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

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 give 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.

## 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; ` `} ` |

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## 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 ` |

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## 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. ` |

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## 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 ` `?> ` |

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**Output :**

5

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