Merge Two Sorted Arrays Without Extra Space in DSA Rust - Time & Space Complexity

Time Complexity: Merge Two Sorted Arrays Without Extra Space

O((n + m) log (n + m))

Understanding Time Complexity

We want to know how long it takes to merge two sorted arrays without using extra space.

How does the time grow as the arrays get bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn merge_without_extra_space(arr1: &mut [i32], arr2: &mut [i32]) {
    let n = arr1.len();
    let m = arr2.len();
    let mut i = 0;
    let mut j = 0;
    while i < n && j < m {
        if arr1[i] > arr2[j] {
            arr1.swap(i, n - 1);
            arr2.swap(j, 0);
            i += 1;
            j += 1;
        } else {
            i += 1;
        }
    }
    arr1.sort();
    arr2.sort();
}
    

This code merges two sorted arrays by swapping elements and then sorting each array again.

Identify Repeating Operations

Look for loops and repeated steps.

• Primary operation: The while loop runs comparing and swapping elements.
• How many times: Up to n times, where n is the size of the first array.
• Additional operations: Sorting both arrays at the end, which involves multiple passes over the data.

How Execution Grows With Input

As the arrays get bigger, the number of comparisons and swaps grows, plus sorting takes more time.

Input Size (n + m)	Approx. Operations
10	About 10 comparisons + sorting small arrays
100	About 100 comparisons + sorting arrays of size 100
1000	About 1000 comparisons + sorting arrays of size 1000

Pattern observation: The comparisons grow linearly, but sorting grows faster, roughly proportional to n log n.

Final Time Complexity

Time Complexity: O((n + m) log (n + m))

This means the time grows a bit faster than the size of the arrays because sorting takes more work as arrays get bigger.

Common Mistake

[X] Wrong: "Swapping elements alone makes the merge fast and linear."

[OK] Correct: Because sorting is still needed after swapping, which takes more time than just swapping.

Interview Connect

Understanding this helps you explain trade-offs between using extra space and time, a common topic in interviews.

Self-Check

"What if we used a gap method to merge without sorting at the end? How would the time complexity change?"

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