Why Merge Two Sorted Arrays Without Extra Space in DSA Rust?

The Big Idea

What if you could merge two sorted lists perfectly without needing any extra space at all?

The Scenario

Imagine you have two sorted lists of numbers written on two separate sheets of paper. You want to combine them into one sorted list, but you only have the space on those two sheets, no extra paper to write on.

The Problem

Trying to merge these lists by writing all numbers on a new sheet is easy but needs extra space. If you try to do it by moving numbers around on the two sheets without extra space, it becomes confusing and slow. You might lose track or overwrite numbers by mistake.

The Solution

This method shows how to cleverly swap and compare numbers between the two lists to keep them sorted without needing extra space. It saves time and avoids the mess of juggling extra sheets.

Before vs After

✗ Before

let mut merged = Vec::new();
merged.extend_from_slice(&array1);
merged.extend_from_slice(&array2);
merged.sort();

✓ After

merge_two_sorted_arrays_without_extra_space(&mut array1, &mut array2);

What It Enables

You can efficiently combine two sorted arrays into one sorted sequence without using extra memory, saving space and improving performance.

Real Life Example

Think of merging two sorted guest lists for a party where you only have two sheets of paper and no extra space to write a new combined list.

Key Takeaways

Manual merging needs extra space and can be slow.

Swapping and comparing elements in place keeps arrays sorted without extra memory.

This technique is useful when memory is limited or performance matters.

Practice

1. What is the main goal when merging two sorted arrays without extra space?

easy

A. Sort each array separately using extra space

B. Combine both arrays into one sorted sequence without using extra memory

C. Copy elements from one array to another without sorting

D. Create a new array to hold merged elements

Solution

1. Step 1: Understand the problem requirement

   The problem asks to merge two sorted arrays into a sorted sequence without using extra space, meaning no new arrays can be created.

2. Step 2: Identify the correct approach

   We must rearrange elements within the given arrays to maintain sorted order combined, without additional memory.

3. Final Answer:

   Combine both arrays into one sorted sequence without using extra memory -> Option B

4. Quick Check:

   Merge without extra space = Combine in place [OK]

Hint: Merge by swapping elements, no new arrays needed [OK]

Common Mistakes:

• Trying to create a new array for merged result
• Sorting arrays separately without merging
• Ignoring the no extra space constraint

2. Which Rust syntax correctly declares two mutable sorted arrays for merging without extra space?

easy

A. let mut arr1 = [1, 3, 5]; let mut arr2 = [2, 4, 6];

B. let arr1 = [1, 3, 5]; let arr2 = [2, 4, 6];

C. mut arr1 = [1, 3, 5]; mut arr2 = [2, 4, 6];

D. let mut arr1 = (1, 3, 5); let mut arr2 = (2, 4, 6);

Solution

1. Step 1: Identify correct Rust mutable array declaration

   In Rust, arrays are declared with let mut for mutability and square brackets for arrays.

2. Step 2: Check each option

   let mut arr1 = [1, 3, 5]; let mut arr2 = [2, 4, 6]; uses let mut with arrays correctly. let arr1 = [1, 3, 5]; let arr2 = [2, 4, 6]; misses mut keyword. mut arr1 = [1, 3, 5]; mut arr2 = [2, 4, 6]; misses let keyword. let mut arr1 = (1, 3, 5); let mut arr2 = (2, 4, 6); uses tuples, not arrays.

