This program calculates the Nth Fibonacci number using a technique called matrix exponentiation.

import java.util.*;
public class NthFibonacci {
static final int MOD = (int) 1e9 + 7;

//Method to perform matrix multiplication

static long[][] multiply(long[][] A, long[][] B) {
int rowA = A.length;
int colA = A[0].length;
int colB = B[0].length;
long[][] C = new long[rowA][colB];
for (int i = 0; i < rowA; i++) {
for (int j = 0; j < colB; j++) {
for (int k = 0; k < colA; k++) {
C[i][j] = (C[i][j] + (A[i][k] * B[k][j]) % MOD) % MOD;
}
}
}
return C;
}

//Method to perform matrix exponentiation

static long[][] power(long[][] M, int n) {
if (n == 1) return M;
long[][] half = power(M, n / 2);
long[][] result = multiply(half, half);
if (n % 2 == 1) result = multiply(result, M);
return result;
}

The power method calculates the matrix raised to the power of N. It uses recursion and optimizes the process by halving the exponent at each step. The method called multiply performs matrix multiplication. This is a key operation in the matrix exponentiation technique.

//Method to find the Nth Fibonacci number using matrix exponentiation

public static int fibonacciNumber(int n) {
if (n <= 1) return n;
long[][] M = {{1, 1}, {1, 0}};
long[][] result = power(M, n - 1);
return (int) result[0][0];
}

public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
System.out.println(fibonacciNumber(n));
scanner.close();
}
}

This program efficiently calculates the nth Fibonacci number using matrix exponentiation, which has a time complexity of O(log n), making it much faster than the traditional recursive approach with O(2^n) time complexity.