Calculate the time complexity and write the appropriate big order 0? notation, and why for the following: Q2 D- Code D int i, j, k = 0; for (i = 1; i <= n; i++) for (i = 1; i <= n; i++) for (i = 2; i <= n; i++) for (j = 1; j <= n; j = j* 2) k = k + n/ 2;
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- 19 The complexity of the following code is int i, j,k = 0; (i = n/ 2; i <= n; i++) { for (j k = k + n / 2; %3D for at of = 2; j <= n; j = j * 2) { uestion Select one: a. O(nlgn) b. O(n) O c. (n'ign) d. O(n)If f(n) is the number of times that y=y+A[i]; is executed in the code segment below, which of the following is the correct big-Oh notation for f? y=0;for(j=0;j<n;j++) { for (i=0;i<n;i++) { y=y+A[i]; }} Group of answer choices O(1) O(n2) O(nlog2n) O(n)Calculate the time complexity and represent by using asymptotic notation for the following code: int p=3, r=2; int f=1; for(int i=0; iConsider the following recursive function to compute factorialint fact (int n) {if (n <= 1)return 1;elsereturn n * fact(n-1);}int main() {cout << fact(5) << endl;return 0;} Show the stack of activation records each time the function fact is called (either directly or recursively). Each activation record needs to be identified by the name of the function and the parameter value, but there is no need to elaborate the specific content of each activation record.Find the space complexity of the following code fib(int n) { if (n <= 1) return n; return fib(n-1) + fib(n-2); }Question 6 What is the time complexity of this recursive function? def foo(x): if x == 0: return 1 else: return foo(x-1) O O O(N) O(N^2) O(log N) 0(1)What is the time complexity of the following code segment in big O notation? for (int i=0; i<n; i++) { for (int k=0; k<n; k++) { if (k < 5) cout << 5*i*k ; else cout <<i*k; } cout << endl; } Question 17 options: O(n3) O(n2) O(nlogn) O(n)Find the time complexity of the following code, for (i=1: iWhat are the complexities of the following code segments in terms of n? Give an upper bound. a) int i=1;while (i<= n) { int j = i; while (j > 0) j = j/2;i++; } b) int i,j s=0; for (i=0; i<n; i++) { i--; s++; if (s == n) { i++; s = 0; } } c) while (n > 0) { for (int i=0; i<n; i++) sum++; n = n/2; }Recommended textbooks for youDatabase System ConceptsComputer ScienceISBN:9780078022159Author:Abraham Silberschatz Professor, Henry F. Korth, S. SudarshanPublisher:McGraw-Hill Education
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