Question

Describe the worst case running time of the following pseudocode functions in Big-Oh notation in terms of the variable n. Show your work

b) void func(int n) { for (int i = 0; i < n; i = i + 10) { for (int j = 0; j < i; ++i) { System.out.println("i = " + i); System.out.println("j = " + j);

Answer 1

Solution: According the give code, the iterator variable i is initialised to 0 and runs until i<n. Here the increment in the i is by 10 at each iteration. Therefore, outer for loop run for n/10 times. Now the second loop inside the first loop i.e. variable j is initialized to 0 and runs until j<i, here increment in j is by 1 at each iteration. Now Iteration i value j runs 10 n/10 Total number of times j runs = 10 + 20+30+ ...........n = (n/10)*2(10+n) On)