Ways to remove one element from a binary string so that XOR becomes zero

0 0 4 minutes read

Given a binary string, task is to erase exactly one integer in the array so that the XOR of the remaining numbers is zero. The task is to count number of ways to remove one element so that XOR of that string become ZERO.

Examples :

Input : 10000
Output : 1
We only have 1 ways to 

Input : 10011
Output : 3
There are 3 ways to make XOR 0. We
can remove any of the three 1's.

Input : 100011100
Output : 5
There are 5 ways to make XOR 0. We
can remove any of the given 0's

A simple solution is to one by one remove an element, then compute XOR of remaining string. And count occurrences where removing an element makes XOR 0.

An efficient solution is based on following fact. If count of 1s is odd, then we must remove a 1 to make count 0 and we can remove any of the 1s. If count of 1s is even, then XOR is 0, We can remove any of the 0s and XOR will remain 0.

Implementation:

C++

// C++ program to count number of ways to
// remove an element so that XOR of remaining
// string becomes 0.
#include <bits/stdc++.h>
using namespace std;
 
// Return number of ways in which XOR become ZERO
// by remove 1 element
int xorZero(string str)
{
    int one_count = 0, zero_count = 0;
    int n = str.length();
 
    // Counting number of 0 and 1
    for (int i = 0; i < n; i++)
        if (str[i] == '1')
            one_count++;
        else
            zero_count++;
     
    // If count of ones is even
    // then return count of zero
    // else count of one
    if (one_count % 2 == 0)
        return zero_count;
    return one_count;
}
 
// Driver Code
int main()
{
    string str = "11111";
    cout << xorZero(str) << endl;
    return 0;
}

Java

// Java program to count number of ways to
// remove an element so that XOR of remaining
// string becomes 0.
import java.util.*;
  
class CountWays
{
    // Returns number of ways in which XOR become
    // ZERO by remove 1 element
    static int xorZero(String s)
    {
        int one_count = 0, zero_count = 0;
        char[] str=s.toCharArray();
        int n = str.length;
      
        // Counting number of 0 and 1
        for (int i = 0; i < n; i++)
            if (str[i] == '1')
                one_count++;
            else
                zero_count++;
          
        // If count of ones is even
        // then return count of zero
        // else count of one
        if (one_count % 2 == 0)
            return zero_count;
        return one_count;
    }
 
    // Driver Code to test above function
    public static void main(String[] args)
    {
        String s = "11111";
        System.out.println(xorZero(s));  
    }
}
 
// This code is contributed by Mr. Somesh Awasthi

Python3

# Python 3 program to count number of 
# ways to remove an element so that 
# XOR of remaining string becomes 0.
 
# Return number of ways in which XOR 
# become ZERO by remove 1 element
def xorZero(str):
    one_count = 0
    zero_count = 0
    n = len(str)
 
    # Counting number of 0 and 1
    for i in range(0, n, 1):
        if (str[i] == '1'):
            one_count += 1
        else:
            zero_count += 1
     
    # If count of ones is even
    # then return count of zero
    # else count of one
    if (one_count % 2 == 0):
        return zero_count
    return one_count
 
# Driver Code
if __name__ == '__main__':
    str = "11111"
    print(xorZero(str))
 
# This code is contributed by
# Surendra_Gangwar

C#

// C# program to count number
// of ways to remove an element 
// so that XOR of remaining 
// string becomes 0.
using System;
 
class GFG
{
    // Returns number of ways 
    // in which XOR become
    // ZERO by remove 1 element
    static int xorZero(string s)
    {
        int one_count = 0, 
            zero_count = 0;
         
        int n = s.Length;
     
        // Counting number of 0 and 1
        for (int i = 0; i < n; i++)
            if (s[i] == '1')
                one_count++;
            else
                zero_count++;
         
        // If count of ones is even
        // then return count of zero
        // else count of one
        if (one_count % 2 == 0)
            return zero_count;
        return one_count;
    }
 
    // Driver Code 
    public static void Main()
    {
        string s = "11111";
        Console.WriteLine(xorZero(s)); 
    }
}
 
// This code is contributed by anuj_67.

PHP

<?php
// PHP program to count number 
// of ways to remove an element 
// so that XOR of remaining
// string becomes 0.
 
// Return number of ways in
// which XOR become ZERO
// by remove 1 element
 
function xorZero($str)
{
    $one_count = 0; $zero_count = 0;
    $n = strlen($str);
 
    // Counting number of 0 and 1
    for ($i = 0; $i < $n; $i++)
        if ($str[$i] == '1')
            $one_count++;
        else
            $zero_count++;
     
    // If count of ones is even
    // then return count of zero
    // else count of one
    if ($one_count % 2 == 0)
        return $zero_count;
    return $one_count;
}
 
// Driver Code
$str = "11111";
echo xorZero($str),"\n";
 
// This code is contributed by aj_36
?>

Javascript

<script>
    // Javascript program to count number
    // of ways to remove an element 
    // so that XOR of remaining 
    // string becomes 0.
     
    // Returns number of ways 
    // in which XOR become
    // ZERO by remove 1 element
    function xorZero(s)
    {
        let one_count = 0, zero_count = 0;
           
        let n = s.length;
       
        // Counting number of 0 and 1
        for (let i = 0; i < n; i++)
            if (s[i] == '1')
                one_count++;
            else
                zero_count++;
           
        // If count of ones is even
        // then return count of zero
        // else count of one
        if (one_count % 2 == 0)
            return zero_count;
        return one_count;
    }
     
    let s = "11111";
      document.write(xorZero(s)); 
 
</script>

Output

Time complexity : O(n) //since one traversal of the String is required to complete all operations hence overall time required by the algorithm is linear
Auxiliary Space : O(1) // since no extra array is used so the space taken by the algorithm is constant

This article is contributed by Sahil Rajput. If you like zambiatek and would like to contribute, you can also write an article using write.zambiatek.com or mail your article to review-team@zambiatek.com. See your article appearing on the zambiatek main page and help other Geeks.

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