What happens when you delete the first element of an array?
The first element is removed, and all other elements shift one position to the left to fill the gap.
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beginner
Why do we need to shift elements after deleting the first element in an array?
Because arrays have fixed positions, shifting keeps the order and fills the empty space left by the deleted element.
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intermediate
What is the time complexity of deleting the first element in an array?
O(n), where n is the number of elements, because all elements must be shifted one position left.
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beginner
Show the array after deleting the first element from [10, 20, 30, 40].
After deletion and shifting, the array becomes [20, 30, 40].
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intermediate
What is a drawback of deleting the first element in a simple array?
It requires shifting all elements, which can be slow for large arrays.
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What must happen to the elements after deleting the first element of an array?
AAll elements shift one position to the left
BAll elements shift one position to the right
CNo elements move
DArray size increases
✗ Incorrect
After deleting the first element, all other elements shift left to fill the empty space.
What is the time complexity of deleting the first element in an array of size n?
AO(n)
BO(1)
CO(log n)
DO(n^2)
✗ Incorrect
Because all elements must be shifted left, the operation takes O(n) time.
If the array is [5, 15, 25, 35], what is the array after deleting the first element?
A[5, 15, 25, 35]
B[5, 15, 25]
C[15, 25, 35]
D[25, 35]
✗ Incorrect
The first element 5 is removed, and the rest shift left.
Why is deleting the first element in an array less efficient than deleting the last element?
ABecause the last element requires shifting
BBecause the first element requires shifting all other elements
CBecause arrays cannot delete elements
DBecause the first element is always the largest
✗ Incorrect
Deleting the first element requires shifting all other elements left, which takes more time.
Which data structure can delete the first element more efficiently than an array?
AStack
BHash Table
CQueue
DLinked List
✗ Incorrect
Linked lists can delete the first element in O(1) time without shifting.
Explain the steps involved in deleting the first element of an array and why shifting is necessary.
Think about how arrays store elements in fixed positions.
You got /4 concepts.
Describe the time complexity of deleting the first element in an array and compare it with deleting the last element.
Consider how many elements move in each case.
You got /4 concepts.
Practice
(1/5)
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
Step 1: Understand deletion at beginning
Deleting the first element means removing the element at index 0.
Step 2: Effect on array elements
All other elements move one position left to fill the gap.
Final Answer:
All elements shift one position to the left, and the first element is removed. -> Option A
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
Step 1: Identify correct loop for shifting left
The loop must start at 0 and assign arr[i] = arr[i+1] to shift left.
Step 2: Check array size update
After shifting, size decreases by 1 (n--).
Final Answer:
for (int i = 0; i < n - 1; i++) arr[i] = arr[i + 1]; n--; -> Option C
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
Step 1: Perform element shifting
Elements shift left: arr[0]=20, arr[1]=30, arr[2]=40, arr[3]=50.
Step 2: Update size and print
Size n becomes 4, so print arr[0] to arr[3]: 20 30 40 50.
Final Answer:
20 30 40 50 -> Option B
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
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.
Step 2: Understand correct shifting for deletion
Deletion requires shifting elements left: arr[i] = arr[i + 1].
Final Answer:
The loop shifts elements right instead of left, causing data overwrite. -> Option D
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
Step 1: Understand cost of shifting on repeated deletions
Shifting elements left every time costs O(n) per deletion, inefficient for many deletions.
Step 2: Use a start index pointer
Keep a start index that moves forward on each deletion, avoiding shifting.
Step 3: Adjust size accordingly
Decrease size or track remaining elements using start index and size.
Final Answer:
Use a variable to track the start index and increment it on each deletion. -> Option A
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
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