int n = 1000;
int cnt = 0;
for (int i = 0; i < n; i++)
cnt++;
Is the above code O(n) or O(1)? Could anyone verify this?
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int n = 1000;
int cnt = 0;
for (int i = 0; i < n; i++)
cnt++;
Is the above code O(n) or O(1)? Could anyone verify this?
Name |
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It's $$$O(n)$$$ and $$$O(1)$$$. $$$f(x)$$$ is $$$O(g(x))$$$ means there exist $$$c, n_0$$$ such that $$$\forall n \geq n_0$$$ $$$f(n) \leq cg(n)$$$
If you dont use
n
as a input and you assignn = 1000
that means then-loops = constant
so the time complexity of the code above isO(1)
. But if just then-loops
itself. It isO(n)
. So I think it is something likeO(n = 1000) = O(1)
complexityCorrect me if I am wrong
There was a blog post on this topic two years ago.