1.) O(n)
time taken: 0 sec.
memory: 1.47 MB
2.) O(n^2)
time taken: 0 sec
memory: 1.4 MB
3.) O(n^3)
time taken: 0 sec
memory: 1.3 MB
4.) O(n)
time taken: 0.6 sec
memory: 1.470496 MB
5.) O(n^2)
time taken: 0.54 sec
memory: 1.470496 MB
6.) O(2^n)
time taken: 0 sec
memory: 1.470496 MB
7.) O(n^4)
time taken: 0 sec
memory: 1.470496 MB
For each of the following six program fragments: a. Give an analysis of the running time...
Data Structures and Algorithms For each of the following program fragments, give an analysis of the running time using Big-Oh notation. Do not give formulas, but analyze every line to calculate the running time, e.g. sum = 0 is equal to 1 unit time ... b. sum = 0; for( i = 0; i < n; i++) for( j = 0; j < i*i; j++) for( k = 0; k < j; k++) sum++; c. sum =...
For each of the following two program fragments: a. Give an analysis of the running time (Big-Oh will do). (1) std::vector<int> my_vect; for( i = 0; i < n; i++ ) my_vect.insert(0, i); (2) std::vector<int> my_vect; for( i = 0; i < n; i++ ) my_vect.push_back(i); (3) std::list<int> my_list; for( i = 1; i < n; i++ ) my_list.insert(i, 0); (4) std::list<int> my_list; for( i = 1; i < n; i++ ) my_list.push_front(i);
4. Big-Oh and Rune time Analysis: describe the worst case running time of the following pseudocode functions in Big-Oh notation in terms of the variable n. howing your work is not required (although showing work may allow some partial t in the case your answer is wrong-don't spend a lot of time showing your work.). You MUST choose your answer from the following (not given in any particular order), each of which could be re-used (could be the answer for...
8. R-4.8 Order the following functions by asymptotic growth rate. 4nlogn + 2n 2^10 2^logn 3n + 100logn 4n 2^n n^2 + 10n n^3 nlogn 9. R-4.9 Give a big-Oh characterization, in terms of n, of the running time of the example 1 method shown in Code Fragment 4.12. 10. R-4.10 Give a big-Oh characterization, in terms of n, of the running time of the example 2 method shown in Code Fragment 4.12. 11. R-4.11 Give a big-Oh characterization, in...
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) for(int i=n-1; i >=0; i--){ for(int k=0; k < i*n; k++){ // do something that takes O(1) time } }
Give a big-Oh characterization, in terms of n,of the running time for each of the following code segments (use the drop-down): - public void func1(int n) { A. @(1). for (int i = n; i > 0; i--) { System.out.println(i); B. follogn). for (int j = 0; j <i; j++) System.out.println(j); c.e(n). System.out.println("Goodbye!"); D.@(nlogn). E.e(n). F.ein). public void func2 (int n) { for (int m=1; m <= n; m++) { system.out.println (m); i = n; while (i >0){ system.out.println(i); i...
Exercises • Determine running time for the following code fragments: (a) a = b + c; d = a + e; (b) sum = 0; for (i=0; i<3; i++) for (j=0; j<n; j++) sum++; (c) sum=0; for (i=0; i<n<n; i++) sum++; (d) for (i=0; i < n-1; i++) for (j=i+1; j <n; j++) { tmp = A[i][j]; A[i][j] = A[j] [i]; A[j][i] = tmp; (e) sum = 0; for (i=1; i<=n; i++) for (j=1; j<=n; j+=2) sum++;
For the following program fragment give a Big-O analysis of the running time. Briefly explain your answer: int t = 0; for(int i=1; i <= n; i++) for(int j=1; j <= i*i; j++) if(j % i == 0) t++; What I have so far, O(1) + O(n) + O(n2) + O(1) + O(1) Drop Low order terms: O(n) + O(n2) And I believe the final answer to be O(n3), but not sure if just drop the O(n) or...
1.4.6 Give the order of growth (as a function of n) of the running times of each of the following code fragments: a, int sum=0; for (int k n: k > 0; k /= 2) for (int i 0; ǐ < k; İ++) sum++; b.int sum 0; for (int i = 1; i < n; i *= 2) for (int j = 0; j < i; j++) sum++; int sum = 0; for (int í = 1; i < n;...
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 ++)...