Given an array of integers arr[] and a number m, count the number of subarrays having XOR of their elements as m.
Examples:
Input : arr[] = {4, 2, 2, 6, 4}, m = 6 Output : 4 Explanation : The subarrays having XOR of their elements as 6 are {4, 2}, {4, 2, 2, 6, 4}, {2, 2, 6}, and {6} Input : arr[] = {5, 6, 7, 8, 9}, m = 5 Output : 2 Explanation : The subarrays having XOR of their elements as 5 are {5} and {5, 6, 7, 8, 9}
A Simple Solution is to use two loops to go through all possible subarrays of arr[] and count the number of subarrays having XOR of their elements as m.
C++
// A simple C++ Program to count all subarrays having // XOR of elements as given value m #include <bits/stdc++.h> using namespace std; // Simple function that returns count of subarrays // of arr with XOR value equals to m long long subarrayXor( int arr[], int n, int m) { long long ans = 0; // Initialize ans // Pick starting point i of subarrays for ( int i = 0; i < n; i++) { int xorSum = 0; // Store XOR of current subarray // Pick ending point j of subarray for each i for ( int j = i; j < n; j++) { // calculate xorSum xorSum = xorSum ^ arr[j]; // If xorSum is equal to given value, // increase ans by 1. if (xorSum == m) ans++; } } return ans; } // Driver program to test above function int main() { int arr[] = { 4, 2, 2, 6, 4 }; int n = sizeof (arr) / sizeof (arr[0]); int m = 6; cout << "Number of subarrays having given XOR is " << subarrayXor(arr, n, m); return 0; } |
Java
// A simple Java Program to count all // subarrays having XOR of elements // as given value m public class GFG { // Simple function that returns // count of subarrays of arr with // XOR value equals to m static long subarrayXor( int arr[], int n, int m) { // Initialize ans long ans = 0 ; // Pick starting point i of // subarrays for ( int i = 0 ; i < n; i++) { // Store XOR of current // subarray int xorSum = 0 ; // Pick ending point j of // subarray for each i for ( int j = i; j < n; j++) { // calculate xorSum xorSum = xorSum ^ arr[j]; // If xorSum is equal to // given value, increase // ans by 1. if (xorSum == m) ans++; } } return ans; } // Driver code public static void main(String args[]) { int [] arr = { 4 , 2 , 2 , 6 , 4 }; int n = arr.length; int m = 6 ; System.out.println( "Number of subarrays" + " having given XOR is " + subarrayXor(arr, n, m)); } } // This code is contributed by Sam007. |
Python3
# A simple Python3 Program to count all subarrays having # XOR of elements as given value m # Simple function that returns count of subarrays # of arr with XOR value equals to m def subarrayXor(arr, n, m): ans = 0 # Initialize ans # Pick starting po i of subarrays for i in range ( 0 ,n): xorSum = 0 # Store XOR of current subarray # Pick ending po j of subarray for each i for j in range (i,n): # calculate xorSum xorSum = xorSum ^ arr[j] # If xorSum is equal to given value, # increase ans by 1. if (xorSum = = m): ans + = 1 return ans # Driver program to test above function def main(): arr = [ 4 , 2 , 2 , 6 , 4 ] n = len (arr) m = 6 print ( "Number of subarrays having given XOR is " , subarrayXor(arr, n, m)) if __name__ = = '__main__' : main() #this code contributed by 29AjayKumar |
C#
// A simple C# Program to count all // subarrays having XOR of elements // as given value m using System; class GFG { // Simple function that returns // count of subarrays of arr // with XOR value equals to m static long subarrayXor( int [] arr, int n, int m) { // Initialize ans long ans = 0; // Pick starting point i of // subarrays for ( int i = 0; i < n; i++) { // Store XOR of current // subarray int xorSum = 0; // Pick ending point j of // subarray for each i for ( int j = i; j < n; j++) { // calculate xorSum xorSum = xorSum ^ arr[j]; // If xorSum is equal to // given value, increase // ans by 1. if (xorSum == m) ans++; } } return ans; } // Driver Program public static void Main() { int [] arr = { 4, 2, 2, 6, 4 }; int n = arr.Length; int m = 6; Console.Write( "Number of subarrays" + " having given XOR is " + subarrayXor(arr, n, m)); } } // This code is contributed by Sam007. |
