Array Insertion at Beginning in DSA C - Execution Trace

Concept Flow - Array Insertion at Beginning

Check if array is full

No↓

Shift elements right by one

↓

Insert new element at index 0

↓

Increase array size

↓

Done

To insert at the beginning, first shift all elements right, then place the new element at index 0, and update the size.

Execution Sample

DSA C

int arr[5] = {10, 20, 30};
int size = 3;
int capacity = 5;
int new_element = 5;

// Insert at beginning
if (size < capacity) {
  for (int i = size; i > 0; i--) arr[i] = arr[i-1];
  arr[0] = new_element;
  size++;
}

This code inserts 5 at the beginning of array {10, 20, 30} shifting elements right.

Execution Table

Step	Operation	Array State	Pointer Changes	Visual State
1	Check if size < capacity	10, 20, 30, _, _	size=3, capacity=5	Array has space, proceed
2	Shift element at index 3 (empty) to index 4	10, 20, 30, _, _	arr[4] = arr[3]	10 -> 20 -> 30 -> _ -> _
3	Shift element at index 2 to index 3	10, 20, 30, 30, _	arr[3] = arr[2]	10 -> 20 -> 30 -> 30 -> _
4	Shift element at index 1 to index 2	10, 20, 20, 30, _	arr[2] = arr[1]	10 -> 20 -> 20 -> 30 -> _
5	Shift element at index 0 to index 1	10, 10, 20, 30, _	arr[1] = arr[0]	10 -> 10 -> 20 -> 30 -> _
6	Insert new element 5 at index 0	5, 10, 20, 30, _	arr[0] = 5	5 -> 10 -> 20 -> 30 -> _
7	Increase size by 1	5, 10, 20, 30, _	size = 4	5 -> 10 -> 20 -> 30 -> _
8	End	5, 10, 20, 30, _	size=4	Insertion complete

💡 Insertion stops after size reaches 4, array updated with new element at start.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	Final
arr	[10,20,30,_,_]	[10,20,30,_,_]	[10,20,30,30,_]	[10,20,20,30,_]	[10,10,20,30,_]	[5,10,20,30,_]	[5,10,20,30,_]
size	3	3	3	3	3	3	4
capacity	5	5	5	5	5	5	5

Key Moments - 3 Insights

Why do we shift elements starting from the end and not from the beginning?

Shifting from the end prevents overwriting elements before they are moved, as shown in steps 2 to 5 in the execution_table.

What happens if the array is already full?

The insertion does not proceed because size is not less than capacity, as checked in step 1.

Why do we increase the size only after inserting the new element?

Increasing size after insertion ensures the new element is counted in the array length, as shown in step 7.

Visual Quiz - 3 Questions

Look at the execution_table, what is the array state after step 4?

A 10, 10, 20, 30, _

B 10, 20, 20, 30, _

C 5, 10, 20, 30, _

D 10, 20, 30, _, _

💡 Hint

Check the 'Array State' column for step 4 in execution_table.

At which step does the new element get inserted at index 0?

A Step 6

B Step 5

C Step 7

D Step 2

💡 Hint

Look for the operation 'Insert new element 5 at index 0' in execution_table.

If the array size was already equal to capacity at start, what would happen?

A Insertion proceeds normally

B Array elements shift but no insertion

C Insertion does not happen at all

D Size increases beyond capacity

💡 Hint

Refer to step 1 where size < capacity is checked.

Concept Snapshot

Array Insertion at Beginning:
- Check if array has space (size < capacity)
- Shift all elements right by one (from end to start)
- Insert new element at index 0
- Increase size by 1
- Stops if array is full

Full Transcript

To insert an element at the beginning of an array, first check if there is space. If yes, shift all existing elements one position to the right starting from the last element to avoid overwriting. Then place the new element at index 0 and increase the size count. If the array is full, insertion does not happen. This process ensures the new element is at the start and the order of other elements is preserved.

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