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)

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 = arr1.Length, m = arr2.Length;
    for (int i = 0; i < n; i++) {
        if (arr1[i] > arr2[0]) {
            int temp = arr1[i];
            arr1[i] = arr2[0];
            arr2[0] = temp;
            // Re-sort arr2's first element to correct position
            int first = arr2[0], k;
            for (k = 1; k < m && arr2[k] < first; k++) {
                arr2[k - 1] = arr2[k];
            }
            arr2[k - 1] = first;
        }
    }
}
    

This code merges two sorted arrays by swapping elements and keeping the second array sorted without extra space.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Outer loop runs through all elements of the first array.
• Nested operation: Inner loop shifts elements in the second array to keep it sorted after each swap.
• How many times: Outer loop runs n times; inner loop can run up to m times in worst case.

How Execution Grows With Input

As the arrays get bigger, the number of operations grows roughly by multiplying their sizes.

Input Size (n, m)	Approx. Operations
10, 10	Up to 100
100, 100	Up to 10,000
1000, 1000	Up to 1,000,000

Pattern observation: Doubling the size roughly squares the number of operations.

Final Time Complexity

Time Complexity: O(n * m)

This means the time needed grows proportionally to the product of the sizes of the two arrays.

Common Mistake

[X] Wrong: "Since both arrays are sorted, merging without extra space is always O(n + m)."

[OK] Correct: The inner shifting loop can run many times, making the total steps multiply, not just add.

Interview Connect

Understanding this helps you explain trade-offs between space and time, a key skill in coding interviews.

Self-Check

"What if we used a gap method to compare and swap elements instead? How would the time complexity change?"

Practice

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

easy

A. Create a new array to hold merged elements

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

C. Copy elements from one array to another without sorting

D. Sort each array separately using extra space

Solution

1. Step 1: Understand the problem requirement

   The task is to merge two sorted arrays so that the combined elements remain sorted, but without using any extra space like a new array.

2. Step 2: Identify the correct approach

   We must rearrange elements within the given arrays themselves to maintain sorted order 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 between arrays without extra space [OK]

Common Mistakes:

• Assuming extra space is allowed
• Sorting arrays separately without merging
• Creating a new array for merged result

2. Which of the following C# code snippets correctly swaps elements between two arrays without extra space?

easy

A. arr1[i] = arr1[i] + arr2[j]; arr2[j] = arr1[i] - arr2[j]; arr1[i] = arr1[i] - arr2[j];

B. arr1[i] = arr2[j]; arr2[j] = arr1[i];

C. if (arr1[i] > arr2[j]) { int temp = arr1[i]; arr1[i] = arr2[j]; arr2[j] = temp; }

D. int temp = arr2[j]; arr2[j] = arr1[i]; arr1[i] = temp + 1;

Solution

1. Step 1: Check swapping logic correctness

   if (arr1[i] > arr2[j]) { int temp = arr1[i]; arr1[i] = arr2[j]; arr2[j] = temp; } uses a temporary variable to swap values correctly without losing data.

2. Step 2: Analyze other options for errors

   arr1[i] = arr2[j]; arr2[j] = arr1[i]; overwrites arr2[j] before saving arr1[i], losing original value. arr1[i] = arr1[i] + arr2[j]; arr2[j] = arr1[i] - arr2[j]; arr1[i] = arr1[i] - arr2[j]; uses arithmetic swap but can cause overflow and is less safe. int temp = arr2[j]; arr2[j] = arr1[i]; arr1[i] = temp + 1; incorrectly adds 1 during swap, changing values.

3. Final Answer:

   if (arr1[i] > arr2[j]) { int temp = arr1[i]; arr1[i] = arr2[j]; arr2[j] = temp; } -> Option C

4. Quick Check:

   Temporary variable swap = Correct swap [OK]

Hint: Use a temp variable to swap safely in C# [OK]

Common Mistakes:

• Overwriting values without temp variable
• Using unsafe arithmetic swaps
• Altering values unintentionally during swap

