Array Deletion at End in DSA C++ - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output after deleting the last element from the array?

Consider the following C++ code that deletes the last element from an array. What will be printed?

DSA C++

int arr[] = {10, 20, 30, 40, 50};
int n = 5;

// Delete last element
n = n - 1;

for (int i = 0; i < n; i++) {
    std::cout << arr[i] << " ";
}

A 10 20 30 40 50

B 50

C 10 20 30 40

D 10 20 30

💡 Hint

Deleting the last element means reducing the size by one and printing only up to the new size.

The array initially has 5 elements. After reducing the size by one, the last element (50) is ignored. The loop prints elements from index 0 to 3, which are 10, 20, 30, and 40.

❓ Predict Output

intermediate

What happens if we delete the last element twice?

Given the code below, what will be the output after deleting the last element two times?

DSA C++

int arr[] = {5, 15, 25, 35, 45};
int n = 5;

// Delete last element twice
n = n - 1;
n = n - 1;

for (int i = 0; i < n; i++) {
    std::cout << arr[i] << " ";
}

A 5 15 25 35

B 5 15 25

C 5 15 25 35 45

D 15 25 35

💡 Hint

Each deletion reduces the size by one, so two deletions reduce size by two.

Starting with 5 elements, deleting twice reduces size to 3. The loop prints elements at indices 0, 1, and 2: 5, 15, and 25.

🔧 Debug

advanced

Why does this code cause an error when deleting the last element?

Identify the error in the code below that tries to delete the last element from an array and explain why it happens.

DSA C++

int arr[] = {1, 2, 3};
int n = 3;

// Delete last element
n = n - 1;

for (int i = 0; i <= n; i++) {
    std::cout << arr[i] << " ";
}

A It causes an out-of-bounds error because the loop runs one index too far.

B It causes a syntax error due to missing semicolon.

C It causes a type error because n is not an integer.

D It causes no error and prints all elements correctly.

💡 Hint

Check the loop condition carefully after reducing the size.

After reducing n to 2, the loop runs from i = 0 to i = 2 inclusive (i <= n). This tries to access arr[2], which is valid, but since n was reduced to 2, the last valid index is 1. Actually, the original array has indices 0,1,2. After deletion, n=2 means size is 2, so valid indices are 0 and 1. Loop condition i <= n runs i=0,1,2 which accesses arr[2] which is out of the new size bounds. This causes out-of-bounds access.

❓ Predict Output

advanced

What is the output after deleting the last element from a dynamic array?

Consider this C++ code using a vector. What will be printed after deleting the last element?

DSA C++

#include <iostream>
#include <vector>

int main() {
    std::vector<int> v = {100, 200, 300, 400};
    v.pop_back();
    for (int x : v) {
        std::cout << x << " ";
    }
    return 0;
}

A 400

B 100 200 300 400

C 100 200

D 100 200 300

💡 Hint

pop_back() removes the last element from the vector.

The vector initially has 4 elements. pop_back() removes the last element (400). The loop prints the remaining elements: 100, 200, 300.

🧠 Conceptual

expert

What is the time complexity of deleting the last element from an array vs a linked list?

Compare the time complexity of deleting the last element from a fixed-size array and a singly linked list. Which statement is correct?

A Deleting last element in an array is O(1), in a singly linked list is O(n).

B Deleting last element in an array is O(n), in a singly linked list is O(1).

C Deleting last element in both array and singly linked list is O(n).

D Deleting last element in both array and singly linked list is O(1).

💡 Hint

Think about how you access the last element in each data structure.

In an array, deleting the last element is just reducing the size, which is O(1). In a singly linked list, to delete the last node, you must traverse from the head to the node before last, which takes O(n) time.

Practice

1. What happens to the size of an array when you delete an element from the end in a dynamic array?

easy

A. The size decreases by one

B. The size remains the same

C. The size increases by one

D. The size becomes zero

Solution

1. Step 1: Understand dynamic array size behavior

   When an element is deleted from the end, the array size reduces by one because one element is removed.

2. Step 2: Confirm size change after deletion

   Deleting from the end does not affect other elements, only reduces size by one.

