Search a sorted array by repeatedly dividing the search interval in half. Begin with an interval covering the whole array. If the value of the search key is less than the item in the middle of the interval, narrow the interval to the lower half. Otherwise narrow it to the upper half. Repeatedly check until the value is found or the interval is empty. It takes a maximum of log(n) searches to search an element from the sorted array.

Binary search in a sorted array
The time to search in an array of ‘n’ elements is equal to the time to search in an array of n/2 elements plus k comparison.
T(n)=T(n/2)+k // k is constant

In the worst case, the element to be searched has to be compared with all elements of linked list. It will take O(n) time to search the element.

In the worst case, the element to be searched has to be compared with all elements of linked list, so the time complexity is O(n)

Total comparisons required
= No. of comparisons if element present in 1^st position + No. of comparisons if element present in 2^ndnd position + ............. + No. of comparisons if element present in n^th position
= 1+2+3+ ... + n
= n(n+1)/2
Since there are n elements in the list, so average no. of comparisons
= Total comparisons/Total no. of elements
= (n(n+1)/2)/n
= n+1/2

A linear search or sequential search is a method for finding an element within a list. It sequentially checks each element of the list until a match is found or the whole list has been searched
A linear search runs in at worst linear time and makes at most n comparisons, where n is the length of the list , whether element is found or not

We can arrange 11 items in the four levels(level-0,1,2,3).
Each level has 2i nodes where i is level number
Level -0 has one node so one comparison(1)
Level -1 has two nodes so for each node two comparisons and total four comparisons (4)
Level -2 has four nodes and total comparisons are 4x3 =12 comparisons
Level-3 has total 8 nodes but in the given question we need to process only 11 items and already we covered 7 nodes so remaining nodes 4 nodes at level -3. Total comparisons at level-3 are 4*4=16
So, total number of comparisons at each level are = 1+4+12+16= 33
Average comparisons required for 11 items = 33/11 = 3

This above code is similar to binary search
First iteration:
I=0,j=9
K=(0+9)/2=4
Elements with duplicate values of “2” the condition if(Y[k] Second iteration :
I=0, j=k=4
K= (0+4)/2=2.
Third iteration:
I=0, j=k=2
K= (0+2)/2=1
Fourth iteration:
I=0, j=k=1
K= (0+1)/2=0
The while condition is Y[k] != x && i < j is true
● The above program doesn’t work for the cases where element to be searched is the last element of Y[ ] or greater than the last element (or maximum element) in Y[ ].
● For such cases, program goes in an infinite loop because i is assigned value as k in all iterations, and i never becomes equal to or greater than j.
● So while condition never becomes false.

This above code is similar to binary search
First iteration:
I=0,j=9
K=(0+9)/2=4
Elements with duplicate values of “2” the condition if(Y[k] Second iteration :
I=0, j=k=4
K= (0+4)/2=2.
Third iteration:
I=0, j=k=2
K= (0+2)/2=1
Fourth iteration:
I=0, j=k=1
K= (0+1)/2=0
The while condition is Y[k] != x && i < j is true
→ The above program doesn’t work for the cases where element to be searched is the last element of Y[ ] or greater than the last element (or maximum element) in Y[ ].
→ For such cases, program goes in an infinite loop because i is assigned value as k in all iterations, and i never becomes equal to or greater than j.
So, while condition never becomes false.

→ Linear search algorithm will traverse all nodes with increment by one element in linear fashion(order).
→ So, it takes time complexity O(n)

Step-1: Binary search we are using formula O(log​ ₂​ n)
Sep-2: The maximum number of comparisons for a particular record among 32 sorted recorder through binary search method will be O(log​ ₂​ 32)=5

Height of the binary search tree = 3

In general, if N is the size of the list to be searched and C is the number of comparisons to do so in the worst case, C = log₂N. In the question two sorted list of elements with the condition A[n]22N

We can easily find solutions based on given options because option D given all numbers in ascending order. In Order will print ascending order only.

Possibility-1:
→ Traversing all the nodes of a binary search tree with n nodes and printing them in order using inorder traversal.
→ Inorder traversal will print all elements in sorted order. It takes worst case time complexity is O(n).
Possibility-2:
Without sorted elements chance to form either left skewed or right skewed tree. It takes worst case time complexity is O(n).

→ A linear search or sequential search is a method for finding an element within a list. It sequentially checks each element of the list until a match is found or the whole list has been searched.
→ A linear search runs in at worst linear time and makes at most n comparisons, where ‘n’ is the length of the list. If each element is equally likely to be searched, then linear search has an average case of n/2 comparisons, but the average case can be affected if the search probabilities for each element vary. → Linear search is rarely practical because other search algorithms and schemes, such as the binary search algorithm and hash tables, allow significantly faster searching for all but short lists.