## Interview Question

Software Engineer / Developers**Country:**Russia

**Interview Type:**In-Person

The solution is suppose to be O(1) space?? am I missing something ... can you give another example like fib of 8

c# implementation.

O(logn) time.

O(1) space.

```
namespace FibonacciInLogNTime {
static class Program {
// time complexity O(log n)
static private long Fib( long n ) {
n--;
long[,] res = { { 1, 0 }, { 0, 1 } }; // Identity matrix (with ones on the main diagonal and zeros elsewhere)
long[,] a = { { 1, 1 }, { 1, 0 } }; // Identity matrix for Fibonacci seq { { F(n+1), F(n) }, { F(n), F(n-1) } }
while ( n > 0 ) {
if ( ( n & 1 ) == 1 ) {
res = res.Multiply( a );
}
n /= 2;
a = a.Multiply( a );
}
return res[ 0, 0 ];
}
private static long[,] Multiply ( this long[,] arrA, long[,] arrB ) {
return new [,] { {
( arrA[ 0, 0 ] * arrB[ 0, 0 ] + arrA[ 0, 1 ] * arrB[ 1, 0 ] ), // [0,0]
( arrA[ 0, 0 ] * arrB[ 0, 1 ] + arrA[ 0, 1 ] * arrB[ 1, 1 ] ) // [0,1]
},{
( arrA[ 1, 0 ] * arrB[ 0, 0 ] + arrA[ 1, 1 ] * arrB[ 1, 0 ] ), // [1,0]
( arrA[ 1, 0 ] * arrB[ 0, 1 ] + arrA[ 1, 1 ] * arrB[ 1, 1 ] ) // [1,1]
}
};
}
static void Main( string[] args ) {
var res = Fib( 5 );
}
}
}
```

Use dynamic programming.

Store the previous fib numbers in an array

and access then at the next fib call

```
#include<stdio.h>
#include<stdlib.h>
#include <string.h>
int check=1,temp =0;
int * array;
int fabonacii(int n)
{
if(check)
{
check=0;
array = (int *) malloc(sizeof(int) * (n+1));
memset(array,0,sizeof(int) * (n+1));
}
if (n==1)
{
array[n]=0;
return 0;
}
else if (n==2)
{
array[n]=1;
return 1;
}
else
{
if(array[n] !=0)
return (array[n]);
else
{
temp=fabonacii(n-2) + fabonacii(n-1);
array[n]=temp;
return temp;
}
}
}
int main(void)
{
printf("%d",fabonacii(100));
free(array);
return 0;
}
```

```
#include<stdio.h>
#include<stdlib.h>
#include <string.h>
int check=1,temp =0;
int * array;
int fabonacii(int n)
{
if(check)
{
check=0;
array = (int *) malloc(sizeof(int) * (n+1));
memset(array,0,sizeof(int) * (n+1));
}
if (n==1)
{
array[n]=0;
return 0;
}
else if (n==2)
{
array[n]=1;
return 1;
}
else
{
if(array[n] !=0)
return (array[n]);
else
{
temp=fabonacii(n-2) + fabonacii(n-1);
array[n]=temp;
return temp;
}
}
}
int main(void)
{
printf("%d",fabonacii(100));
free(array);
return 0;
}
```

Here is a hint:

- ninhnnsoc January 03, 2016Represent Fibonacci numbers in matrix form:

(F[n+1], F[n]

F[n], F[n-1])

= A^n, where

A =

((1,1),

(1,0))

Perform fast exponential (repeated squaring) to compute A^n.