Question 1 (25 pts)
Find the running time complexity for the following code fragments. Express your answers using either the Big-O or Big-Θ notations, and the tightest bound possible. Justify your answers.

First loop is running for n*n times. Second for loop is running for i times. Max value of i is n*n. So, Total time complexity = O(n*n*n*n) = O(n^4)
Question 1 (25 pts) Find the running time complexity for the following code fragments. Express yo...
Question 3: Given the following two
code fragments [2 Marks]
(i)Find T(n), the time complexity (as
operations count) in the worst case?
(ii)Express the growth rate of the
function in asymptotic notation in the closest bound possible.
(iii)Prove that T(n) is Big O (g(n)) by
the definition of Big O
(iv)Prove that T(n) is (g(n)) by using
limits
1(5 pts): For each code fragment below, give the complexity of the algorithm (O or Θ). Give the tightest possible upper bound as the input size variable increases. The input size variable in these questions is exclusively n. Complexity Code public static int recursiveFunction (int n)f f( n <= 0 ) return 0; return recursiveFunction (n - 1) 1; for(int i 0i <n; i+) j=0; for ( int j k=0; i; k < < j++) for (int j; m <...
(10') 6. For each of the following code blocks, write the best (tightest) big-o time complexity i) for (int i = 0; ǐ < n/2; i++) for (int j -0: ni j++) count++ i) for (int í = 0; i < n; i++) for (int ni j0 - for (int k j k ni kt+) count++ İİİ) for (int í ー 0; i < n; i++) for(int j = n; j > 0; j--) for (int k = 0; k...
Using C++ please explain
What is the Big-O time complexity of the following code: for (int i=0; i<N; i+=2) { ... constant time operations... Select one: o a. O(n^2) O b. O(log n) c. O(n) O d. 0(1) What is the Big-O time complexity of the following code: for(int i=1; i<N; i*=2) { ... constant time operations... Select one: O O a. O(n^2) b. 0(1) c. O(n) d. O(log n) O What is the Big-O time complexity of the following...
Compute the total running time for each of the functions in the following code. First, compute the running time in terms of some constants a, b, c, d, e, etc. Show your work and give the answer in the text box. Then give the tightest big-O bound you can for each. (All of the functions are O(n!), but you can give a more informative bound for each.) void f(int n) { int i = 1; while (i <= sqrt(n)) {...
Which big-O expression best characterizes the worst case time complexity of the following code? public static int foo(int N) ( int count = 0; int i1; while (i <N) C for (int j = 1; j < N; j=j+2) { count++ i=i+2; return count; A. O(log log N) B. O(log N2) C. O(N log N) D. O(N2)
In Big-Θ notation, analyze the running time of the following pieces of code/pseudo-code. Describe the running time as a function of the input size (here, n) int *a = new int [10]; // new is O(1) int size = 10; for (int i = 0; i < n; i ++) { if (i == size) { int newsize = 3*size/2; int *b = new int [newsize]; // new is O(1) for (int j = 0; j < size; j ++)...
discrete math
(1) (15 pts) Time Complexity Analysis 1) (5 pts) What is the time complexity of the following code segment? Explain your answer; otherwise, you can't get full mark from this question. for(int i=1; i<n; i*=2) { sum-0; sum++; Answer: 2) (5 pts) What is the time complexity of the following code segment? Explain your answer; otherwise, you can't get full mark from this question. for(int j=0; j<n; j++){ for (int k=0; k<n; k++) { for (int =0; i<n;...
find complexity Problem 1 Find out the computational complexity (Big-Oh notation) of the code snippet: Code 1: for (int i = n; i > 0; i /= 2) { for (int j = 1; j < n; j *= 2) { for (int k = 0; k < n; k += 2) { // constant number of operations here } } } Code 2: Hint: Lecture Note 5, Page 7-8 void f(int n) { if (n...
Show your work Count the number of operations and the big-O time complexity in the worst-case and best-case for the following code int small for ( i n t i = 0 ; i < n ; i ++) { i f ( a [ i ] < a [ 0 ] ) { small = a [ i ] ; } } Show Work Calculate the Big-O time complexity for the following code and explain your answer by showing...