Java Program to Count number of binary strings without consecutive 1's

Last Updated : 10 Nov, 2023

Write a Java program for a given positive integer N, the task is to count all possible distinct binary strings of length N such that there are no consecutive 1s.

Examples:

Input: N = 2
Output: 3
// The 3 strings are 00, 01, 10

Input: N = 3
Output: 5
// The 5 strings are 000, 001, 010, 100, 101

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Java Program to Count a number of binary strings without consecutive 1's using Dynamic Programming:

Let a[i] be the number of binary strings of length i that do not contain any two consecutive 1s and which end in 0. Similarly, let b[i] be the number of such strings which end in 1. We can append either 0 or 1 to a string ending in 0, but we can only append 0 to a string ending in 1. This yields the recurrence relation:

a[i] = a[i - 1] + b[i - 1]
b[i] = a[i - 1]

The base cases of above recurrence are a[1] = b[1] = 1. The total number of strings of length i is just a[i] + b[i].

Below is the implementation of the above approach:

Java

import java.io.*;
class Subset_sum
{
    static int countStrings(int n)
    {
        int a[] = new int [n];
        int b[] = new int [n];
        a[0] = b[0] = 1;
        for (int i = 1; i < n; i++)
        {
            a[i] = a[i-1] + b[i-1];
            b[i] = a[i-1];
        }
        return (a[n-1] + b[n-1])%1000000007;
    }
    /* Driver program to test above function */
    public static void main (String args[])
    {
        System.out.println(countStrings(3));
    }
}/* This code is contributed by Rajat Mishra */

Output

Time Complexity: O(N)
Auxiliary Space: O(N)

Another method:

From the above method it is clear that we only want just previous value in the for loop which we can also do by replacing the array with the variable.

Below is the implementation of the above approach:

Java

import java.io.*;
class Subset_sum {
    static int countStrings(int n)
    {
        int a = 1;
        int b = 1;
        for (int i = 1; i < n; i++) {
            // Here we have used the temp variable because
            // we want to assign the older value of a to b
            int temp = a + b;
            b = a;
            a = temp;
        }
        return (a + b)%1000000007;
    }
    /* Driver program to test above function */
    public static void main(String args[])
    {
        System.out.println(countStrings(3));
    }
} 

// This code is contributed by Aditya Kumar (adityakumar129)

Output

Time Complexity: O(N)
Auxiliary Space: O(1)

Fibonacci in Log (n) without using Matrix Exponentiation:

As for calculating the n we can express it as the product of n/2 and n/2 ((n/2+1) for n is odd) as they are independent ,now as there are the cases when we combining these two half values to get full length there may be occurrence of consecutive 1’s at the joining. So we have to subtract those cases to get the final result.

f(n)=f(n/2)*f(n/2) - (those values which has consecutive 1's at the joining part)

Below is the implementation of the above approach:

Java

import java.util.*;

class Solution {
    // Storing the pre-calculated value
    // using a HashMap for DP
    Map<Long, Long> mp = new HashMap<>();

    long countStrings(long N) {
        // Base cases
        if (N == 0) return 1;
        if (N == 1) return 2;
        if (N == 2) return 3;
        
        long mod = (long)1e9 + 7;
        
        // Check if the value is already present in the map
        if (mp.containsKey(N)) {
            return mp.get(N);
        }
        
        // Recursive calls and memoization
        long a = mp.getOrDefault(N/2 - 1, countStrings(N/2 - 1)) % mod;
        long b = mp.getOrDefault(N/2, countStrings(N/2)) % mod;
        long c = mp.getOrDefault(N/2 + 1, countStrings(N/2 + 1)) % mod;
        
        // Calculations
        long result;
        if (N % 2 == 1)
            result = ((b * c % mod) - ((c - b) * (b - a) % mod) + mod) % mod;
        else
            result = ((b * b % mod) - ((b - a) * (b - a) % mod) + mod) % mod;
        
        // Store the result in the map for future use
        mp.put(N, result);
        
        return result;
    }
}

public class Main {
    public static void main(String[] args) {
        long N = 10;
        Solution obj = new Solution();
        System.out.println(obj.countStrings(N));
    }
}