Time complexity of this for-loop

Asked Dec 28 '21 at 16:17

Active Dec 28 '21 at 20:57

Viewed 70 times

I know, the inner loop here is O(log n). But, I get confused by the multiplication in the outer loop part.

for (int i = 1; i <= 5*N; i++) {
  for (int j = 1; j <= 1000; j = 2*j) {
     doSomething(i, j);
   }
}

The correct answer to the question is, O(N) with regard to: how many times does the function doSomething() run? But, I want to know, how this is calculated.

T(n) = O(log n), according to: https://www.quora.com/Can-anyone-tell-the-time-complexity-of-for-i-1-to-n-i-i*2-and-how

edited Dec 28 '21 at 20:57

D.W.

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asked Dec 28 '21 at 16:17

John Smith

2

because $\log 1000 = O(1)$ and not $O(\log N)$. – Inuyasha Yagami Dec 28 '21 at 16:20
j=2*j don't you see it? – John Smith Dec 28 '21 at 16:20
2

$\log 1000$ is a constant. Even if it is j=j+1 it is $O(1)$. It is a constant (i.e. fixed) number of iterations, regardless of the value of $N$. – zyl1024 Dec 28 '21 at 16:24
It's NOT addition. It's doubling with multiplication. j=2 * j – John Smith Dec 28 '21 at 16:25
I understand that its a constant. What about the outer loop, which actually goes: i<=5*N, that part is my concern, what time complexity is it? I am confused by the growth rate when multiplying. – John Smith Dec 28 '21 at 16:27
So, even though, it's multiplying 5*N, it too remains constant, hmm. – John Smith Dec 28 '21 at 16:30
2

It is true that $O(5n)=O(n)$. Can you see why the outer loop works in $O(5n)$? – nir shahar Dec 28 '21 at 17:39
@nirshahar, I guess, because, it's constantly growing by 5x ? – John Smith Dec 28 '21 at 17:44
https://cs.stackexchange.com/q/23593/755, https://cs.stackexchange.com/q/192/755 – D.W. Dec 28 '21 at 20:59

Time complexity of this for-loop

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