Array Deletion at Beginning in DSA C++ - Build from Scratch

Array Deletion at Beginning

📖 Scenario: You have a list of daily temperatures recorded in a week. You want to remove the oldest temperature (the first one) to keep the list updated.

🎯 Goal: Learn how to delete the first element from an array by shifting all other elements to the left.

📋 What You'll Learn

Create an array with exact values

Create a variable for the array size

Write code to delete the first element by shifting

Print the updated array after deletion

💡 Why This Matters

🌍 Real World

Managing lists of data where the oldest entry needs to be removed, like daily temperatures or recent messages.

💼 Career

Understanding array manipulation is fundamental for programming tasks involving fixed-size data and memory management.

1

Create the temperature array

Create an integer array called temperatures with these exact values: 30, 32, 31, 29, 35, 33, 34.

DSA C++

// Your code here

Hint

Use curly braces to list the values inside the array.

2

Create the size variable

Create an integer variable called size and set it to 7 to represent the number of elements in temperatures.

DSA C++

int temperatures[7] = {30, 32, 31, 29, 35, 33, 34};
// Your code here

Hint

Use a simple integer variable to store the size.

3

Delete the first element by shifting

Use a for loop with variable i from 0 to size - 2 to shift each element in temperatures one position to the left, overwriting the first element.

DSA C++

int temperatures[7] = {30, 32, 31, 29, 35, 33, 34};
int size = 7;
// Your code here

Hint

Remember to reduce the size by 1 after shifting.

4

Print the updated array

Use a for loop with variable i from 0 to size - 1 to print each element of temperatures separated by spaces on one line.

DSA C++

int temperatures[7] = {30, 32, 31, 29, 35, 33, 34};
int size = 7;
for (int i = 0; i < size - 1; i++) {
    temperatures[i] = temperatures[i + 1];
}
size--;
// Your code here

Hint

Print all elements from index 0 to size - 1 separated by spaces.

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