PHP
<?php // A simple PHP Program to // count all subarrays having // XOR of elements as given value m // Simple function that returns // count of subarrays of arr // with XOR value equals to m function subarrayXor( $arr , $n , $m ) { // Initialize ans $ans = 0; // Pick starting point // i of subarrays for ( $i = 0; $i < $n ; $i ++) { // Store XOR of // current subarray $xorSum = 0; // Pick ending point j of // subarray for each i for ( $j = $i ; $j < $n ; $j ++) { // calculate xorSum $xorSum = $xorSum ^ $arr [ $j ]; // If xorSum is equal // to given value, // increase ans by 1. if ( $xorSum == $m ) $ans ++; } } return $ans ; } // Driver Code $arr = array (4, 2, 2, 6, 4); $n = count ( $arr ); $m = 6; echo "Number of subarrays having given XOR is " , subarrayXor( $arr , $n , $m ); // This code is contributed by anuj_67. ?> |
Output:
Number of subarrays having given XOR is 4
The Time Complexity of the above solution is O(n2).
Efficient Approach:
An Efficient Solution solves the above problem in O(n) time. Let us call the XOR of all elements in the range [i+1, j] as A, in the range [0, i] as B, and in the range [0, j] as C. If we do XOR of B with C, the overlapping elements in [0, i] from B and C zero out, and we get XOR of all elements in the range [i+1, j], i.e. A. Since A = B XOR C, we have B = A XOR C. Now, if we know the value of C and we take the value of A as m, we get the count of A as the count of all B satisfying this relation. Essentially, we get the count of all subarrays having XOR-sum m for each C. As we take the sum of this count overall C, we get our answer.
1) Initialize ans as 0. 2) Compute xorArr, the prefix xor-sum array. 3) Create a map mp in which we store count of all prefixes with XOR as a particular value. 4) Traverse xorArr and for each element in xorArr (A) If m^xorArr[i] XOR exists in map, then there is another previous prefix with same XOR, i.e., there is a subarray ending at i with XOR equal to m. We add count of all such subarrays to result. (B) If xorArr[i] is equal to m, increment ans by 1. (C) Increment count of elements having XOR-sum xorArr[i] in map by 1. 5) Return ans.
C++
// C++ Program to count all subarrays having // XOR of elements as given value m with // O(n) time complexity. #include <bits/stdc++.h> using namespace std; // Returns count of subarrays of arr with XOR // value equals to m long long subarrayXor( int arr[], int n, int m) { long long ans = 0; // Initialize answer to be returned // Create a prefix xor-sum array such that // xorArr[i] has value equal to XOR // of all elements in arr[0 ..... i] int * xorArr = new int [n]; // Create map that stores number of prefix array // elements corresponding to a XOR value unordered_map< int , int > mp; // Initialize first element of prefix array xorArr[0] = arr[0]; // Computing the prefix array. for ( int i = 1; i < n; i++) xorArr[i] = xorArr[i - 1] ^ arr[i]; // Calculate the answer for ( int i = 0; i < n; i++) { // Find XOR of current prefix with m. int tmp = m ^ xorArr[i]; // If above XOR exists in map, then there // is another previous prefix with same // XOR, i.e., there is a subarray ending // at i with XOR equal to m. ans = ans + (( long long )mp[tmp]); // If this subarray has XOR equal to m itself. if (xorArr[i] == m) ans++; // Add the XOR of this subarray to the map mp[xorArr[i]]++; } // Return total count of subarrays having XOR of // elements as given value m return ans; } // Driver program to test above function int main() { int arr[] = { 4, 2, 2, 6, 4 }; int n = sizeof (arr) / sizeof (arr[0]); int m = 6; cout << "Number of subarrays having given XOR is " << subarrayXor(arr, n, m); return 0; } |
