Formas de eliminar un elemento de una string binaria para que XOR se convierta en cero

Dada una string binaria, la tarea es borrar exactamente un entero en la array para que el XOR de los números restantes sea cero. La tarea es contar el número de formas de eliminar un elemento para que XOR de esa string se convierta en CERO.

Ejemplos: 

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 give 0's

Una solución simple es eliminar un elemento uno por uno, luego calcular XOR de la string restante. Y cuente las ocurrencias en las que eliminar un elemento hace que XOR sea 0.

Una solución eficiente se basa en el siguiente hecho. Si la cuenta de 1 es impar, entonces debemos quitar un 1 para hacer la cuenta 0 y podemos quitar cualquiera de los 1. Si el conteo de 1 es par, entonces XOR es 0, podemos eliminar cualquiera de los 0 y XOR seguirá siendo 0.

Implementación:

// 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 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

# 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# 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 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
?>

<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>

5

Complejidad temporal : O(n) 
Espacio auxiliar : O(1)
 

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