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 elementint 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 Codeint 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 elementdef 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 Codeif __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 elementfunction 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> |
5
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
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