Array Insertion at Beginning in DSA C - Time & Space Complexity

Time Complexity: Array Insertion at Beginning

O(n)

Understanding Time Complexity

We want to understand how long it takes to add a new item at the start of an array.

Specifically, how the time grows as the array gets bigger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void insertAtBeginning(int arr[], int *size, int capacity, int value) {
    if (*size >= capacity) return; // no space
    for (int i = *size; i > 0; i--) {
        arr[i] = arr[i - 1];
    }
    arr[0] = value;
    (*size)++;
}
    

This code inserts a new value at the start of the array by shifting all existing elements one step to the right.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for-loop that shifts elements to the right.
• How many times: It runs once for each existing element in the array (size times).

How Execution Grows With Input

As the array size grows, the number of shifts grows the same way.

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

Pattern observation: The work grows directly with the number of elements; doubling the array size doubles the work.

Final Time Complexity

Time Complexity: O(n)

This means the time to insert at the beginning grows linearly with the array size.

Common Mistake

[X] Wrong: "Inserting at the beginning is always fast because it's just one step."

[OK] Correct: Because all existing elements must move one step to make space, so the time depends on how many elements there are.

Interview Connect

Understanding this helps you explain why arrays are not always the best choice for adding items at the start, 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 insertion at the beginning change?"

Practice

1. What happens when you insert an element at the beginning of an array in C?

easy

A. The new element replaces the first element without shifting others.

B. All existing elements shift one position to the right to make space.

C. The array size automatically increases without shifting elements.

D. The last element is removed to insert the new element at the start.

Solution

1. Step 1: Understand array memory layout

   Arrays store elements in continuous memory. To insert at the start, existing elements must move to free space.

2. Step 2: Shift elements to the right

   Each element moves one position right to open index 0 for the new element.

3. Final Answer:

   All existing elements shift one position to the right to make space. -> Option B

4. Quick Check:

   Insert at start = shift right [OK]

Hint: Insert at start means shift all elements right [OK]

Common Mistakes:

• Assuming new element overwrites first element
• Thinking array size grows automatically
• Removing last element without reason

2. Which of the following is the correct syntax to insert an element at the beginning of an array in C (assuming space is available)?

easy

A. arr[0] = new_element; for(int i = 1; i <= n; i++) arr[i] = arr[i-1];

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

C. for(int i = n; i > 0; i--) arr[i] = arr[i-1]; arr[0] = new_element;

D. arr[n] = new_element; for(int i = n-1; i >= 0; i--) arr[i] = arr[i+1];

Solution

1. Step 1: Identify correct loop direction for shifting

   To avoid overwriting, shift elements from end to start (i = n down to 1).

2. Step 2: Assign new element at index 0

   After shifting, place new_element at arr[0].

3. Final Answer:

   for(int i = n; i > 0; i--) arr[i] = arr[i-1]; arr[0] = new_element; -> Option C

4. Quick Check:

   Shift right from end, then insert [OK]

Hint: Shift from end to start to avoid overwrite [OK]

Common Mistakes:

• Shifting from start to end causing overwrite
• Assigning new element before shifting
• Incorrect loop boundaries

3. What will be the content of the array after executing this code?

int arr[5] = {10, 20, 30, 40, 0};
int n = 4;
int new_element = 5;
for(int i = n; i > 0; i--) {
  arr[i] = arr[i-1];
}
arr[0] = new_element;

medium

A. [0, 10, 20, 30, 40]

B. [10, 5, 20, 30, 40]

C. [5, 10, 20, 30, 0]

D. [5, 10, 20, 30, 40]

Solution

1. Step 1: Shift elements right from index 4 to 1

   arr[4]=arr[3]=40, arr[3]=arr[2]=30, arr[2]=arr[1]=20, arr[1]=arr[0]=10

2. Step 2: Insert new_element at arr[0]

   arr[0] = 5

3. Final Answer:

   [5, 10, 20, 30, 40] -> Option D

4. Quick Check:

   Shift right + insert 5 at start = [5,10,20,30,40] [OK]

Hint: Trace shifts from end, then insert new element at start [OK]

Common Mistakes:

• Not shifting last element correctly
• Overwriting elements during shift
• Forgetting to insert new element at index 0

4. Identify the error in this code snippet that tries to insert an element at the beginning of an array:

int arr[5] = {1, 2, 3, 4, 0};
int n = 4;
int new_element = 0;
for(int i = 0; i < n; i++) {
  arr[i+1] = arr[i];
}
arr[0] = new_element;

medium

A. Loop direction causes overwriting of elements during shift.

B. Array size is too small to insert new element.

C. New element is assigned after shifting elements.

D. Loop condition should be i <= n instead of i < n.

Solution

1. Step 1: Analyze loop direction and assignments

   Loop shifts elements from start to end, causing arr[i+1] to overwrite arr[i+1] before it is copied.

2. Step 2: Correct approach is to shift from end to start

   Shifting from end to start prevents overwriting original values during shift.

3. Final Answer:

   Loop direction causes overwriting of elements during shift. -> Option A

4. Quick Check:

   Shift must go backward to avoid overwrite [OK]

Hint: Shift elements backward to avoid overwriting [OK]

Common Mistakes:

• Shifting forward overwrites data
• Ignoring array size limits
• Incorrect loop boundaries

5. Given an array int arr[7] = {2, 4, 6, 8, 10, 0, 0}; with n = 5 elements, write the steps to insert 1 at the beginning and then 0 at the beginning again. What will be the final array content?

hard

A. [0, 1, 2, 4, 6, 8, 10]

B. [0, 1, 2, 4, 6, 8]

C. [0, 1, 2, 4, 6, 8, 10, 0]

D. [1, 0, 2, 4, 6, 8]

Solution

1. Step 1: Insert 1 at beginning

   Shift elements right: arr[5]=arr[4]=10, arr[4]=arr[3]=8, arr[3]=arr[2]=6, arr[2]=arr[1]=4, arr[1]=arr[0]=2; then arr[0]=1. Array: [1, 2, 4, 6, 8, 10]

2. Step 2: Insert 0 at beginning

   Shift elements right again: arr[6]=arr[5]=10, arr[5]=arr[4]=8, arr[4]=arr[3]=6, arr[3]=arr[2]=4, arr[2]=arr[1]=2, arr[1]=arr[0]=1; then arr[0]=0. Array: [0, 1, 2, 4, 6, 8, 10]

3. Final Answer:

   [0, 1, 2, 4, 6, 8, 10] -> Option A

4. Quick Check:

   Two shifts + inserts = [0,1,2,4,6,8,10] [OK]

Hint: Shift right twice, insert each new element at index 0 [OK]

Common Mistakes:

• Not updating array size after insertions
• Overwriting elements during second insertion
• Assuming array can hold more than 6 elements