Count number of binary strings without consecutive 1’s

Given a positive integer N, count all possible distinct binary strings of length N such that there are no consecutive 1’s.

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

This problem can be solved using Dynamic Programming. Let a[i] be the number of binary strings of length i which do not contain any two consecutive 1’s 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].
Following is the implementation of above solution. In the following implementation, indexes start from 0. So a[i] represents the number of binary strings for input length i+1. Similarly, b[i] represents binary strings for input length i+1.

Implementation:

C++

// C++ program to count all distinct binary strings
// without two consecutive 1's
#include <iostream>
using namespace std;
 
int countStrings(int n)
{
    int a[n], b[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 functions
int main()
{
    cout << countStrings(3) << endl;
    return 0;
}

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 */

Python3

# Python program to count
# all distinct binary strings
# without two consecutive 1's
 
def countStrings(n):
 
    a=[0 for i in range(n)]
    b=[0 for i in range(n)]
    a[0] = b[0] = 1
    for i in range(1,n):
        a[i] = a[i-1] + b[i-1]
        b[i] = a[i-1]
     
    return a[n-1] + b[n-1]
 
# Driver program to test
# above functions
 
print(countStrings(3))
 
# This code is contributed
# by Anant Agarwal.

C#

// C# program to count all distinct binary
// strings without two consecutive 1's
using System;
 
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 Code
    public static void Main ()
    {
        Console.Write(countStrings(3));
    }
}
 
// This code is contributed by nitin mittal

Javascript

<script>
 
// JavaScript program to count all
// distinct binary strings
// without two consecutive 1's
function countStrings(n)
{
    let a = [];
    let b = [];
    a[0] = b[0] = 1;
     
    for(let 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 code
document.write(countStrings(3));
 
// This code is contributed by rohan07
 
</script>

PHP

<?php
// PHP program to count all distinct
// binary stringswithout two
// consecutive 1's
 
function countStrings($n)
{
    $a[$n] = 0;
    $b[$n] = 0;
    $a[0] = $b[0] = 1;
    for ($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 Code
    echo countStrings(3) ;
 
// This code is contributed by nitin mittal
?>

Output

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

Source:

If we take a closer look at the pattern, we can observe that the count is actually (n+2)’th Fibonacci number for n >= 1. The Fibonacci Numbers are 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 141, ….

n = 1, count = 2  = fib(3)
n = 2, count = 3  = fib(4)
n = 3, count = 5  = fib(5)
n = 4, count = 8  = fib(6)
n = 5, count = 13 = fib(7)
................

Therefore we can count the strings in O(Log n) time also using the method 5 here.

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:-

C++

// C++ program to count all distinct binary strings
// without two consecutive 1's
#include <bits/stdc++.h>
using namespace std;
 
int countStrings(int n)
{
    int a = 1, 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 functions
int main()
{
    cout << countStrings(3) << endl;
    return 0;
}
 
// This code is contributed by Aditya Kumar (adityakumar129)

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)

Python3

class Subset_sum :
    @staticmethod
    def  countStrings( n) :
        a = 1
        b = 1
        i = 1
        while (i < n) :
            # Here we have used the temp variable because
            # we want to assign the older value of a to b
            temp = a + b
            b = a
            a = temp
            i += 1
        return (a + b)%1000000007
       
    # Driver program to test above function
    @staticmethod
    def main( args) :
        print(Subset_sum.countStrings(3))
     
 
if __name__=="__main__":
    Subset_sum.main([])
     
    # This code is contributed by aadityaburujwale.

C#

// Include namespace system
using System;
 
public class Subset_sum
{
    public static int countStrings(int n)
    {
        var a = 1;
        var 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
            var temp = a + b;
            b = a;
            a = temp;
        }
        return (a + b)%1000000007;
    }
   
    // Driver program to test above function
    public static void Main(String[] args)
    {
        Console.WriteLine(Subset_sum.countStrings(3));
    }
}
 
// This code is contributed by aadityaburujwale.

Javascript

function countStrings(n){
    var a = 1;
    var b = 1;
    for(let i=1;i<n;i++){
        // Here we have used the temp variable because
        // we want to assign the older value of a to b
        var temp = a + b;
        b = a;
        a = temp;
    }
    return (a + b)%1000000007;
}
 
console.log(countStrings(3));
 
// This code is contributed by lokesh

Output

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

Another Useful Approach:-

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)

To find the values of consecutive 1’s at the intersection

As moving from n to n+1 ,the difference between them is the only value which have 0’s at the end as 0 values only double the values . And 0’s values at the end will only give two results as 0 and 1 because if last element is 1 then we can’t get 1 after it as then they become two consecutive.

