Rotations of a Binary String with Odd Value

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 the last bit is 2^0.Hence, since we are doing the circular rotation. 

Implementation:

// CPP program to find count of rotations
// with odd value.
#include <bits/stdc++.h>
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 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 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# 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 
// PHP program to find count 
// of rotations with odd value. 

// function to calculate total 
// odd decimal equivalent 
function oddEquivalent($s, $n) 
{ 
    $count = 0; 
    for ($i = 0; $i < $n; $i++) 
    { 
        if ($s[$i] == '1') 
            $count++; 
    } 
    return $count; 
} 

// Driver code 
$s = "1011011"; 
$n = strlen($s); 
echo(oddEquivalent($s, $n)); 

// This code is contributed 
// by Smitha
?>

<script>

// Javascript program to find count of rotations
// with odd value.

// function to calculate total odd decimal
// equivalent
function oddEquivalent(s, n)
{
    var count = 0; 
    for (var i = 0; i < n; i++) {
        if (s[i] == '1')
            count++;
    }
    return count;
}

// Driver code
var s = "1011011";
var n = s.length;
document.write( oddEquivalent(s, n));

</script>    
    

5

Time Complexity: O(n), where n is the length of the given string.
Auxiliary Space: O(1), no extra space is required, so it is a constant.