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.

- Compute the number of binary strings
**without**consecutive 1’s using the approach discussed here. Let this count be**c**. - Count of all possible binary strings with consecutive 1’s is 2^n where n is input length.
- Total binary strings with consecutive 1 is 2^n – c.

Below is the implementation of the above steps.

## C++

`// C++ program to count all distinct` `// binary strings with two consecutive 1's` `#include <iostream>` `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 << n) - a[n - 1] - b[n - 1];` `}` `// Driver code` `int` `main()` `{` ` ` `cout << countStrings(5) << endl;` ` ` `return` `0;` `}` |

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## 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 code` ` ` `public` `static` `void` `main(String args[])` ` ` `{` ` ` `System.out.println(countStrings(` `5` `));` ` ` `}` `}` `// This code is contributed by Nikita tiwari.` |

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## 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 code` `print` `(countStrings(` `5` `))` `# This code is contributed` `# by Nikita tiwari.` |

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## 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 code` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `Console.WriteLine(countStrings(5));` ` ` `}` `}` `// This code is contributed by Anant Agarwal.` |

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## PHP

`<?php` `// PHP program to count all` `// distinct binary strings` `// with two consecutive 1's` `// Returns count of n length binary ` `// strings with consecutive 1's` `function` `countStrings(` `$n` `)` `{` ` ` ` ` `// Count binary strings without consecutive 1's.` ` ` `// See the approach discussed on be` ` ` `// ( http://goo.gl/p8A3sW )` ` ` `$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` `[` `$i` `- 1];` ` ` `}` ` ` `// Subtract a[n-1]+b[n-1] from 2^n` ` ` `return` `(1 << ` `$n` `) - ` `$a` `[` `$n` `- 1] - ` ` ` `$b` `[` `$n` `- 1];` `}` ` ` `// Driver Code` ` ` `echo` `countStrings(5), ` `"\n"` `;` `// This Code is contributed by Ajit` `?>` |

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**Output**

19

**Optimization:**

The time complexity of the above solution is O(n). We can optimize the 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 = 2^{1}- fib(3) n = 2, count = 1 = 2^{2}- fib(4) n = 3, count = 3 = 2^{3}- fib(5) n = 4, count = 8 = 2^{4}- fib(6) n = 5, count = 19 = 2^{5}- fib(7) ................

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

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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