How we can tell about this without value of c

#1

void pat(){
//executes log (input size) times
for (int i = 2; i <=n; i = pow(i, c)) {
System.out.print("")
}
//Here fun is sqrt or cuberoot or any other constant root
//executes log n times
for (int i = n; i > 1; i = fun(i)) {
System.out.print("")
}
}

#2

hello @guptanikhil898
time complexity of the loop is O(log(log(n)))
for ( int i = 2; i <=n; i = pow (i, k))

{

// some O(1) expressions or statements

}

image

The last term must be less than or equal to n, and we have image , which completely agrees with the value of our last term. So there are in total logk(log(n)) many iterations, and each iteration takes a constant amount of time to run, therefore the total time complexity is O(log(log(n))).

#3

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