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; }

## 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

Output:

5

This article is contributed by **Sahil Rajput**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.