Count Binary Strings With No Consecutive

//{ Driver Code Starts
// Initial Template for Java
// Initial Template for Java

import java.util.*;
import java.lang.*;
import java.io.*;

class GFG {
    public static void main(String[] args) throws IOException {
        BufferedReader br =
            new BufferedReader(new InputStreamReader(System.in));
        int t =
            Integer.parseInt(br.readLine().trim()); // Inputting the testcases
        while (t-- > 0) {

            long N = Long.parseLong(br.readLine().trim());

            Solution ob = new Solution();
            System.out.println(ob.countStrings(N));
        }
    }
}

// } Driver Code Ends


// User function Template for Java

class Solution {
    
        // Code here
int modulus = 1000000007;


    // This function takes a long n as input and returns the number of binary strings of length n

public int countStrings(long N) {

    // We use matrix power to find the value of F(n+1) where F(i) is the ith Fibonacci number

    /* The number of binary strings of length n without consecutive 1's is just the

    (n+1)th Fibonacci number

    We use matrix power to compute F(n+1) since the matrix representation of Fibonacci numbers

    has a property that can be used to compute them efficiently*/

    return matrixPower(new int[][]{{1, 1}, {1, 0}}, N + 1)[0][0];

}


// This function takes two 2x2 matrices a and b and returns their product

int[][] multiplyMatrices(int[][] a, int[][] b) {

    int[][] result = new int[2][2];    // Initialize the result matrix to 0

    for (int i = 0; i < 2; i++) {

        for (int j = 0; j < 2; j++) {

            for (int k = 0; k < 2; k++) {

                result[i][j] = (int)((result[i][j] + (long)a[i][k] * b[k][j]) % modulus);   // Compute the (i, j)th entry of the product matrix

            }

        }

    }

    return result;   // Return the product matrix

}


// This function takes a 2x2 matrix a and a long n and returns the matrix power a^n

int[][] matrixPower(int[][] a, long n) {

    int[][] result = new int[][]{{1, 0}, {0, 1}};  // Initialize the result matrix to the identity matrix

    while (n > 0) {

        if (n % 2 == 1) {

            result = multiplyMatrices(result, a);   // If n is odd, multiply result by a

        }

        a = multiplyMatrices(a, a);   // Square a

        n /= 2;   // Divide n by 2

    }

    return result;   // Return the matrix power a^n

    
        
    }
}