. Result: [2,1,3].

4. Step 4: Check code logic carefully

   Code breaks loop early if i=0 and condition met, so i=0 is pivot. But code returns false and reverses if no pivot found. Here pivot found, so returns true and prints new nums.

5. Final Answer:

   [2, 1, 3] -> Option A

6. Quick Check:

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

Hint: Find pivot i, swap with j, reverse suffix [OK]

Common Mistakes:

• Not reversing suffix after swap
• Wrong pivot index calculation
• Confusing when to return false

4. The following Rust function attempts to implement next permutation but has a bug. What is the bug?

fn next_permutation(nums: &mut Vec<i32>) -> bool {
    let n = nums.len();
    if n < 2 { return false; }
    let mut i = n - 2;
    while nums[i] >= nums[i + 1] {
        i -= 1;
    }
    if nums[i] >= nums[i + 1] {
        nums.reverse();
        return false;
    }
    let mut j = n - 1;
    while nums[j] <= nums[i] {
        j -= 1;
    }
    nums.swap(i, j);
    nums[i + 1..].reverse();
    true
}

medium

A. The suffix reversal is missing.

B. The loop misses checking when i == 0, causing panic on underflow.

C. The swap method is used incorrectly.

D. The function does not return a boolean.

Solution

1. Step 1: Analyze the while loop condition

   The loop uses while nums[i] >= nums[i + 1] and decrements i. When i reaches 0 and condition holds, next decrement causes underflow (usize wraps to large number).

2. Step 2: Check consequences

   This causes panic or incorrect behavior. The correct approach is to use i != usize::MAX or check before decrementing to avoid underflow.

3. Final Answer:

   The loop misses checking when i == 0, causing panic on underflow. -> Option B

4. Quick Check:

   Loop underflow bug = missing i == 0 check [OK]

Hint: Check loop index underflow when decrementing unsigned integers [OK]

Common Mistakes:

• Ignoring unsigned integer underflow
• Incorrect swap usage
• Forgetting to reverse suffix

5. Given a vector nums = vec![3, 2, 1], which is the highest permutation, what will be the state of nums after applying the next permutation function that returns false when no next permutation exists?

hard

A. [3, 2, 1]

B. Original vector unchanged

C. Empty vector

D. [1, 2, 3]

Solution

1. Step 1: Identify if next permutation exists

   For [3, 2, 1], the array is in descending order, which is the highest permutation. No next permutation exists.

2. Step 2: Apply algorithm behavior when no next permutation

   The algorithm reverses the entire array to the lowest permutation and returns false.

3. Final Answer:

   [1, 2, 3] -> Option D

4. Quick Check:

   Highest permutation reversed = lowest permutation [OK]

Hint: No next permutation means reverse to lowest order [OK]

Common Mistakes:

• Expecting no change when no next permutation
• Returning empty vector incorrectly
• Confusing highest and lowest permutations