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# 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```
Recommended Practice

Naive Approach

We will follow the following steps-

• we made a variable “ans” which will tell how many strings have 2 consecutive ones
• We will find all strings with 0 and 1 of a given length using the pick and non-pick concept of recursion
• After that, if any string has maximum consecutive ones greater than or equal to 2 then increment “ans” by 1
• Then finally print the value stored in “ans” variable.

## C++

 `// C++ program to count all distinct` `// binary strings with two consecutive 1's` `#include ` `using` `namespace` `std;`   `// Gives count of n length binary` `// strings with consecutive 1's` `void` `countStrings(``int` `n,``int` `ind,string str,``int` `&ans)` `{` `    ``if``(ind==n){` `        ``int` `count=0;` `        ``int` `temp=0;` `        ``for``(``int` `i=0;i=2){ans++;}` `        ``return``;` `    ``}` `    `  `    ``countStrings(n,ind+1,str+``"0"``,ans);` `    ``countStrings(n,ind+1,str+``"1"``,ans);` `}`   `// Driver code` `int` `main()` `{` `    ``int` `ans=0;` `    ``countStrings(5,0,``""``,ans) ;` `    ``cout <

## Java

 `// Java program to count all distinct` `// binary strings with two consecutive 1's` `import` `java.util.*;`   `public` `class` `GFG {`   `    ``// Gives count of n length binary` `    ``// strings with consecutive 1's` `    ``public` `static` `void` `countStrings(``int` `n, ``int` `ind,` `                                    ``String str, ``int``[] ans)` `    ``{` `        ``if` `(ind == n) {` `            ``int` `count = ``0``;` `            ``int` `temp = ``0``;` `            ``for` `(``int` `i = ``0``; i < n; i++) {` `                ``if` `(str.charAt(i) == ``'1'``) {` `                    ``temp++;` `                ``}` `                ``else` `{` `                    ``temp = ``0``;` `                ``}` `                ``count = Math.max(count, temp);` `            ``}` `            ``if` `(count >= ``2``) {` `                ``ans[``0``]++;` `            ``}` `            ``return``;` `        ``}`   `        ``countStrings(n, ind + ``1``, str + ``"0"``, ans);` `        ``countStrings(n, ind + ``1``, str + ``"1"``, ans);` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int``[] ans = { ``0` `};` `        ``countStrings(``5``, ``0``, ``""``, ans);` `        ``System.out.println(ans[``0``]);` `    ``}` `}` `// This code is contributed by Susobhan Akhuli`

## Python3

 `# Python program to count all distinct` `# binary strings with two consecutive 1's`   `# Gives count of n length binary` `# strings with consecutive 1's` `def` `countStrings(n, ind, s, ans):` `    ``if` `ind ``=``=` `n:` `        ``count ``=` `0` `        ``temp ``=` `0` `        ``for` `i ``in` `range``(n):` `            ``if` `s[i] ``=``=` `'1'``:` `                ``temp ``+``=` `1` `            ``else``:` `                ``temp ``=` `0` `            ``count ``=` `max``(count, temp)` `        ``if` `count >``=` `2``:` `            ``ans[``0``] ``+``=` `1` `        ``return` `    `  `    ``countStrings(n, ind ``+` `1``, s ``+` `"0"``, ans)` `    ``countStrings(n, ind ``+` `1``, s ``+` `"1"``, ans)`   `# Driver code` `if` `__name__ ``=``=` `'__main__'``:` `    ``ans ``=` `[``0``]` `    ``countStrings(``3``, ``0``, "", ans)` `    ``print``(ans[``0``])`   `# This code is contributed by Susobhan Akhuli`

## C#

 `// C# program to count all distinct` `// binary strings with two consecutive 1's` `using` `System;`   `public` `class` `GFG {` `    ``// Gives count of n length binary` `    ``// strings with consecutive 1's` `    ``public` `static` `void` `CountStrings(``int` `n, ``int` `ind,` `                                    ``string` `str, ``ref` `int` `ans)` `    ``{` `        ``if` `(ind == n) {` `            ``int` `count = 0;` `            ``int` `temp = 0;` `            ``for` `(``int` `i = 0; i < n; i++) {` `                ``if` `(str[i] == ``'1'``) {` `                    ``temp++;` `                ``}` `                ``else` `{` `                    ``temp = 0;` `                ``}` `                ``count = Math.Max(count, temp);` `            ``}` `            ``if` `(count >= 2) {` `                ``ans++;` `            ``}` `            ``return``;` `        ``}`   `        ``CountStrings(n, ind + 1, str + ``"0"``, ``ref` `ans);` `        ``CountStrings(n, ind + 1, str + ``"1"``, ``ref` `ans);` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `Main()` `    ``{` `        ``int` `ans = 0;` `        ``CountStrings(5, 0, ``""``, ``ref` `ans);` `        ``Console.WriteLine(ans);` `    ``}` `}`   `// This code is contributed by Susobhan Akhuli`

## Javascript

 `// Javascript code to count all distinct ` `// binary strings with two consecutive 1's`   `// Gives count of n length binary strings with consecutive 1's` `function` `countStrings(n, ind, str, ans){` `  ``if` `(ind == n){` `    ``let count = 0;` `    ``let temp = 0;` `    ``for` `(let i = 0; i < n; i++){` `      ``if` `(str.charAt(i) == ``'1'``){` `temp++;` `      ``}` `      ``else` `{` `temp = 0;` `      ``}` `      ``count = Math.max(count, temp);` `    ``}` `    ``if` `(count >= 2) {` `      ``ans++;` `    ``}` `    ``return``;` `  ``}`   `  ``countStrings(n, ind + 1, str + ``"0"``, ans);` `  ``countStrings(n, ind + 1, str + ``"1"``, ans);` `}`   `// Driver code` `let ans = ;` `countStrings(5, 0, ``""``, ans);` `console.log(ans);`   `// This code is contributed by Susobhan Akhuli.`

Output

`19`

Time Complexity: O((2^n) * n), “2^n” for generating all strings, and “n” for traversing each string to count the maximum number of consecutive ones.
Auxiliary Space: O(n),Recursion Stack Space

Optimized Approach

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 the 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 of length n is 2^n.
3. 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 ` `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 = b = 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;` `}`

## 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.`

## 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.`

## 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 = b = 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.`

## PHP

 ``

## Javascript

 ``

Output

`19`

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

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  = 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 = 25 - fib(7)
................```

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