Question: Find Nth fibonacci number in O(logN) time complexity.

Answer: We all know the Fibonacci recurrence as F(n+1) = F(n) + F(n-1) but we can represent this in the form a matrix as shown below:

Look at the matrix A = [ [ 1 1 ] [ 1 0 ] ] . Multiplying A with [ F(n) F(n-1) ] gives us [ F(n+1) F(n) ] , so we say that

A* [ F(n) F(n-1) ] = [ F(n+1) F(n) ]

start with [ F(1) F(0) ] , multiplying it with A gives us [ F(2) F(1) ]; again multiplying [ F(2) F(1) ] with A gives us [ F(3) F(2) ] and so on...

A* [ F(1) F(0) ] = [ F(2) F(1) ]
A* [ F(2) F(1) ] = [ F(3) F(2) ] = A^2 * [ F(1) F(0) ]
A* [ F(3) F(2) ] = [ F(4) F(3) ] = A^3 * [ F(1) F(0) ]
..
..
..
..
A* [ F(n) F(n-1) ] = [ F(n+1) F(n) ] = A^n * [ F(1) F(0) ]

So all that is left is finding the nth power of the matrix A. Well, this can be computed in O(log n) time, by recursive doubling. The idea is, to find A^n , we can do R = A^(n/2) * A^(n/2) and if n is odd, we need do multiply with an A at the end. The following pseudo code shows the same.

Matrix findNthPower( Matrix M , power n )
{
if( n == 1 ) return M;
Matrix R = findNthPower ( M , n/2 );
R = RxR; // matrix multiplication
if( n%2 == 1 ) R = RxM; // matrix multiplication
return R;
}
In this manner, by using the magic of DP, we can get the Nth fibonacci number in O(logN).

PS: This question was asked me in Amazon Interview and I could not replied it that time. I got this answer here and sharing the same.

Subscribe - To get an automatic feed of all future posts subscribe here, or to receive them via email go here and enter your email address in the box. You can also like us on facebook and follow me on Twitter @akashag1001.