Complexity of a function with 1 loop - c++

Question

Can anyone tell me what's the complexity of the below function? And how to calculate the complexity?
I am suspecting that it's O(log(n)) or O(sqrt(N)).
My reasoning was based on taking examples of n=4, n=8, n=16 and I found that the loop will take log(n) but I don't think it'll be enough since sqrt also will give the same values so I need to work on bigger values of n, so I am not sure how to approach this.
I had this function in the exam today.
void f(int n){
int i=1;
int j=1;
while(j <= n){
i += 1;
j += i;
}
}

Answer 1

The sequence j goes through is 1 3 6 10 15 21, aka the triangular numbers, aka n*(n+1)/2.
Expanded, this is ( n^2 + n ) / 2. We can ignore the scaling ( / 2) and linear ( + n) factors, which leaves us with n^2.
j grows as a n^2 polynomial, so the loop will stop after the inverse of that growth:
The time complexity is O(sqrt(n))

Answer 2

For what it's worth, I wrote a small program that attempts to illustrate whether this is O(log(N)) or O(sqrt(N)) by actually counting how many iterations of your loop execute. This seemed a reasonable approximation given that the body of the loop is largely negligible (simply incrementing two integer variables).
#include <stdio.h>
#include <math.h>
int f(int n)
{
int i=1;
int j=1;
int count = 0;
while(j <= n){
i += 1;
j += i;
count++;
}
return count;
}
int main()
{
for (int ii = 0; ii < 10; ii++) {
int count = pow(10, ii);
int rc = f(count);
char *fmt = "N=%d^%-2d -> %d, log(N)=%.2f, sqrt(N)=%.2f\n";
printf(fmt, 10, ii, rc, log(count), sqrt(count));
}
return 0;
}
Running this code results in the following output:
N=10^0 -> 1, log(N)=0.00, sqrt(N)=1.00
N=10^1 -> 4, log(N)=2.30, sqrt(N)=3.16
N=10^2 -> 13, log(N)=4.61, sqrt(N)=10.00
N=10^3 -> 44, log(N)=6.91, sqrt(N)=31.62
N=10^4 -> 140, log(N)=9.21, sqrt(N)=100.00
N=10^5 -> 446, log(N)=11.51, sqrt(N)=316.23
N=10^6 -> 1413, log(N)=13.82, sqrt(N)=1000.00
N=10^7 -> 4471, log(N)=16.12, sqrt(N)=3162.28
N=10^8 -> 14141, log(N)=18.42, sqrt(N)=10000.00
N=10^9 -> 44720, log(N)=20.72, sqrt(N)=31622.78
So, for example, you can see that when N=10^9, the number of iterations is 44720, which is much greater than log(N) (20.72) but quite close to sqrt(N) (31622.78).

Answer 3

It is depended your condition. In other words, the time complexity is O(log n).
How many statements are executed, relative to input size n? Often,
but NOT always, we can get an idea from the number of times a loop
iterates. The loop body executes for i= 2^0 + 2^1 + 2^2 + .... + 2^n; and this
sequence has O(log n) values.
Check the "Introduction to Algorithms" book about more details.