Given an array A of n integers. The task is to count the number of ways to split given array elements into two disjoint groups, such that XOR of elements of each group is equal.
Examples:
Input : A[] = { 1, 2, 3 }
Output : 3
{(1), (2, 3)}, {(2), (1, 3)}, {(3), (1, 2)} are three ways with equal XOR value of two groups.Input : A[] = { 5, 2, 3, 2 }
Output : 0
Let’s denote XOR between all elements in the first group as G1 and XOR between all elements in the second group as G2. Now, the following relation is always correct: G1 ⊕ G2 = A1 ⊕ A2 ⊕ …. ⊕ An.
So for G1 = G2, xor between all elements of array A is equal to 0. So, in that case, answer will be (2n – 2)/2 = (2n-1 – 1). In second case, when XOR between all elements isn’t 0, we can not split array. Answer will be 0.
Implementation:
C++
// CPP Program to count number of ways to split // array into two groups such that each group // has equal XOR value #include<bits/stdc++.h>using namespace std;// Return the count number of ways to split // array into two groups such that each group // has equal XOR value.int countgroup(int a[], int n){ int xs = 0; for (int i = 0; i < n; i++) xs = xs ^ a[i]; // We can split only if XOR is 0. Since // XOR of all is 0, we can consider all // subsets as one group. if (xs == 0) return (1 << (n-1)) - 1; return 0;}// Driver Programint main(){ int a[] = { 1, 2, 3 }; int n = sizeof(a)/sizeof(a[0]); cout << countgroup(a, n) << endl; return 0;} |
Java
// Java Program to count number of ways // to split array into two groups such // that each group has equal XOR valueimport java.io.*;import java.util.*;class GFG {// Return the count number of ways to split// array into two groups such that each group// has equal XOR value.static int countgroup(int a[], int n) { int xs = 0; for (int i = 0; i < n; i++) xs = xs ^ a[i]; // We can split only if XOR is 0. Since // XOR of all is 0, we can consider all // subsets as one group. if (xs == 0) return (1 << (n - 1)) - 1; return 0;}// Driver programpublic static void main(String args[]) { int a[] = {1, 2, 3}; int n = a.length; System.out.println(countgroup(a, n));}}// This code is contributed by Nikita Tiwari. |
Python3
# Python3 code to count number of ways # to split array into two groups such# that each group has equal XOR value # Return the count of number of ways# to split array into two groups such# that each group has equal XOR value.def countgroup(a, n): xs = 0 for i in range(n): xs = xs ^ a[i] # We can split only if XOR is 0. # Since XOR of all is 0, we can # consider all subsets as one group. if xs == 0: return (1 << (n-1)) - 1 return 0 # Driver Programa = [1, 2, 3]n = len(a)print(countgroup(a, n)) # This code is contributed by "Sharad_Bhardwaj". |
C#
// C# Program to count number of ways// to split array into two groups such// that each group has equal XOR valueusing System;class GFG { // Return the count number of ways to split // array into two groups such that each group // has equal XOR value. static int countgroup(int[] a, int n) { int xs = 0; for (int i = 0; i < n; i++) xs = xs ^ a[i]; // We can split only if XOR is 0. Since // XOR of all is 0, we can consider all // subsets as one group. if (xs == 0) return (1 << (n - 1)) - 1; return 0; } // Driver program public static void Main() { int[] a = { 1, 2, 3 }; int n = a.Length; Console.WriteLine(countgroup(a, n)); }}// This code is contributed by vt_m. |
PHP
<?php// PHP Program to count number// of ways to split array into // two groups such that each // group has equal XOR value // Return the count number of// ways to split array into // two groups such that each // grouphas equal XOR value.function countgroup($a, $n){ $xs = 0; for ($i = 0; $i < $n; $i++) $xs = $xs ^ $a[$i]; // We can split only if XOR is 0. Since // XOR of all is 0, we can consider all // subsets as one group. if ($xs == 0) return (1 << ($n - 1)) - 1; return 0;}// Driver Code$a = array(1, 2, 3);$n = count($a);echo countgroup($a, $n);// This code is contributed by anuj_67.?> |
Javascript
<script> // JavaScript Program to count number of ways to split // array into two groups such that each group // has equal XOR value // Return the count number of ways to split // array into two groups such that each group // has equal XOR value. function countgroup(a, n) { var xs = 0; for (var i = 0; i < n; i++) xs = xs ^ a[i]; // We can split only if XOR is 0. Since // XOR of all is 0, we can consider all // subsets as one group. if (xs == 0) return (1 << (n - 1)) - 1; } // Driver Program var a = [1, 2, 3]; var n = a.length; document.write(countgroup(a, n) + "<br>"); </script> |
Output
3
Time Complexity: O(n)
Auxiliary Space: O(1)
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