By using binet's theorem we can solve this in O(1) time complexity &amp; O(1) space complexity.
<blockquote>
#include 
#include 
int fib(int n){
 double sqrt5 = sqrt(5);
 int F_n = ( pow((1 + sqrt5), n) - pow((1 - sqrt5), n) ) / ( pow(2, n) * sqrt5 );
 return F_n;
}
</blockquote>

By using binet's theorem we can solve this in O(1) time complexity & O(1) space complexity.

> #include <math.h>
> #include <stdio.h>
> 
> int fib(int n){
> double sqrt5 = sqrt(5);
> int F_n = ( pow((1 + sqrt5), n) - pow((1 - sqrt5), n) ) / ( pow(2, n) * sqrt5 );
> return F_n;
> }

Fibonacci sequence by using binets formula