Array Insertion at Beginning in DSA Python

Mental Model

To insert a new item at the start of an array, we shift 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 back to free the first chair.

Index: 0   1   2   3   4
Array: [1]->[2]->[3]->[4]->null

Dry Run Walkthrough

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

Goal: Add 0 at the start so the array becomes [0, 1, 2, 3, 4]

Step 1: Shift element at index 3 (4) to index 4

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

Why: Make space at the start by moving last element one step right

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

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

Why: Continue shifting elements right to free index 0

Step 3: Shift element at index 1 (2) to index 2

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

Why: Keep shifting elements right

Step 4: Shift element at index 0 (1) to index 1

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

Why: Last shift to free index 0

Step 5: Insert new value 0 at index 0

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

Why: Place the new value at the beginning

Result:

Final array: [0]->[1]->[2]->[3]->[4]

Annotated Code

DSA Python

class Array:
    def __init__(self, capacity):
        self.capacity = capacity
        self.size = 0
        self.data = [None] * capacity

    def insert_at_beginning(self, value):
        if self.size == self.capacity:
            raise Exception("Array is full")
        # Shift elements right to make space at index 0
        for i in range(self.size - 1, -1, -1):
            self.data[i + 1] = self.data[i]
        self.data[0] = value
        self.size += 1

    def __str__(self):
        return ' -> '.join(str(self.data[i]) for i in range(self.size)) + ' -> null'

# Driver code
arr = Array(10)
arr.data[0] = 1
arr.data[1] = 2
arr.data[2] = 3
arr.data[3] = 4
arr.size = 4

arr.insert_at_beginning(0)
print(arr)

for i in range(self.size - 1, -1, -1):

Shift elements right starting from the last to avoid overwriting

self.data[i + 1] = self.data[i]

Move element at i to i+1 to create space at index 0

self.data[0] = value

Insert new value at the beginning

self.size += 1

Increase size after insertion

Output

0 -> 1 -> 2 -> 3 -> 4 -> null

Complexity Analysis

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

Space: O(1) since insertion is done in place without extra storage

vs Alternative: Compared to appending at the end which is O(1), insertion at beginning is slower due to shifting

Edge Cases

Empty array

Inserts value at index 0 without shifting

DSA Python

if self.size == self.capacity: raise Exception("Array is full")

Array at full capacity

Raises exception to prevent overflow

DSA Python

if self.size == self.capacity: raise Exception("Array is full")

When to Use This Pattern

When you need to add an element at the start of a fixed-size list, use the shifting pattern to make space.

Common Mistakes

Mistake: Shifting elements from left to right causing overwriting
Fix: Shift elements from right to left to preserve data

Mistake: Not checking for full array before insertion
Fix: Add capacity check before shifting and inserting

Summary

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

Use when you need to add items at the beginning of an array with fixed size.

Shift elements from the end backward to avoid overwriting data.

Practice

1. What does inserting an element at the beginning of an array do to the existing elements?

easy

A. Shifts all existing elements one position to the right

B. Deletes the last element of the array

C. Replaces the first element with the new element

D. Leaves the array unchanged

Solution

1. Step 1: Understand array insertion at beginning

   When we insert at the start, the new element takes index 0.

2. Step 2: Effect on existing elements

   All existing elements move one position right to make space.

3. Final Answer:

   Shifts all existing elements one position to the right -> Option A

4. Quick Check:

   Insert at start = shift right [OK]

Hint: New start means shift all right by one [OK]

Common Mistakes:

• Thinking insertion replaces first element
• Assuming array size shrinks
• Believing array stays unchanged

2. Which of the following Python statements correctly inserts the value 10 at the beginning of list arr?

easy

A. arr.add(10)

B. arr.append(10)

C. arr[0] = 10

D. arr.insert(0, 10)

Solution

1. Step 1: Identify correct method for insertion at index

   Python list method insert(index, value) inserts at given index.

2. Step 2: Check each option

   arr.insert(0, 10) uses insert(0, 10) which inserts 10 at start. Others either append at end, replace element, or invalid method.

3. Final Answer:

   arr.insert(0, 10) -> Option D

4. Quick Check:

   Use insert(0, value) to add at start [OK]

Hint: Use insert(0, value) to add at beginning [OK]

Common Mistakes:

• Using append() adds at end, not start
• Assigning arr[0] replaces element, no shift
• Using non-existent method add()

3. What is the output of the following code?

arr = [2, 3, 4]
arr.insert(0, 1)
print(arr)

medium

A. [1, 3, 4]

B. [2, 3, 4, 1]

C. [1, 2, 3, 4]

D. [2, 3, 4]

Solution

1. Step 1: Initial array and insertion

   Original array is [2, 3, 4]. Using insert(0, 1) adds 1 at index 0.

2. Step 2: Resulting array after insertion

   All elements shift right. New array is [1, 2, 3, 4].

3. Final Answer:

   [1, 2, 3, 4] -> Option C

4. Quick Check:

   Insert at start shifts elements right [OK]

Hint: Insert at 0 puts element first, shifts others [OK]

Common Mistakes:

• Thinking insert adds at end
• Replacing first element instead of shifting
• Ignoring shift effect on elements

4. Find the error in the code below that tries to insert 5 at the beginning of the list nums:

nums = [10, 20, 30]
nums.add(5)
print(nums)

medium

A. Method add() does not exist for lists

B. The list nums is empty

C. The print statement is incorrect

D. The value 5 cannot be inserted into a list

Solution

1. Step 1: Check method used for insertion

   Python lists do not have an add() method; this causes an error.

2. Step 2: Correct method to insert at beginning

   Use insert(0, 5) to add 5 at start of list.

3. Final Answer:

   Method add() does not exist for lists -> Option A

4. Quick Check:

   Use insert() not add() for list insertion [OK]

Hint: Use insert() for lists, add() is invalid [OK]

Common Mistakes:

• Assuming add() works like append()
• Thinking print statement causes error
• Believing list cannot hold new values

5. Given the list data = [4, 5, 6], which code snippet correctly inserts the elements of another list new_items = [1, 2, 3] at the beginning so that the final list is [1, 2, 3, 4, 5, 6]?

hard

A. for item in new_items:
data.insert(0, item)

B. for item in reversed(new_items):
data.insert(0, item)

C. data.extend(new_items)

D. data.append(new_items)

Solution

1. Step 1: Understand insertion order impact

   Inserting each item at index 0 shifts existing elements right. To keep order, insert reversed list.

2. Step 2: Analyze each option

   for item in new_items: data.insert(0, item) inserts 1, then 2 at front, reversing order. for item in reversed(new_items): data.insert(0, item) inserts reversed list, preserving order. Options C and D add at end or nested list.

3. Final Answer:

   for item in reversed(new_items): data.insert(0, item) -> Option B

4. Quick Check:

   Insert reversed to keep order at start [OK]

Hint: Insert reversed list at 0 to keep order [OK]

Common Mistakes:

• Inserting forward order reverses new items
• Using extend() adds at end, not start
• Appending list nests it inside