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

Time Complexity: Merge Two Sorted Arrays Without Extra Space

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

Understanding Time Complexity

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

How does the time needed grow as the arrays get bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void merge(int arr1[], int arr2[], int n, int m) {
  int i = 0, j = 0, k = n - 1;
  while (i <= k && j < m) {
    if (arr1[i] <= arr2[j]) {
      i++;
    } else {
      std::swap(arr2[j++], arr1[k--]);
    }
  }
  std::sort(arr1, arr1 + n);
  std::sort(arr2, arr2 + m);
}
    

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that compares and swaps elements.
• How many times: Up to n + m times combined, since i and j move through the arrays.
• Secondary operations: Two sorting calls on arrays of sizes n and m.

How Execution Grows With Input

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

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

Pattern observation: The sorting steps dominate and grow faster than the initial swaps.

Final Time Complexity

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

This means the time grows roughly proportional to the total size times the log of the total size, mainly due to sorting.

Common Mistake

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

[OK] Correct: Because after swapping, sorting is still needed, 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. Copy elements from one array to another without sorting

B. Sort each array separately using extra space

C. Create a new array to hold merged elements

D. Combine both arrays into one sorted array without using extra memory

Solution

1. Step 1: Understand the problem requirement

   The problem asks to merge two sorted arrays into a single sorted sequence without using extra space, meaning no additional arrays or large buffers.

2. Step 2: Identify the correct approach

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

3. Final Answer:

   Combine both arrays into one sorted array without using extra memory -> Option D

4. Quick Check:

   Merge without extra space = Combine in-place [OK]

Hint: Think in-place rearrangement, no new arrays needed [OK]

Common Mistakes:

• Assuming extra space is allowed
• Trying to merge by creating a new array
• Sorting arrays separately without merging

2. Which of the following C++ code snippets correctly swaps two elements without using extra space?

easy

A. a = a + b; b = a - b; a = a - b;

B. temp = a; a = b; b = temp;

C. a = b; b = a;

D. a = a * b; b = a / b; a = a / b;

Solution

1. Step 1: Analyze swapping without extra variable

   a = a + b; b = a - b; a = a - b; uses arithmetic operations to swap values without a temporary variable.

2. Step 2: Check correctness of each option

   temp = a; a = b; b = temp; uses a temporary variable, so it uses extra space. a = b; b = a; assigns a to b and then b to a, which loses original value. a = a * b; b = a / b; a = a / b; uses multiplication/division which can cause overflow or errors if b is zero.

3. Final Answer:

   a = a + b; b = a - b; a = a - b; -> Option A

4. Quick Check:

   Swap without temp = arithmetic trick [OK]

Hint: Use arithmetic swap to avoid extra variable [OK]

Common Mistakes:

• Using a temporary variable (extra space)
• Overwriting values without saving
• Using multiplication/division causing errors

3. What will be the content of arrays arr1 and arr2 after merging without extra space using the gap method?

arr1 = {1, 5, 9, 10}
arr2 = {2, 3, 8, 13}

medium

A. arr1 = {1, 2, 3, 5}, arr2 = {8, 9, 10, 13}

B. arr1 = {1, 3, 5, 8}, arr2 = {2, 9, 10, 13}

C. arr1 = {1, 2, 5, 8}, arr2 = {3, 9, 10, 13}

D. arr1 = {1, 2, 3, 8}, arr2 = {5, 9, 10, 13}

Solution

1. Step 1: Apply gap method to merge arrays

   Start with gap = ceil((4+4)/2) = 4. Compare elements at gap distance and swap if needed, then reduce gap until 0.

2. Step 2: Final sorted arrays after gap method

   After applying the method, arr1 becomes {1, 2, 3, 5} and arr2 becomes {8, 9, 10, 13}.

3. Final Answer:

   arr1 = {1, 2, 3, 5}, arr2 = {8, 9, 10, 13} -> Option A

4. Quick Check:

   Gap method merges arrays in-place correctly [OK]

Hint: Use gap method: reduce gap, swap if needed [OK]

Common Mistakes:

• Not reducing gap properly
• Swapping wrong elements
• Confusing array indices across two arrays

4. Identify the error in the following code snippet that attempts to merge two sorted arrays without extra space:

for (int i = 0; i < n; i++) {
  if (arr1[i] > arr2[0]) {
    swap(arr1[i], arr2[0]);
    sort(arr2, arr2 + m);
  }
}

medium

A. Swapping elements without checking bounds causes runtime error

B. Loop should run till m instead of n

C. Sorting arr2 inside the loop causes high time complexity

D. No error, code is optimal

Solution

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

   The code sorts arr2 inside the loop for every swap, which is inefficient and increases time complexity to O(n*m*log m).

2. Step 2: Identify the performance issue

   Sorting arr2 repeatedly is costly; better methods avoid sorting inside the loop.

3. Final Answer:

   Sorting arr2 inside the loop causes high time complexity -> Option C

4. Quick Check:

   Sorting inside loop = inefficient [OK]

Hint: Avoid sorting inside loops for better performance [OK]

Common Mistakes:

• Ignoring time complexity impact
• Assuming swap causes runtime error
• Confusing loop bounds

5. Given two sorted arrays arr1 = {10, 12, 15, 18} and arr2 = {2, 5, 20, 25}, how can you merge them without extra space so that after merging, arr1 contains the first four smallest elements and arr2 contains the remaining elements sorted?

hard

A. Swap elements only if arr1[i] is greater than arr2[j] without sorting arr2

B. Use the gap method to compare and swap elements across arrays until fully merged

C. Sort both arrays separately and then concatenate

D. Copy all elements to a new array, sort it, then copy back

Solution

1. Step 1: Understand the gap method for merging

   The gap method reduces the gap between compared elements and swaps if needed, working across both arrays without extra space.

2. Step 2: Apply gap method to given arrays

   By repeatedly comparing and swapping elements at the gap distance, arr1 will hold the smallest elements and arr2 the larger ones, both sorted.

3. Final Answer:

   Use the gap method to compare and swap elements across arrays until fully merged -> Option B

4. Quick Check:

   Gap method merges sorted arrays in-place [OK]

Hint: Use gap method for in-place merge without extra space [OK]

Common Mistakes:

• Using extra arrays violating constraints
• Sorting arrays separately without merging
• Swapping without sorting arr2 leads to wrong order