Question

Question No.1 [CLO 1][7 marks] 1. Consider the following pseudocode: Algorithm IterativeFunction (a, b) // a and b are integers while (a>0) B- a/2 A a-2 end while return b; i. What is the time complexity of the IterativeFunction pseudocode shown above? ii. What is the space complexity of the IterativeFunction pseudocode shown above? 2. What is the time complexity of the following algorithm (Note that n(n+1) 2,2 n(n+1)(2n+1) 2 and ): Provide both T(n) and order, e(f(n)). int A=0; for (int i=0;i<n;i++) for (int j-i;j<n; i++) for (int k=0; k<n; k++) A+=i+j+k; } 3. Starting from an unsorted array of n items, what is the time complexity for the most time efficient algorithm for finding the lowest valued item and moving it to the first location of the array? (The resulting order of the other elements is unimportant.) {

Answer 1

(1):

Time complexity of IterativeFunctioin is O(n/2) which is equivalent to O(n) because every time we decrease the value of "a" by 2, so it takes (n/2) time to complete the code.

Space complexity of IterativeFunctioin is O(1) because we have used constant space like only variable a and b .

(2):

In the given code,

for i=0, j loop iterates from 0 to n-1, which is for n times and k loop iterates for n times. So for i=0, it takes n*n iterations.

for i=1, j loop iterates from 1 to n-1, which is for n-1 times and k loop iterates for n times. So for i=1, it takes n*(n-1)iterations.

.

.

for i=n-1, j loop iterates from n-1 to n-1, which is for 1 time and k loop iterates for n times. So for i=0, it takes niterations.

So the answer is n + n*2 + n*3 + .... + n*n = n[1+2+...+n] = n[n(n+1)/2] = O(n³)

So T(n) = O(n³)

Time complexity of given code is $\theta$ (n³)

(3):

Algorithm is as follows:

• Find the index of minimum value in array. (It takes O(n) time to complete the task)
• Next swap values present at index 0 and minimum value index which takes O(1) time

So the time complexity of this efficient algorith is O(n)+O(1) = O(n)

Time complexity : O(n)

Mention in comments if any mistakes or errors are found. Thank you.