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 know how the time needed to merge two sorted arrays without extra space changes as the arrays get bigger.

How does the number of steps grow when the input size increases?

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 {
            int temp = arr2[j];
            arr2[j] = arr1[k];
            arr1[k] = temp;
            j++;
            k--;
        }
    }
    // Sort both arrays after swapping
    qsort(arr1, n, sizeof(int), compare);
    qsort(arr2, m, sizeof(int), compare);
}

This code merges two sorted arrays by swapping elements to avoid extra space, then sorts both arrays again.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that compares and swaps elements between the two arrays.
• How many times: Up to n + m times, since i moves forward and k moves backward in arr1, and j moves forward in arr2.
• Secondary operations: Sorting both arrays again using qsort, which internally uses a sorting algorithm like quicksort.

How Execution Grows With Input

As the arrays get bigger, the number of comparisons and swaps in the while loop grows roughly with the total size of both arrays.

Input Size (n + m)	Approx. Operations
10	About 10 to 20 operations in the loop, plus sorting small arrays.
100	About 100 to 200 operations in the loop, plus sorting arrays of size 100.
1000	About 1000 to 2000 operations in the loop, plus sorting arrays of size 1000.

Pattern observation: The loop grows linearly with input size, but sorting adds a factor that grows faster than linear.

Final Time Complexity

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

This means the time needed grows a bit faster than the total size of the arrays because of the sorting step after swapping.

Common Mistake

[X] Wrong: "The merging without extra space is just a simple linear pass, so it must be O(n + m)."

[OK] Correct: After swapping, the arrays are not fully sorted, so sorting both arrays again takes extra time, which dominates the total time.

Interview Connect

Understanding this helps you explain how to handle merging without extra space and why sorting after swapping affects performance. It shows you can analyze combined steps clearly.

Self-Check

"What if we used a gap method (Shell sort style) 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. Combine both arrays into one sorted array without using extra memory

B. Sort each array separately 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 task is 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 without allocating extra memory.

3. Final Answer:

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

4. Quick Check:

   Merge without extra space = Combine in place [OK]

Hint: Merge by swapping elements, no extra array needed [OK]

Common Mistakes:

• Thinking extra array is allowed
• Sorting arrays separately without merging
• Copying elements without sorting

2. Which of the following C code snippets correctly declares two sorted arrays for merging without extra space?

easy

A. int arr1[] = (1, 3, 5); int arr2[] = (2, 4, 6);

B. int arr1 = {1, 3, 5}; int arr2 = {2, 4, 6};

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

D. int arr1[] = {1, 3, 5}; int arr2[] = {2, 4, 6};

Solution

1. Step 1: Check array declaration syntax in C

   Arrays in C are declared using curly braces {} with type and square brackets [].

2. Step 2: Validate each option

   int arr1[] = {1, 3, 5}; int arr2[] = {2, 4, 6}; uses correct syntax: int arr[] = {elements}; Options A, B, and D use invalid syntax for arrays in C.

3. Final Answer:

   int arr1[] = {1, 3, 5}; int arr2[] = {2, 4, 6}; -> Option D

4. Quick Check:

   C arrays use {} for initialization [OK]

Hint: Use curly braces {} to initialize arrays in C [OK]

Common Mistakes:

• Using parentheses () or square brackets [] for initialization
• Missing data type or brackets
• Assigning arrays without braces

3. Given two sorted arrays:
arr1 = {1, 4, 7, 8, 10}
arr2 = {2, 3, 9}
After merging without extra space, what will be the content of arr1?

medium

A. [1, 2, 3, 4, 8]

B. [1, 2, 3, 4, 7]

C. [1, 2, 3, 7, 8]

D. [1, 3, 4, 7, 8]

Solution

1. Step 1: Understand merging without extra space

   We merge by comparing elements and swapping to keep smallest elements in arr1.

2. Step 2: Merge process

   Initial arr1: [1,4,7,8,10], arr2: [2,3,9]
   Compare 10 (arr1 last) and 2 (arr2 first), swap -> arr1: [1,4,7,8,2], arr2: [10,3,9]
   Sort arr2: [3,9,10]
   Repeat swaps and sorts until arr1 holds smallest 5 elements:
   Final arr1: [1,2,3,4,7]

3. Final Answer:

   [1, 2, 3, 4, 7] -> Option B

4. Quick Check:

   Smallest 5 elements in arr1 after merge = [1,2,3,4,7] [OK]

Hint: Keep smallest elements in first array by swapping [OK]

Common Mistakes:

• Not sorting arr2 after swaps
• Assuming arr1 remains unchanged
• Mixing elements without sorting

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

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

medium

A. Using sort() without including <algorithm> header

B. Swapping elements incorrectly between arrays

C. Sorting arr2 inside the loop causes inefficiency

D. Loop should run from 0 to m instead of n

Solution

1. Step 1: Analyze the loop and sorting

   The code swaps elements and sorts arr2 inside the loop for every iteration.

2. Step 2: Identify inefficiency

   Sorting arr2 in every iteration is costly and inefficient; better to optimize sorting or use a different approach.

3. Final Answer:

   Sorting arr2 inside the loop causes inefficiency -> Option C

4. Quick Check:

   Sorting inside loop = inefficient [OK]

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

Common Mistakes:

• Ignoring performance impact of repeated sorting
• Confusing array sizes for loop limits
• Assuming swap logic is incorrect

5. You have two sorted arrays:
arr1 = {1, 5, 9, 10, 15, 20}
arr2 = {2, 3, 8, 13}
Using the gap method to merge without extra space, what is the final state of arr2 after merging?

hard

A. [10, 13, 15, 20]

B. [8, 9, 10, 15]

C. [8, 13, 15, 20]

D. [9, 10, 13, 15]

Solution

1. Step 1: Understand the gap method

   The gap method compares elements at a gap distance and swaps if out of order, reducing gap until 1.

2. Step 2: Apply gap method to arrays

   Initial arr1: [1,5,9,10,15,20], arr2: [2,3,8,13]
   After merging, arr1 holds smallest elements: [1,2,3,5,8,9]
   arr2 holds remaining elements: [10,13,15,20]

3. Step 3: Final arr2 state

   arr2 after merge is [10, 13, 15, 20].

4. Final Answer:

   [10, 13, 15, 20] -> Option A

5. Quick Check:

   Gap method final arr2 = [10, 13, 15, 20] [OK]

Hint: Gap method sorts by comparing distant elements [OK]

Common Mistakes:

• Confusing final positions of elements in arr2
• Assuming arr2 remains unchanged
• Misapplying gap calculation