[Type B] Chapter– 7 Class 12 CS – Sumita Arora Assignment | Q/A

Solution:

O(n)

Due to the while loop having upper bound input i.e. n

Solution:

O(n^3)

Due to the three nested while loop each having n and product of three loop i.e. n^3.

Solution:

O(5n^2)

Due to the nesting of the doIt() and having 5n in product with n for the while loop.

Solution:

O(n) = n^4

Due to the nested while loop having upper bound input i.e. n hence it gives n^2 in product with nested function doIt() having O(n) = n^2.

Solution:

O(n) = n^3

Due to the while loop having upper bound input i.e. n and the nested function doIt() having O(n) = n^2. Hence O(n) =n*n^2

Solution:

O(n) = n

As the while loop will execute for the n times because i is initialized with n in the starting.

Solution:

O(n) = n*p*m

The all three nested loop will be get product with upper bound to provide the complexity.

Solution:

O(n) = logn

As of the mid positioning property similar to binary search.

Solution:

O(n) = 2n

The two concatenate for loop of each upper bound n will be added to result the complexity.
O(n) = n+n

Solution:

O(n) = n^2

Solution:

O(n) = n
 
The while loop will execute for the n times as if bound has at most limit is n.

Solution:

O(n) = logn

As of the mid positioning property similar to binary search.

Solution:

O(n) = 2n

The two concatenate for loop of each upper bound n will be added to result the complexity.
O(n) = n+n

(i) Insertion Sort
O(n^2)

(ii) Binary Search
O(logn)

(iii) Linear Search
O(n)

Binary sort will be a good option as of having O(logn) time complexity than the other mentioned options.