Array Insertion at Middle Index in DSA C++ - Time & Space Complexity

Time Complexity: Array Insertion at Middle Index

O(n)

Understanding Time Complexity

We want to understand how the time needed changes when inserting an element in the middle of an array.

How does the work grow as the array gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void insertMiddle(int arr[], int &size, int capacity, int value) {
    int mid = size / 2;
    if (size == capacity) return; // no space
    for (int i = size; i > mid; i--) {
        arr[i] = arr[i - 1];
    }
    arr[mid] = value;
    size++;
}
    

This code inserts a new value in the middle of the array by shifting elements to the right.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for-loop that shifts elements one position to the right.
• How many times: It runs from the end of the array down to the middle index, roughly n/2 times where n is the array size.

How Execution Grows With Input

As the array size grows, the number of shifts grows roughly half as fast.

Input Size (n)	Approx. Operations (shifts)
10	5
100	50
1000	500

Pattern observation: The number of operations grows linearly with the input size, about half the size.

Final Time Complexity

Time Complexity: O(n)

This means the time to insert grows directly with the size of the array.

Common Mistake

[X] Wrong: "Inserting in the middle is always fast because we just place the element there."

[OK] Correct: We must move all elements after the middle to make space, which takes time proportional to the number of elements shifted.

Interview Connect

Understanding this helps you explain why arrays are slower for middle insertions compared to linked lists, showing your grasp of data structure trade-offs.

Self-Check

"What if we changed the array to a linked list? How would the time complexity of inserting in the middle change?"

Practice

1. What does inserting an element at the middle index of an array mean?

easy

A. Adding a new element exactly in the center, shifting others right

B. Replacing the middle element with a new one

C. Adding a new element at the end of the array

D. Removing the middle element from the array

Solution

1. Step 1: Understand middle index insertion

   Inserting at the middle means placing a new element at the center position, pushing existing elements right.

2. Step 2: Differentiate from other operations

   Replacing or removing elements is not insertion; adding at the end is not middle insertion.

3. Final Answer:

   Adding a new element exactly in the center, shifting others right -> Option A

4. Quick Check:

   Middle insertion means insert and shift right [OK]

Hint: Middle insertion shifts elements right from center [OK]

Common Mistakes:

• Thinking insertion replaces the middle element
• Confusing insertion with appending at the end
• Assuming middle means removing element

2. Which of the following C++ code snippets correctly inserts an element `x` at the middle index of array `arr` of size `n`?

easy

A. arr[n] = x; for(int i = n/2; i < n; i++) arr[i] = arr[i+1];

B. for(int i = 0; i < n/2; i++) arr[i] = arr[i+1]; arr[n/2] = x;

C. arr[n/2] = x; for(int i = n/2; i < n; i++) arr[i+1] = arr[i];

D. for(int i = n; i > n/2; i--) arr[i] = arr[i-1]; arr[n/2] = x;

Solution

1. Step 1: Shift elements right from middle index

   Loop from end (n) down to middle (n/2 + 1) shifting elements right to make space.

2. Step 2: Insert element at middle index

   Assign new element x at index n/2 after shifting.

3. Final Answer:

   for(int i = n; i > n/2; i--) arr[i] = arr[i-1]; arr[n/2] = x; -> Option D

4. Quick Check:

   Shift right then insert at middle [OK]

Hint: Shift elements right from end to middle before insert [OK]

Common Mistakes:

• Shifting elements left instead of right
• Inserting before shifting elements
• Incorrect loop direction causing overwrite

3. What is the output array after inserting 99 at the middle index of arr = {10, 20, 30, 40, 50} using the correct insertion method?

medium

A. [10, 20, 30, 40, 99, 50]

B. [10, 20, 99, 30, 40, 50]

C. [10, 20, 30, 99, 40, 50]

D. [10, 20, 30, 40, 50, 99]

Solution

1. Step 1: Calculate middle index

   Array size n=5, middle index = n/2 = 2 (0-based indexing).

2. Step 2: Shift elements right and insert 99

   Shift elements from index 4 to 2 right by one position, then insert 99 at index 2.

3. Final Answer:

   [10, 20, 99, 30, 40, 50] -> Option B

4. Quick Check:

   Insert at index 2 shifts right [OK]

Hint: Middle index = size/2; shift right, then insert [OK]

Common Mistakes:

• Inserting at wrong index (like 3 or 4)
• Not shifting elements before insertion
• Appending at the end instead of middle

4. Identify the error in the following code snippet that inserts element `x` at the middle index of array `arr` of size `n`:

for(int i = n/2; i < n; i++) {
  arr[i+1] = arr[i];
}
arr[n/2] = x;

medium

A. Loop shifts elements left instead of right

B. Index out of bounds when accessing arr[i+1]

C. Loop direction causes overwriting elements

D. No error, code works correctly

Solution

1. Step 1: Analyze loop direction and effect

   The loop moves forward from middle to end, copying arr[i] to arr[i+1], overwriting elements before they are moved.

2. Step 2: Identify correct approach

   Elements must be shifted right starting from the end backward to avoid overwriting data.

3. Final Answer:

   Loop direction causes overwriting elements -> Option C

4. Quick Check:

   Shift right must go backward to avoid overwrite [OK]

Hint: Shift elements right backward from end to middle [OK]

Common Mistakes:

• Shifting elements forward causing data loss
• Ignoring array bounds when shifting
• Assuming forward loop works for shifting right

5. Given an array arr = {5, 10, 15, 20, 25, 30} of size 6, after inserting element 99 at the middle index, what will be the final array?

hard

A. [5, 10, 15, 99, 20, 25, 30]

B. [5, 10, 99, 15, 20, 25, 30]

C. [5, 10, 15, 20, 99, 25, 30]

D. [5, 10, 15, 20, 25, 30, 99]

Solution

1. Step 1: Calculate middle index for even-sized array

   Array size n=6, middle index = n/2 = 3 (0-based indexing).

2. Step 2: Shift elements right from index 5 to 3

   Shift elements at indices 5,4,3 one position right to make space at index 3.

3. Step 3: Insert 99 at middle index

   Place 99 at index 3 after shifting.

4. Final Answer:

   [5, 10, 15, 99, 20, 25, 30] -> Option A

5. Quick Check:

   Insert at index 3 shifts right [OK]

Hint: For even size, middle = size/2; shift right, then insert [OK]

Common Mistakes:

• Inserting at index 2 instead of 3
• Not shifting elements before insertion
• Appending at the end instead of middle