3. What will be the content of arrays after running this code?

int[] arr1 = {1, 5, 9, 10, 15, 20};
int[] arr2 = {2, 3, 8, 13};
int n = arr1.Length, m = arr2.Length;
for (int i = 0; i < n; i++) {
  if (arr1[i] > arr2[0]) {
    int temp = arr1[i];
    arr1[i] = arr2[0];
    arr2[0] = temp;
    int first = arr2[0];
    int k;
    for (k = 1; k < m && arr2[k] < first; k++) {
      arr2[k - 1] = arr2[k];
    }
    arr2[k - 1] = first;
  }
}

medium

A. arr1 = [1, 2, 3, 5, 9, 10], arr2 = [8, 13, 15, 20]

B. arr1 = [1, 3, 5, 8, 9, 10], arr2 = [2, 13, 15, 20]

C. arr1 = [1, 2, 5, 8, 9, 10], arr2 = [3, 13, 15, 20]

D. arr1 = [1, 2, 3, 5, 8, 9], arr2 = [10, 13, 15, 20]

Solution

1. Step 1: Trace the swapping and insertion in arr2

   For each element in arr1, if it is greater than arr2[0], swap and then insert arr2[0] in sorted order in arr2. This pushes larger elements to arr2.

2. Step 2: Final arrays after all iterations

   After all swaps and insertions, arr1 becomes [1, 2, 3, 5, 8, 9] and arr2 becomes [10, 13, 15, 20].

3. Final Answer:

   arr1 = [1, 2, 3, 5, 8, 9], arr2 = [10, 13, 15, 20] -> Option D

4. Quick Check:

   In-place merge with insertion sort = arr1 and arr2 sorted [OK]

Hint: Swap if arr1[i] > arr2[0], then insert arr2[0] sorted [OK]

Common Mistakes:

• Not inserting arr2[0] correctly after swap
• Overwriting arr2 elements incorrectly
• Confusing array indices during insertion

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;
    int first = arr2[0];
    int k = 1;
    while (k < m && arr2[k] < first) {
      arr2[k - 1] = arr2[k];
      k++;
    }
    arr2[k] = first;
  }
}

medium

A. The insertion index in arr2 is incorrect; should be arr2[k - 1] = first

B. The swap logic between arr1[i] and arr2[0] is wrong

C. The loop should run from i = 1 instead of 0

D. The condition arr2[k] < first should be arr2[k] > first

Solution

1. Step 1: Analyze insertion logic in arr2

   The code shifts elements left to make space for 'first'. The correct insertion index after shifting is k - 1, not k.

2. Step 2: Identify the error impact

   Assigning arr2[k] = first overwrites the next element incorrectly, causing wrong order.

3. Final Answer:

   The insertion index in arr2 is incorrect; should be arr2[k - 1] = first -> Option A

4. Quick Check:

   Correct insertion index = k - 1 [OK]

Hint: Insert shifted element at k-1 index after loop ends [OK]

Common Mistakes:

• Using wrong index for insertion after shifting
• Incorrect swap logic
• Wrong loop start index

5. Given two sorted arrays:
arr1 = [1, 4, 7, 8, 10]
arr2 = [2, 3, 9]
How would you merge them without extra space so that after merging, arr1 contains the first 5 smallest elements and arr2 contains the remaining sorted elements? Choose the correct final arrays.

hard

A. arr1 = [1, 2, 3, 4, 7], arr2 = [8, 9, 10]

B. arr1 = [1, 3, 4, 7, 8], arr2 = [2, 9, 10]

C. arr1 = [1, 2, 4, 7, 8], arr2 = [3, 9, 10]

D. arr1 = [1, 2, 3, 4, 8], arr2 = [7, 9, 10]

Solution

1. Step 1: Compare and swap elements between arr1 and arr2

   Start from beginning of arr1, if arr1[i] > arr2[0], swap and then insert arr2[0] in sorted order in arr2.

2. Step 2: After all swaps and insertions

   arr1 becomes [1, 2, 3, 4, 7] and arr2 becomes [8, 9, 10], maintaining sorted order without extra space.

3. Final Answer:

   arr1 = [1, 2, 3, 4, 7], arr2 = [8, 9, 10] -> Option A

4. Quick Check:

   In-place merge keeps first array smallest elements [OK]

Hint: Swap if arr1[i] > arr2[0], then insert arr2[0] sorted [OK]

Common Mistakes:

• Not maintaining sorted order after swap
• Incorrect insertion in second array
• Assuming extra space is allowed