C/C++ Program to Count number of binary strings without consecutive 1’s

Last Updated : 10 Nov, 2023

Write a C/C++ 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.

C/C++ 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:

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;
}

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:

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)

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:

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