So,

f(n)-f(n-1) = 0's value at the last position of f(n-1) = 1's value at the last position of f(n)

Finally ,

If n is even:

f(n)=f(n/2)*f(n/2) - (f(n/2)-f(n/2-1))(f(n/2)-f(n/2-1))

If n is odd:

f(n)=f(n/2)*f(n/2+1) - (f(n/2)-f(n/2-1))(f(n/2+1)-f(n/2))

And we will store pre-calculated value using map.

C++

// Fibonacci in log(n)
// Easy code
// Without using matrix exponentiation
#include <bits/stdc++.h>
using namespace std;
 
class Solution {
  public:
  //storing the pre-calculated value
  //what we called DP
    map<long long int,long long int>mp;
    int countStrings(long long int N) {
        // Code here
        if(N==0)return 1;
        if(N==1)return 2;
        if(N==2)return 3;
        long long int mod=1e9+7;
        long long int a=mp[N/2-1]=((mp[N/2-1])?mp[N/2-1]:countStrings(N/2-1))%mod,b=mp[N/2]=((mp[N/2])?mp[N/2]:countStrings(N/2))%mod,c=mp[N/2+1]=((mp[N/2+1])?mp[N/2+1]:countStrings(N/2+1))%mod;
        if(N%2)
        return mp[N]=((b*c%mod)-((c-b)*(b-a)%mod)+mod)%mod;
        return mp[N]=((b*b%mod)-((b-a)*(b-a)%mod)+mod)%mod;
    }
};
 
int main() {
    int t;
    cin >> t;
    while (t--) {
        long long int N;
        cin >> N;
        Solution obj;
        cout << obj.countStrings(N) << endl;
    }
}
// Article contributed by Chetan Chaudhary

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));
    }
}

Python3

class Solution:
    def __init__(self):
        self.mp = {}
 
    def countStrings(self, N):
        if N == 0:
            return 1
        if N == 1:
            return 2
        if N == 2:
            return 3
 
        mod = 10**9 + 7
 
        a = self.mp.get(N//2-1, self.countStrings(N//2-1)) % mod
        b = self.mp.get(N//2, self.countStrings(N//2)) % mod
        c = self.mp.get(N//2+1, self.countStrings(N//2+1)) % mod
 
        if N % 2:
            return (b*c - (c-b)*(b-a) + mod) % mod
        return (b*b - (b-a)*(b-a) + mod) % mod
 
 
# t = int(input())
t=1
while t > 0:
    # N = int(input())
    N=10
    obj = Solution()
    print(obj.countStrings(N))
    t -= 1
# Code contributed by Chetan Chaudhary

C#

using System;
using System.Collections.Generic;
 
class Solution {
    // Dictionary to store memoized values
    static Dictionary<long, long> mp = new Dictionary<long, long>();
 
    // Recursive function to count strings
    public static 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 for N is already memoized
        long a = mp.ContainsKey(N / 2 - 1) ? mp[N / 2 - 1] : countStrings(N / 2 - 1);
        long b = mp.ContainsKey(N / 2) ? mp[N / 2] : countStrings(N / 2);
        long c = mp.ContainsKey(N / 2 + 1) ? mp[N / 2 + 1] : countStrings(N / 2 + 1);
         
        // Calculations based on whether N is even or odd
        if (N % 2 == 1) {
            // Memoize and return result for odd N
            return mp[N] = ((b * c % mod) - ((c - b) * (b - a) % mod) + mod) % mod;
        } else {
            // Memoize and return result for even N
            return mp[N] = ((b * b % mod) - ((b - a) * (b - a) % mod) + mod) % mod;
        }
    }
}
 
class Program {
    static void Main(string[] args) {
        long N = 10;
        Console.WriteLine(Solution.countStrings(N));
    }
}

Javascript

class Solution {
  constructor() {
    // Storing the pre-calculated value
    // using a Map for DP
    this.mp = new Map();
  }
 
  countStrings(N) {
    // Base cases
    if (N === 0n) return 1n;
    if (N === 1n) return 2n;
    if (N === 2n) return 3n;
     
    const mod = 1000000007n; // Use 1000000007n instead of 1e9+7n
     
    // Check if the value is already present in the map
    if (this.mp.has(N)) {
      return this.mp.get(N);
    }
     
    // Recursive calls and memoization
    const a = this.mp.has(N / 2n - 1n) ? this.mp.get(N / 2n - 1n) : this.countStrings(N / 2n - 1n);
    const b = this.mp.has(N / 2n) ? this.mp.get(N / 2n) : this.countStrings(N / 2n);
    const c = this.mp.has(N / 2n + 1n) ? this.mp.get(N / 2n + 1n) : this.countStrings(N / 2n + 1n);
     
    // Calculations
    let result;
    if (N % 2n === 1n)
      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
    this.mp.set(N, result);
     
    return result;
  }
}
 
const N = 10n; // Use BigInt literal for N
const obj = new Solution();
console.log(obj.countStrings(N).toString());
 
 
// This code is contributed by rambabuguphka