3. Final Answer:

   The size decreases by one -> Option A

4. Quick Check:

   Size after deletion = original size - 1 [OK]

Hint: Deleting at end reduces size by one [OK]

Common Mistakes:

• Thinking size remains same after deletion
• Assuming size increases after deletion
• Confusing deletion at end with deletion at start

2. Which of the following is the correct syntax to delete the last element from a vector named vec in C++?

easy

A. vec.erase_end();

B. vec.delete_last();

C. vec.remove_end();

D. vec.pop_back();

Solution

1. Step 1: Recall vector deletion syntax

   The standard method to remove the last element from a vector in C++ is pop_back().

2. Step 2: Verify other options

   Other methods like delete_last(), remove_end(), and erase_end() do not exist in C++ STL vector.

3. Final Answer:

   vec.pop_back(); -> Option D

4. Quick Check:

   Use pop_back() to delete last element [OK]

Hint: Use pop_back() to remove last element in vector [OK]

Common Mistakes:

• Using non-existent methods like delete_last()
• Confusing erase() with erase_end()
• Trying to use remove_end() which is invalid

3. What will be the output of the following C++ code?

std::vector<int> v = {10, 20, 30, 40};
v.pop_back();
for(int x : v) std::cout << x << " ";

medium

A. 10 20 30 40

B. 20 30 40

C. 10 20 30

D. 10 20 40

Solution

1. Step 1: Understand pop_back effect on vector

   pop_back() removes the last element, which is 40 in this vector.

2. Step 2: Print remaining elements

   After removal, vector contains 10, 20, 30. Loop prints these with spaces.

3. Final Answer:

   10 20 30 -> Option C

4. Quick Check:

   Last element removed, output = 10 20 30 [OK]

Hint: pop_back() removes last element before printing [OK]

Common Mistakes:

• Printing original vector without deletion
• Confusing which element is removed
• Forgetting pop_back() modifies vector

4. Identify the error in the following code that attempts to delete the last element of an array:

int arr[3] = {1, 2, 3};
arr.pop_back();

medium

A. pop_back() deletes first element, not last

B. pop_back() is not a member of built-in arrays

C. Array size is zero, cannot delete

D. No error, code is correct

Solution

1. Step 1: Check array type and methods

   Built-in arrays in C++ do not have member functions like pop_back().

2. Step 2: Understand correct usage

   pop_back() is a method of std::vector, not raw arrays.

3. Final Answer:

   pop_back() is not a member of built-in arrays -> Option B

4. Quick Check:

   pop_back() only works with vectors, not arrays [OK]

Hint: pop_back() works only on vectors, not raw arrays [OK]

Common Mistakes:

• Assuming pop_back() works on raw arrays
• Ignoring that arrays have fixed size
• Thinking pop_back() deletes first element

5. You have a vector v with elements {5, 10, 15, 20, 25}. You want to delete the last two elements and then print the vector. Which code snippet correctly does this?

hard

A. v.erase(v.end()-2, v.end()); for(auto x : v) std::cout << x << ' ';

B. v.pop_back(); for(auto x : v) std::cout << x << ' ';

C. v.remove_end(); v.remove_end(); for(auto x : v) std::cout << x << ' ';

D. v.pop_back(2); for(auto x : v) std::cout << x << ' ';

Solution

1. Step 1: Understand methods to delete multiple elements

   pop_back() removes only one element at a time. To remove two, it must be called twice. erase() can remove a range efficiently.

2. Step 2: Analyze the snippets

   The snippet v.erase(v.end()-2, v.end()); for(auto x : v) std::cout << x << ' '; correctly removes the last two elements. The snippet with a single v.pop_back(); removes only one element. remove_end() does not exist. pop_back(2) uses invalid syntax.

3. Final Answer:

   v.erase(v.end()-2, v.end()); for(auto x : v) std::cout << x << ' '; -> Option A

4. Quick Check:

   erase() with range removes last two elements efficiently [OK]

Hint: Use erase(end()-2, end()) to delete last two elements [OK]

Common Mistakes:

• Using pop_back() only once instead of twice
• Using non-existent remove_end() method
• Not knowing erase() can remove ranges