Related Articles
Rotations of a Binary String with Odd Value
• Last Updated : 11 May, 2021

Given a binary string. We are allowed to do circular rotation of the string without changing the relative order of the bits in the string.
For Example, all possible circular rotation of string “011001” are:

```101100
010110
001011
100101
110010```

We are required to tell total number of distinct odd decimal equivalent possible of binary string, by doing circular rotation.
Examples:

```Input : 011001
Output : 3
Explanation:
All odd possible binary representations are:
["011001", "001011", "100101"]

Input : 11011
Output : 4
Explanation:
All odd possible binary representations are:
["11011", "01111", "10111", "11101"]```

Concept:
It can be observed that a binary string can only be odd if it’s last bit is 1, because the value of last bit is 2^0.Hence, since we are doing circular rotation.

## C++

 `// CPP program to find count of rotations``// with odd value.``#include ``using` `namespace` `std;` `// function to calculate total odd decimal``// equivalent``int` `oddEquivalent(string s, ``int` `n)``{``    ``int` `count = 0;``    ``for` `(``int` `i = 0; i < n; i++) {``        ``if` `(s[i] == ``'1'``)``            ``count++;``    ``}``    ``return` `count;``}` `// Driver code``int` `main()``{``    ``string s = ``"1011011"``;``    ``int` `n = s.length();``    ``cout << oddEquivalent(s, n);``    ``return` `0;``}`

## Java

 `// Java program to find count of rotations``// with odd value.` `class` `solution``{``static` `int` `oddEquivalent(String s, ``int` `n)``{` `int` `count = ``0``;``// function to calculate total odd decimal``// equivalent``for` `(``int` `i = ``0``; i < n; i++)``    ``{``        ``if``(s.charAt(i) == ``'1'``)``            ``count++;``    ``}``    ``return` `count;``}` `// Driver code``public` `static` `void` `main(String ar[])``{` `String s = ``"1011011"``;``int` `n = s.length();``System.out.println(oddEquivalent(s, n));` `}``}``//This code is contributed``//By Surendra_Gangwar`

## Python3

 `# Python3 program to find count``# of rotations with odd value` `#function to calculate total odd equivalent``def` `oddEquivalent(s, n):``    ``count``=``0``    ``for` `i ``in` `range``(``0``,n):``        ``if` `(s[i] ``=``=` `'1'``):``            ``count ``=` `count ``+` `1``    ``return` `count``    ` `#Driver code``if` `__name__``=``=``'__main__'``:``    ``s ``=` `"1011011"``    ``n ``=` `len``(s)``    ``print``(oddEquivalent(s, n))` `# this code is contributed by Shashank_Sharma`

## C#

 `// C# program to find count of``// rotations with odd value.``using` `System;` `class` `GFG``{``static` `int` `oddEquivalent(String s, ``int` `n)``{``    ``int` `count = 0;``    ` `    ``// function to calculate total``    ``// odd decimal equivalent``    ``for` `(``int` `i = 0; i < n; i++)``        ``{``            ``if``(s[i] == ``'1'``)``                ``count++;``        ``}``    ``return` `count;``}` `// Driver code``public` `static` `void` `Main()``{``    ``String s = ``"1011011"``;``    ``int` `n = s.Length;``    ``Console.WriteLine(oddEquivalent(s, n));``}``}` `// This code is contributed``// by Subhadeep`

## PHP

 ``

## Javascript

 `   `
Output:
`5`

My Personal Notes arrow_drop_up