Register

Time Complexity

483 views
0 votes
0 votes

Why ?

Isnt Option A true using "Binary search" 

User Avatar

1 Answer

Best answer
1 votes
1 votes
Because O(logn) time is required only to find the element not to delete it.. To delete it we require to move the other elements to their new position which would require additional time. if it would have been the last elemnt then it would be O(logn) as we search the elemnt and delete it.

Option c is correct As in a sorted array to find the element at index i it requires only O(1) time..
• selected by
User Avatar
Voting & Accepting an answer helps improve the community
← PreviousNext →
← Previous in categoryNext in category →

Related questions

0 votes
0 votes
1 answer
1
Rackson asked Jan 12, 2019
963 views
Algorithms Time complexity
User Avatar
0 votes
0 votes
1 answer
2
srestha asked Nov 21, 2018
2,001 views
Programming (Time Complexity)
What will be time complexity: main() { int sum=0; for(int bound=1;bound<=n;bound*=2) { for(int i=0;i<bound;i++) { for(j=0;j<n;j+=2) { sum+=j; } for(int j=1;j<n;j*=2) { sum*=j; } } } } I think every for loop is participating to find T.C. something*2 means why will it run upto $log n$ and not upto $\frac{n}{2}$
What will be time complexity:main() { int sum=0; for(int bound=1;bound<=n;bound*=2) { for(int i=0;i<bound;i++) { for(j=0;j<n;j+=2) { sum+=j; } for(int j=1;j<n;j*=2) { sum...
User Avatar
0 votes
0 votes
0 answers
3
sidlewis asked Sep 1, 2018
423 views
Testbook Test Series :Algorithms - TIme Complexity
User Avatar
0 votes
0 votes
1 answer
4
pradeepchaudhary asked Aug 19, 2018
813 views
Time Complexity
What would be the asymptotic time complexity to add a node at the end of singly linked list, if the pointer is initially pointing to the head of the list? a) O(1) b) O(n) c) θ(n) d) θ(1)
What would be the asymptotic time complexity to add a node at the end of singly linked list, if the pointer is initially pointing to the head of the list?a) O(1)b) O(n)c)...
User Avatar
Link Copied!