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:
C++
// CPP program to find count of rotations// with odd value.#include <bits/stdc++.h>using namespace std;// function to calculate total odd decimal// equivalentint oddEquivalent(string s, int n){ int count = 0; for (int i = 0; i < n; i++) { if (s[i] == '1') count++; } return count;}// Driver codeint 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 codepublic 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 equivalentdef oddEquivalent(s, n): count=0 for i in range(0,n): if (s[i] == '1'): count = count + 1 return count #Driver codeif __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
<?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?> |
Javascript
<script>// Javascript program to find count of rotations// with odd value.// function to calculate total odd decimal// equivalentfunction oddEquivalent(s, n){ var count = 0; for (var i = 0; i < n; i++) { if (s[i] == '1') count++; } return count;}// Driver codevar s = "1011011";var n = s.length;document.write( oddEquivalent(s, n));</script> |
Output
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.
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