Question

g) Func7(n) while (i 5n3) do 10n3 while 3) do s i j return 8 (h) Func8(n) for 1 to n do for i to n2 do for k j to n do s i -t- j return s (i) Func9(n) for i- 1 to n/2 do for j i to i do return s G) Func10 (m) while (i n) do for j 1 to n do i 1 s i+ j return s
Find asymptotic running time , find expression for the running time as a function of n, then find valid upper and lower bound which differ by only a constant factor

Answer 1

(g)

3ⁱ = n^2 for outer loop
=> 2log ₃n

For outer loop

1 = n^3 / 5ⁱ for outer loop
=> 3log ₅n

So overall time complexity is O(log ₃n3log ₅n)

(h) Outer for loop n times ,
the next for loop n^2 times itself and n times for outer loop so n^3
the next for loop n^3 times itself and n^3 times for outer loop so n^6

Time complexity is O(n⁶)


(i) Outer loop runs for n/2 time, Inner loop runs for n^2 - n + 1 times.
So overall time complexity is O(n^3)

(j) n times i.e O(n)



Thanks, let me know if there is any concern.