Question
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
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 terms of n, of the running time of the example 3 method shown in Code Fragment 4.12.
12. R-4.12 Give a big-Oh characterization, in terms of n, of the running time of the example 4 method shown in Code Fragment 4.12.
13. R-4.13 Give a big-Oh characterization, in terms of n, of the running time of the example 5 method shown in Code Fragment 4.12.
/?? Returns the sum of the integers in given array. ?/
public static int example1(int[ ] arr) {
int n = arr.length, total = 0;
for (int j=0; j < n; j++)
total += arr[j];
return total; // loop from 0 to n-1
/?? Returns the sum of the integers with even index in given array. ?/
public static int example2(int[ ] arr) {
int n = arr.length, total = 0;
for (int j=0; j < n; j+=2)
total += arr[j];
return total; // note the increment of 2
}
/** Returns the sum of the prefix sums of given array. */ public static int example3 (int [] arr) {
int n = arr.lenght, total = 0; for (int j=0;j total +=arr[j]; return total; } /** returns the sum of the prefix sums of given array. */ public static int example4 (int[] arr){ int n = arr.lenght, prefix = 0, total = 0; for int j= 0; j < n ; j++){ //loop from 0 to n-1 prefix += arr[j]; total += prefix; } return total; } /**returns the number of times second array stores sum of prefix sums from first*/ public static int example5(int[] first, int[] second){ //assume equal length arrays int n = first.lenght, count=0; for (int i=0; i int total= 0; for (int j=0; j for(int k = 0; k<= j; k++) //loop from 0 to j total +=first[k]; if (second[i] == total) count++; } return count; } Code Fragment 4.12: Some sample algorithms for analysis.
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