So you want the number of steps k such that n/2k≤1. That’s the smallest k for which 2k≥n. The definition of the logarithm says that k is about log2(n), so binary search has that complexity. So basically, in this case log2( ) is simplified as log n in the lecture.

What is time complexity of Binary Search with iteration?

O(n2)

Is O 1 faster than O logN?

Sometimes, O(log n) will outperform O(1) but as the input size ‘n’ increases, O(log n) will take more time than the execution of O(1).

Which is better O N or O Nlogn?

Yes for Binary search the time complexity in Log(n) not nlog(n). So it will be less than O(n). But N*Log(N) is greater than O(N).

What is the time complexity recurrence relation of binary search?

Recurrence relation is T(n) = T(n/2) + 1, where T(n) is the time required for binary search in an array of size n.

What is binary search recursively?

Binary search is a recursive algorithm. The high level approach is that we examine the middle element of the list. The value of the middle element determines whether to terminate the algorithm (found the key), recursively search the left half of the list, or recursively search the right half of the list.

Is Logn faster than Nlogn?

Usually the base is less than 4. So for higher values n, n*log(n) becomes greater than n. And that is why O(nlogn) > O(n).

Is Nlogn the same as Logn?

I understand O(Logn) but I haven’t understood O(nlogn). It’s the same as the difference between O(1) and O(n) or the difference between O(n) and O(n^2). You still need to study a lot. O(..) describes the complexity of your algorithm.