3. Final Answer:

   let mut arr1 = [1, 3, 5]; let mut arr2 = [2, 4, 6]; -> Option A

4. Quick Check:

   Mutable arrays use let mut and [] [OK]

Hint: Use let mut and square brackets for mutable arrays [OK]

Common Mistakes:

• Forgetting mut keyword for mutable arrays
• Using tuples instead of arrays
• Missing let keyword

3. What will be the content of arrays after merging arr1 = [1, 4, 7] and arr2 = [2, 5, 6] without extra space using the gap method?

medium

A. ]7 ,6 ,5[ = 2rra ,]4 ,2 ,1[ = 1rra

B. arr1 = [1, 4, 5], arr2 = [2, 6, 7]

C. rr1 = [1, 2, 4], arr2 = [5, 6, 7]

D. arr1 = [1, 2, 4], arr2 = [5, 6, 7]

Solution

1. Step 1: Apply gap method to merge arrays

   Start with gap = ceil((3+3)/2) = 3. Compare and swap elements at gap distance across arrays until gap reduces to 0.

2. Step 2: Final arrays after merging

   After applying gap method, arr1 becomes [1, 2, 4] and arr2 becomes [5, 6, 7], maintaining sorted order without extra space.

3. Final Answer:

   arr1 = [1, 2, 4], arr2 = [5, 6, 7] -> Option D

4. Quick Check:

   Gap method merges arrays in place [OK]

Hint: Use gap method to compare and swap elements at decreasing gaps [OK]

Common Mistakes:

• Not reducing gap correctly
• Swapping elements incorrectly across arrays
• Confusing array indices during merge

4. Identify the error in this Rust code snippet for merging two sorted arrays without extra space:

let mut arr1 = [1, 5, 9];
let mut arr2 = [2, 6, 8];
for i in 0..arr1.len() {
    if arr1[i] > arr2[0] {
        let temp = arr1[i];
        arr1[i] = arr2[0];
        arr2[0] = temp;
    }
    arr2.sort();
}

medium

A. Using let temp causes ownership error in Rust

B. Swapping elements between arr1 and arr2 is not allowed

C. Sorting arr2 inside the loop is inefficient and incorrect

D. The loop range should be arr2.len() instead of arr1.len()

Solution

1. Step 1: Analyze the loop and sorting inside it

   The code sorts arr2 inside the loop for every iteration, which is inefficient and unnecessary.

2. Step 2: Correct approach for sorting arr2

   Sorting arr2 should be done once after all swaps, not inside the loop, to improve performance and correctness.

3. Final Answer:

   Sorting arr2 inside the loop is inefficient and incorrect -> Option C

4. Quick Check:

   Sort arr2 once after loop [OK]

Hint: Sort arr2 once after loop, not inside it [OK]

Common Mistakes:

• Sorting arr2 repeatedly inside loop
• Confusing loop ranges
• Misunderstanding Rust ownership with temp variable

5. Given two sorted arrays arr1 = [1, 3, 5, 7] and arr2 = [0, 2, 6, 8, 9], how can you merge them without extra space so that after merging, arr1 contains the first 4 smallest elements and arr2 contains the remaining elements sorted? Choose the correct final arrays.

hard

A. arr1 = [0, 1, 2, 3], arr2 = [5, 6, 7, 8, 9]

B. arr1 = [1, 2, 3, 5], arr2 = [0, 6, 7, 8, 9]

C. arr1 = [0, 1, 3, 5], arr2 = [2, 6, 7, 8, 9]

D. arr1 = [0, 1, 3, 6], arr2 = [2, 5, 7, 8, 9]

Solution

1. Step 1: Understand merging without extra space for unequal sizes

   We must rearrange elements so arr1 holds smallest 4 elements and arr2 holds the rest, both sorted.

2. Step 2: Apply gap method or swapping approach

   By comparing and swapping elements between arrays and sorting arr2, final arrays become arr1 = [0, 1, 2, 3] and arr2 = [5, 6, 7, 8, 9].

3. Final Answer:

   arr1 = [0, 1, 2, 3], arr2 = [5, 6, 7, 8, 9] -> Option A

4. Quick Check:

   First 4 smallest in arr1, rest in arr2 [OK]

Hint: Place smallest elements in arr1 by swapping, then sort arr2 [OK]

Common Mistakes:

• Not handling different array sizes correctly
• Failing to keep both arrays sorted after merge
• Assuming extra space is allowed