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# 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 = b = 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 ``using` `namespace` `std;` `int` `countStrings(``int` `n)``{``    ``int` `a[n], b[n];``    ``a = b = 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 = b = 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`

## PHP

 ``

## Javascript

 ``

Output

`5`

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 ``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

Output

`5`

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 ``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`

## 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`

Output

`144`

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

For n=10 , the output is 144.

We can calculate Fibonacci in log(n) using above result .

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