Why Next Permutation of Array in DSA Rust?

The Big Idea

Discover how to jump to the next number sequence without trying every single one!

The Scenario

Imagine you have a list of numbers representing a combination lock. You want to find the next combination in order without trying every possible combination manually.

The Problem

Trying all combinations one by one is slow and tiring. It's easy to make mistakes and miss the next correct combination, especially if the list is long.

The Solution

The next permutation method quickly finds the next ordered combination by rearranging the list just a little. It saves time and avoids errors by following clear steps.

Before vs After

✗ Before

let mut arr = vec![1, 2, 3];
// Manually try all permutations until next found
// Very slow and error-prone

✓ After

fn next_permutation(arr: &mut Vec<i32>) {
    // Efficiently find next permutation
}

What It Enables

This lets you jump directly to the next sequence without checking all previous ones, making tasks like generating combinations fast and easy.

Real Life Example

When generating all possible passwords in order, next permutation helps move to the next guess quickly without repeating or missing any.

Key Takeaways

Manual checking of permutations is slow and error-prone.

Next permutation finds the next sequence efficiently.

It helps in tasks needing ordered combinations or sequences.

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