# Count strings with consecutive 1’s

Given a number n, count number of n length strings with consecutive 1’s in them.

Examples:

```Input  : n = 2
Output : 1
There are 4 strings of length 2, the
strings are 00, 01, 10 and 11. Only the
string 11 has consecutive 1's.

Input  : n = 3
Output : 3
There are 8 strings of length 3, the
strings are 000, 001, 010, 011, 100,
101, 110 and 111.  The strings with
consecutive 1's are 011, 110 and 111.

Input : n = 5
Output : 19
```

The reverse problem of counting strings without consecutive 1’s can be solved using Dynamic Programming (See the solution here). We can use that solution and find the required count using below steps.

1) Compute number of binary strings without consecutive 1’s using the approach discussed here. Let this count be c.

2) Count of all possible binary strings with consecutive 1’s is 2^n where n is input length.

3) Total binary strings with consecutive 1 is 2^n – c.

Below is implementation of above steps.

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## C++

 `// C++ program to count all distinct  ` `// binary strings with two consecutive 1's ` `#include ` `using` `namespace` `std; ` ` `  `// Returns count of n length binary  ` `// strings with consecutive 1's ` `int` `countStrings(``int` `n) ` `{ ` `  ``// Count binary strings without consecutive 1's. ` `  ``// See the approach discussed on be ` `  ``// ( http://goo.gl/p8A3sW ) ` `    ``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]; ` `    ``} ` ` `  `    ``// Subtract a[n-1]+b[n-1] from 2^n ` `    ``return` `(1<

## Java

 `// Java program to count all distinct  ` `// binary strings with two consecutive 1's ` ` `  `class` `GFG { ` ` `  `    ``// Returns count of n length binary  ` `    ``// strings with consecutive 1's ` `    ``static` `int` `countStrings(``int` `n) ` `    ``{ ` `     ``// Count binary strings without consecutive 1's. ` `     ``// See the approach discussed on be ` `     ``// ( http://goo.gl/p8A3sW ) ` `        ``int` `a[] = ``new` `int``[n], 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``]; ` `        ``} ` ` `  `       ``// Subtract a[n-1]+b[n-1] ` ` ``from ``2``^n ` `        ``return` `(``1` `<< n) - a[n - ``1``] - b[n - ``1``]; ` `    ``} ` ` `  `// Driver program to test above functions ` `   ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``System.out.println(countStrings(``5``)); ` `    ``} ` `} ` ` `  `//This code is contributed by Nikita tiwari. `

## Python 3

 `# Python 3 program to count all  ` `# distinct binary strings with  ` `# two consecutive 1's ` ` `  ` `  `# Returns count of n length ` `# binary strings with  ` `# consecutive 1's ` `def` `countStrings(n) : ` `     `  `    ``# Count binary strings without  ` `    ``# consecutive 1's. ` `    ``# See the approach discussed on be ` `    ``# ( http://goo.gl/p8A3sW ) ` `    ``a ``=` `[``0``] ``*` `n ` `    ``b ``=` `[``0``] ``*` `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``] ` `     `  `     `  `    ``# Subtract a[n-1]+b[n-1] from 2^n ` `    ``return` `(``1` `<< n) ``-` `a[n ``-` `1``] ``-` `b[n ``-` `1``] ` `     `  `# Driver program  ` `print``(countStrings(``5``)) ` ` `  ` `  `# This code is contributed ` `# by Nikita tiwari. `

## C#

 `// program to count all distinct ` `// binary strings with two ` `// consecutive 1's ` `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``// Returns count of n length ` `    ``// binary strings with ` `    ``// consecutive 1's ` `    ``static` `int` `countStrings(``int` `n) ` `    ``{ ` `        ``// Count binary strings without ` `        ``// consecutive 1's. ` `        ``// See the approach discussed on ` `        ``// ( http://goo.gl/p8A3sW ) ` `        ``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]; ` `        ``} ` ` `  `        ``// Subtract a[n-1]+b[n-1] ` `        ``// from 2^n ` `        ``return` `(1 << n) - a[n - 1] - b[n - 1]; ` `    ``} ` ` `  `    ``// Driver program ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``Console.WriteLine(countStrings(5)); ` `    ``} ` `} ` ` `  `// This code is contributed by Anant Agarwal. `

## PHP

 ` `

Output :

`19`

Optimization:
Time complexity of above solution is O(n). We can optimize above solution to work in O(Logn).

If we take a closer look at the pattern of counting strings without consecutive 1’s, 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 = 0  = 21 - fib(3)
n = 2, count = 1  = 22 - fib(4)
n = 3, count = 3  = 23 - fib(5)
n = 4, count = 8  = 24 - fib(6)
n = 5, count = 19 = 24 - fib(7)
................
```

We can find n’th Fibonacci Number in O(Log n) time (See method 4 here).

## tags:

Bit Magic binary-string Bit Magic