Space efficient iterative method to Fibonacci number

Given a number n, find n-th Fibonacci Number. Note that F0 = 0, F1 = 1, F2 = 2, …..

Examples :

Input : n = 5
Output : 5

Input :  n = 10
Output : 89



We have discussed below recursive solution in method 4 of Program for Fibonacci numbers.

F[2][2] = |1, 1|
          |1, 0|

M[2][2] = |1, 1|
          |1, 0|
F[n][n] = fib(n)  | fib(n-1)
          ------------------
          fib(n-1)| fib(n-2)

In this post an iterative method is discussed that avoids extra recursion call stack space. We have also used bitwise operators to further optimize. In the previous method, we divide the number with 2 so that at the end we get 1 and then we start the multiplication process
In this method we get the second MSB then start to multiply with FxF matrix then if bit is set then multiply again FxM matrix and so on. then we get the final result.

Approach :
1. First get the MSB of a number.
2. while (MSB > 0)
      multiply(F, F);
      if (n & MSB)
      multiply(F, M);
      and then shift MSB till MSB != 0

C++

// CPP code to find nth fibonacci
#include <bits/stdc++.h>
using namespace std;
  
// get second MSB
int getMSB(int n)
{
    // consectutively set all the bits
    n |= n >> 1;
    n |= n >> 2;
    n |= n >> 4;
    n |= n >> 8;
    n |= n >> 16;
  
    // returns the second MSB
    return ((n + 1) >> 2);
}
  
// Multiply function
void multiply(int F[2][2], int M[2][2])
{
    int x = F[0][0] * M[0][0] + F[0][1] * M[1][0];
    int y = F[0][0] * M[0][1] + F[0][1] * M[1][1];
    int z = F[1][0] * M[0][0] + F[1][1] * M[1][0];
    int w = F[1][0] * M[0][1] + F[1][1] * M[1][1];
  
    F[0][0] = x;
    F[0][1] = y;
    F[1][0] = z;
    F[1][1] = w;
}
  
// Function to calculate F[][]
// raise to the power n
void power(int F[2][2], int n)
{
    // Base case
    if (n == 0 || n == 1)
        return;
  
    // take 2D array to store number's
    int M[2][2] = { 1, 1, 1, 0 };
  
    // run loop till MSB > 0
    for (int m = getMSB(n); m; m = m >> 1) {
        multiply(F, F);
  
        if (n & m) {
            multiply(F, M);
        }
    }
}
  
// To return fibonacci numebr
int fib(int n)
{
    int F[2][2] = { { 1, 1 }, { 1, 0 } };
    if (n == 0)
        return 0;
    power(F, n - 1);
    return F[0][0];
}
  
// Driver Code
int main()
{
    // Given n
    int n = 6;
  
    cout << fib(n) << " ";
  
    return 0;
}

Java

// Java code to 
// find nth fibonacci
  
class GFG
{
      
// get second MSB
static int getMSB(int n)
{
    // consectutively set
    // all the bits
    n |= n >> 1;
    n |= n >> 2;
    n |= n >> 4;
    n |= n >> 8;
    n |= n >> 16;
  
    // returns the
    // second MSB
    return ((n + 1) >> 2);
}
  
// Multiply function
static void multiply(int F[][], 
                     int M[][])
{
    int x = F[0][0] * M[0][0] +
            F[0][1] * M[1][0];
    int y = F[0][0] * M[0][1] +
            F[0][1] * M[1][1];
    int z = F[1][0] * M[0][0] + 
            F[1][1] * M[1][0];
    int w = F[1][0] * M[0][1] + 
            F[1][1] * M[1][1];
  
    F[0][0] = x;
    F[0][1] = y;
    F[1][0] = z;
    F[1][1] = w;
}
  
// Function to calculate F[][]
// raise to the power n
static void power(int F[][], 
                  int n)
{
    // Base case
    if (n == 0 || n == 1)
        return;
  
    // take 2D array to
    // store number's
    int[][] M ={{1, 1}, 
                {1, 0}};
  
    // run loop till MSB > 0
    for (int m = getMSB(n);
             m > 0; m = m >> 1
    {
        multiply(F, F);
  
        if ((n & m) > 0
        {
            multiply(F, M);
        }
    }
}
  
// To return 
// fibonacci numebr
static int fib(int n)
{
    int[][] F = {{1, 1},    
                 {1, 0}};
    if (n == 0)
        return 0;
    power(F, n - 1);
    return F[0][0];
}
  
// Driver Code
public static void main(String[] args)
{
    // Given n
    int n = 6;
  
    System.out.println(fib(n));
}
}
  
// This code is contributed 
// by mits

Output:

8

Time Complexity :- O(logn) and space complexity :- O(1).



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