Array Insertion at Beginning in DSA C

Mental Model

To insert a new item at the start of an array, we move all existing items one step to the right to make space.

Analogy: Imagine a row of chairs where you want to add a new person at the front. Everyone must stand up and move one chair to the right to free the first chair.

Index: 0   1   2   3   4
Array: [1]->[2]->[3]->[4]->[5]
Insert new value 0 at index 0
After shift:
       [ ]->[1]->[2]->[3]->[4]->[5]

Dry Run Walkthrough

Input: array: [1, 2, 3, 4, 5], insert value 0 at beginning

Goal: Insert 0 at index 0 and shift all elements right

Step 1: Shift element at index 4 (5) to index 5

[1]->[2]->[3]->[4]->[5]->[5]

Why: Make space at the end to avoid overwriting data

Step 2: Shift element at index 3 (4) to index 4

[1]->[2]->[3]->[4]->[4]->[5]

Why: Continue shifting elements right to free index 0

Step 3: Shift element at index 2 (3) to index 3

[1]->[2]->[3]->[3]->[4]->[5]

Why: Keep shifting elements right

Step 4: Shift element at index 1 (2) to index 2

[1]->[2]->[2]->[3]->[4]->[5]

Why: Keep shifting elements right

Step 5: Shift element at index 0 (1) to index 1

[1]->[1]->[2]->[3]->[4]->[5]

Why: Last shift to free index 0

Step 6: Insert new value 0 at index 0

[0]->[1]->[2]->[3]->[4]->[5]

Why: Place new element at the beginning

Result:

[0]->[1]->[2]->[3]->[4]->[5]

Annotated Code

DSA C

#include <stdio.h>

void insertAtBeginning(int arr[], int *size, int capacity, int value) {
    if (*size >= capacity) {
        printf("Array is full, cannot insert\n");
        return;
    }
    // Shift elements right to make space at index 0
    for (int i = *size - 1; i >= 0; i--) {
        arr[i + 1] = arr[i]; // move element one step right
    }
    arr[0] = value; // insert new value at beginning
    (*size)++; // increase size
}

void printArray(int arr[], int size) {
    for (int i = 0; i < size; i++) {
        printf("%d", arr[i]);
        if (i != size - 1) printf(" -> ");
    }
    printf("\n");
}

int main() {
    int capacity = 10;
    int arr[10] = {1, 2, 3, 4, 5};
    int size = 5;

    insertAtBeginning(arr, &size, capacity, 0);
    printArray(arr, size);

    return 0;
}

for (int i = *size - 1; i >= 0; i--) { arr[i + 1] = arr[i]; }

Shift all elements one position right to free index 0

arr[0] = value;

Insert new value at the beginning of the array

(*size)++;

Increase the size of the array after insertion

Output

0 -> 1 -> 2 -> 3 -> 4 -> 5

Complexity Analysis

Time: O(n) because we must shift all existing elements one step to the right

Space: O(1) because insertion is done in place without extra arrays

vs Alternative: Compared to inserting at the end (O(1)), inserting at beginning requires shifting all elements, making it slower for large arrays

Edge Cases

Empty array

Insertion simply places the new value at index 0 without shifting

DSA C

for (int i = *size - 1; i >= 0; i--) { arr[i + 1] = arr[i]; }

Array at full capacity

Insertion is rejected with a message to avoid overflow

DSA C

if (*size >= capacity) { printf("Array is full, cannot insert\n"); return; }

Single element array

The single element is shifted right and new value placed at index 0

DSA C

for (int i = *size - 1; i >= 0; i--) { arr[i + 1] = arr[i]; }

When to Use This Pattern

When you need to add an element at the start of a fixed-size list and must keep order, use array insertion with shifting to make space.

Common Mistakes

Mistake: Shifting elements from left to right causing overwriting data
Fix: Shift elements from right to left (end to start) to avoid overwriting

Mistake: Not checking if array has capacity before insertion
Fix: Add a capacity check before shifting and inserting

Summary

Inserts a new element at the start of an array by shifting existing elements right.

Use when you want to add an item at the beginning of a fixed-size array while preserving order.

Shift elements from the end towards the start to avoid overwriting data during insertion.

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