The worst-case time complexity of inserting n elements into an empty linked list, if the linked list needs to be maintained in sorted order, is Θ(n^2).

Here's why:

1. For a linked list, insertion at the beginning or at the end takes constant time, i.e., O(1). However, if the linked list needs to be maintained in sorted order, we cannot simply insert at the beginning or the end.

2. We need to find the correct position for each element to maintain the sorted order. This requires traversing the list until we find the correct position. In the worst case, we may have to traverse the entire list.

3. If we are inserting n elements, for the first element, we just insert it as the list is empty. For the second element, we may have to traverse 1 node to find its correct position. For the third element, we may have to traverse 2 nodes, and so on.

4. So, the total number of operations can be calculated as 0 + 1 + 2 + 3 + ... + (n-1), which is equal to n(n-1)/2.

5. Therefore, the worst-case time complexity is O(n^2).

So, the correct answer is c. Θ(n^2).

Question

The worst-case time complexity of inserting n elements into an empty linked list, if the linked list needs to be maintained in sorted order, is Θ(n^2).

Here's why:

1. For a linked list, insertion at the beginning or at the end takes constant time, i.e., O(1). However, if the linked list needs to be maintained in sorted order, we cannot simply insert at the beginning or the end.

2. We need to find the correct position for each element to maintain the sorted order. This requires traversing the list until we find the correct position. In the worst case, we may have to traverse the entire list.

3. If we are inserting n elements, for the first element, we just insert it as the list is empty. For the second element, we may have to traverse 1 node to find its correct position. For the third element, we may have to traverse 2 nodes, and so on.

4. So, the total number of operations can be calculated as 0 + 1 + 2 + 3 + ... + (n-1), which is equal to n(n-1)/2.

5. Therefore, the worst-case time complexity is O(n^2).

So, the correct answer is c. Θ(n^2).

Knowee AI · Accepted Answer

The worst-case time complexity of inserting n elements into an empty linked list, if the linked list needs to be maintained in sorted order, is Θ(n^2).

Here's why:

1. For a linked list, insertion at the beginning or at the end takes constant time, i.e., O(1). However, if the linked list needs to be maintained in sorted order, we cannot simply insert at the beginning or the end.

2. We need to find the correct position for each element to maintain the sorted order. This requires traversing the list until we find the correct position. In the worst case, we may have to traverse the entire list.

3. If we are inserting n elements, for the first element, we just insert it as the list is empty. For the second element, we may have to traverse 1 node to find its correct position. For the third element, we may have to traverse 2 nodes, and so on.

4. So, the total number of operations can be calculated as 0 + 1 + 2 + 3 + ... + (n-1), which is equal to n(n-1)/2.

5. Therefore, the worst-case time complexity is O(n^2).

So, the correct answer is c. Θ(n^2).

What is the worst-case time complexity of inserting n elements into an empty linked list, if the linked list needs to be maintained in sorted order? More than one answer may be correct. [GATE CSE 2020]Question 5Answera.Θ(1)b.Θ(n log n)c.Θ(n^2)d.Θ

Question

Solution

Similar Questions

Upgrade your grade with Knowee