determine the time complexity of the algorithems - time-complexity

Question

I have just started to learn time complexity, but I don't really get the idea, could you help with those questions and explain the way of thinking:
int Fun1(int n)
{
for (i = 0; i < n; i += 1) {
for (j = 0; j < i; j += 1) {
for (k = j; k < i; i += 1) {
// do something
}
}
}
}
void Fun2(int n){
i=o
while(i<n){
for (j = 0; j < i; j += 1) {
k=n
while(k>j){
k=k/2
}
k=j
while(k>1){
k=k/2
}
}
}
int Fun3(int n){
for (i = 0; i < n; i += 1) {
print("*")
}
if(n<=1){
print("*")
return
}
if (n%2 != 0){
Fun3(n-1)
}
else{
Fun3(n/2)
}
}
for function 1, I think its Theta(n^3) because it runs at most
n*n*n times but I am not sure how to prove this.
for the second I think its Theta (n^2log(n))
I am not sure
Could you help, please?

Answer 1

First a quick note, in Fun2(n) there should be a i++ before closing the while loop, anyway, time complexity is important in order to understand the efficiency of your algorithms. In this case you have these 3 functions:
Fun1(n)
In this function you have three nested for loops, each for loops iterates n times over a given input, we know that the complexity of this iteration is O(n). Since there are three nested for loops, the second for loop will iterate n times over each iteration of the outer for loop. The same will do the most inner loop. The resulting complexity, as you correctly said, is O(n) * O(n) * O(n) = O(n^3)
Fun2(n)
This function has a while loop that iterates n times over a given input. So the outer loop complexity is O(n). Same as before we have an inner for loop that iterates n times on each cycle of the outer loop. So far we have O(n) * O(n) which is O(n^2) as complexity. Inside the for loop we have a while loop, that differs from the other loops, since does not iterate on each element in a specific range, but it divides the range by 2 at each iteration. For example from 0 to 31 we have 0-31 -> 0-15 -> 0-7 -> 0-3 -> 0-1
As you know the number of iteration is the result of the logarithmic function, log(n), so we end up with O(n^2) * O(log(n)) = O(n^2(log(n)) as time complexity
Fun3(n)
In this function we have a for loop with no more inner loops, but then we have a recursive call. The complexity of the for loop as we know is O(n), but how many times will this function be called?
If we take a small number (like 6) as example we have a first loop with 6 iteration, then we call again the function with n = 6-1 since 6 mod 2 = 0
Now we have a call to Fun3(5), we do 5 iteration and the recursively we call Fun3(2) since 5 mod 2 != 0
What are we having here? We having a recursive call that in the worst case will call itself n times
The complexity result is O(n!)
Note that when we calculate time complexity we ignore the coefficients since are not relevant, usually the function we consider, especially in CS, are:
O(1), O(n), O(log(n)), O(n^a) with a > 1, O(n!)
and we combine and simplify them in order to know who has the best (lowest) time complexity to have an idea of which algorithm could be used