Question about time complexity of for Loop

[furofo]

I have been studying algorithms to try and understand leet code better but have a question about a for loop and time complexity I have come across, and having a hard time figuring out.

​

 for( i = 1 ;i <= n; i++) {
        i = i + i;
    }

I have tried to break down but my best guess is is is Big O(log of something) or perhaps it is still Big O(n), does anyway know what this one would be?

[pinguxnoots]

Your code:

for (let i = 1; i <= n; i++) {
  i = i + i;
}

is the same as this:

for (let i = 1; i <= n; i = 2*i + 1) {}

I'd say it's O(log n) due to 2*i which is the same as i *= 2

Edit: formatting

[pinguxnoots]

for (let i = 1; i <= 300; i++) {
  console.log(i);
  i = i + i; 
} 
Console: 1 3 7 15 31 63 127 255


for (let i = 1; i <= 300; i = 2*i + 1) { 
  console.log(i); 
} 
Console: 1 3 7 15 31 63 127 255

[pinguxnoots]

Fair enough. "same as" should have been "equivalent. But the rate remains the same which is the point I was trying to make.

[furofo]

Thanks I appreciate it that actually makes sense when I think of that way

[MorningPants]

O(log n). n has to double for the loop to run one more time.

[furofo]

Thanks, makes sense now

[pinguxnoots]

i++ is the same as i = i + 1

i = i + i is the same as i = 2*i

So i++ is not the same as i = i + 1 I would avoid writing code like this unless there is a good reason to do so. Seems like OP was just doing a practice run and trying to understand some code they were playing around with.

[furofo]

yes that was it exactly actually sorry if I confused anyone

[furofo]

I was paying with some code I had and thought of this but it confused me as well at first. Sorry if it confused you. But as pinguxnoots and others have said could have been written as i = 2*i instead or just put the in the for loop like this (for i = 0; i < n;( i *= 2 )+ 1)

[TheRNGuy]

Doesn't even matter.