Array Deletion at Beginning in DSA C++ - Time & Space Complexity

Time Complexity: Array Deletion at Beginning

O(n)

Understanding Time Complexity

When we delete an element from the start of an array, the time it takes depends on how many elements need to move.

We want to know how the work grows as the array gets bigger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void deleteAtBeginning(int arr[], int &n) {
    for (int i = 1; i < n; i++) {
        arr[i - 1] = arr[i];
    }
    n--;
}
    

This code removes the first element by shifting all other elements one position to the left.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for-loop that shifts elements left.
• How many times: It runs (n - 1) times, where n is the array size.

How Execution Grows With Input

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

Input Size (n)	Approx. Operations (shifts)
10	9
100	99
1000	999

Pattern observation: The operations increase roughly in a straight line with input size.

Final Time Complexity

Time Complexity: O(n)

This means the time to delete the first element grows directly with the number of elements in the array.

Common Mistake

[X] Wrong: "Deleting the first element is always fast and takes constant time."

[OK] Correct: Because all other elements must move to fill the gap, the work depends on the array size, not just one step.

Interview Connect

Understanding this helps you explain why some data structures are better for front deletions, showing your grasp of efficiency.

Self-Check

"What if we used a linked list instead of an array? How would the time complexity of deleting the first element change?"

Practice

1. What happens to the elements of an array when you delete the first element (at the beginning)?

easy

A. All elements shift one position to the left, and the first element is removed.

B. All elements shift one position to the right, and the last element is removed.

C. The array size increases by one.

D. The array remains unchanged.

Solution

1. Step 1: Understand deletion at beginning

   Deleting the first element means removing the element at index 0.

2. Step 2: Effect on array elements

   All other elements move one position left to fill the gap.

3. Final Answer:

   All elements shift one position to the left, and the first element is removed. -> Option A

4. Quick Check:

   Deletion at start shifts elements left [OK]

Hint: Deleting start means shift left all elements [OK]

Common Mistakes:

• Thinking elements shift right instead of left
• Assuming array size increases
• Believing array stays unchanged

2. Which of the following C++ code snippets correctly deletes the first element of an array arr of size n by shifting elements left?

easy

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

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

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

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

Solution

1. Step 1: Identify correct loop for shifting left

   The loop must start at 0 and assign arr[i] = arr[i+1] to shift left.

2. Step 2: Check array size update

   After shifting, size decreases by 1 (n--).

3. Final Answer:

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

4. Quick Check:

   Shift left loop from 0 to n-2, then n-- [OK]

Hint: Shift left from i=0 to n-2, then reduce size [OK]

Common Mistakes:

• Looping backwards instead of forwards
• Increasing size instead of decreasing
• Copying elements incorrectly

3. What is the output of the following C++ code?

int arr[5] = {10, 20, 30, 40, 50};
int n = 5;
for (int i = 0; i < n - 1; i++) {
    arr[i] = arr[i + 1];
}
n--;
for (int i = 0; i < n; i++) {
    std::cout << arr[i] << " ";
}

medium

A. 50 40 30 20

B. 20 30 40 50

C. 10 20 30 40 50

D. 30 40 50

Solution

1. Step 1: Perform element shifting

   Elements shift left: arr[0]=20, arr[1]=30, arr[2]=40, arr[3]=50.

2. Step 2: Update size and print

   Size n becomes 4, so print arr[0] to arr[3]: 20 30 40 50.

3. Final Answer:

   20 30 40 50 -> Option B

4. Quick Check:

   First element removed, rest shifted left [OK]

Hint: After deletion, print first n-1 elements shifted left [OK]

Common Mistakes:

• Printing original array without shift
• Printing one extra element
• Confusing order of elements

4. Identify the error in the following code that attempts to delete the first element of an array arr of size n:

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

medium

A. The loop should start from i = 0 and go to n - 1.

B. The loop correctly shifts elements left, no error.

C. The array size should increase, not decrease.

D. The loop shifts elements right instead of left, causing data overwrite.

Solution

1. Step 1: Analyze loop direction and assignment

   The loop assigns arr[i] = arr[i - 1], shifting elements right, which is wrong for deletion at start.

2. Step 2: Understand correct shifting for deletion

   Deletion requires shifting elements left: arr[i] = arr[i + 1].

3. Final Answer:

   The loop shifts elements right instead of left, causing data overwrite. -> Option D

4. Quick Check:

   Deletion needs left shift, not right [OK]

Hint: Deletion shifts left; right shift is wrong [OK]

Common Mistakes:

• Shifting elements in wrong direction
• Not updating array size
• Starting loop at wrong index

5. Given an array arr of size n, you want to delete the first element multiple times until the array is empty. Which approach is the most efficient to handle this repeated deletion without shifting elements every time?

hard

A. Use a variable to track the start index and increment it on each deletion.

B. Shift all elements left on every deletion and decrease size.

C. Create a new array each time with one less element.

D. Reverse the array and delete from the end.

Solution

1. Step 1: Understand cost of shifting on repeated deletions

   Shifting elements left every time costs O(n) per deletion, inefficient for many deletions.

2. Step 2: Use a start index pointer

   Keep a start index that moves forward on each deletion, avoiding shifting.

3. Step 3: Adjust size accordingly

   Decrease size or track remaining elements using start index and size.

4. Final Answer:

   Use a variable to track the start index and increment it on each deletion. -> Option A

5. Quick Check:

   Track start index to avoid shifting [OK]

Hint: Track start index to delete efficiently [OK]

Common Mistakes:

• Shifting elements every time causing slow code
• Creating new arrays repeatedly wasting memory
• Reversing array unnecessarily