Java
// Java Program to count all subarrays having // XOR of elements as given value m with // O(n) time complexity. import java.util.*; class GFG { // Returns count of subarrays of arr with XOR // value equals to m static long subarrayXor( int arr[], int n, int m) { long ans = 0 ; // Initialize answer to be returned // Create a prefix xor-sum array such that // xorArr[i] has value equal to XOR // of all elements in arr[0 ..... i] int [] xorArr = new int [n]; // Create map that stores number of prefix array // elements corresponding to a XOR value HashMap<Integer, Integer> mp = new HashMap<Integer, Integer>(); // Initialize first element of prefix array xorArr[ 0 ] = arr[ 0 ]; // Computing the prefix array. for ( int i = 1 ; i < n; i++) xorArr[i] = xorArr[i - 1 ] ^ arr[i]; // Calculate the answer for ( int i = 0 ; i < n; i++) { // Find XOR of current prefix with m. int tmp = m ^ xorArr[i]; // If above XOR exists in map, then there // is another previous prefix with same // XOR, i.e., there is a subarray ending // at i with XOR equal to m. ans = ans + (mp.containsKey(tmp) == false ? 0 : (( long )mp.get(tmp))); // If this subarray has XOR equal to m itself. if (xorArr[i] == m) ans++; // Add the XOR of this subarray to the map if (mp.containsKey(xorArr[i])) mp.put(xorArr[i], mp.get(xorArr[i]) + 1 ); else mp.put(xorArr[i], 1 ); } // Return total count of subarrays having XOR of // elements as given value m return ans; } // Driver code public static void main(String[] args) { int arr[] = { 4 , 2 , 2 , 6 , 4 }; int n = arr.length; int m = 6 ; System.out.print( "Number of subarrays having given XOR is " + subarrayXor(arr, n, m)); } } // This code is contributed by PrinciRaj1992 |
Python3
# Python3 Program to count all subarrays # having XOR of elements as given value m # with O(n) time complexity. # Returns count of subarrays of arr # with XOR value equals to m def subarrayXor(arr, n, m): ans = 0 # Initialize answer to be returned # Create a prefix xor-sum array such that # xorArr[i] has value equal to XOR # of all elements in arr[0 ..... i] xorArr = [ 0 for _ in range (n)] # Create map that stores number of prefix array # elements corresponding to a XOR value mp = dict () # Initialize first element # of prefix array xorArr[ 0 ] = arr[ 0 ] # Computing the prefix array. for i in range ( 1 , n): xorArr[i] = xorArr[i - 1 ] ^ arr[i] # Calculate the answer for i in range (n): # Find XOR of current prefix with m. tmp = m ^ xorArr[i] # If above XOR exists in map, then there # is another previous prefix with same # XOR, i.e., there is a subarray ending # at i with XOR equal to m. if tmp in mp.keys(): ans = ans + (mp[tmp]) # If this subarray has XOR # equal to m itself. if (xorArr[i] = = m): ans + = 1 # Add the XOR of this subarray to the map mp[xorArr[i]] = mp.get(xorArr[i], 0 ) + 1 # Return total count of subarrays having # XOR of elements as given value m return ans # Driver Code arr = [ 4 , 2 , 2 , 6 , 4 ] n = len (arr) m = 6 print ( "Number of subarrays having given XOR is" , subarrayXor(arr, n, m)) # This code is contributed by mohit kumar |
C#
// C# Program to count all subarrays having // XOR of elements as given value m with // O(n) time complexity. using System; using System.Collections.Generic; class GFG { // Returns count of subarrays of arr with XOR // value equals to m static long subarrayXor( int [] arr, int n, int m) { long ans = 0; // Initialize answer to be returned // Create a prefix xor-sum array such that // xorArr[i] has value equal to XOR // of all elements in arr[0 ..... i] int [] xorArr = new int [n]; // Create map that stores number of prefix array // elements corresponding to a XOR value Dictionary< int , int > mp = new Dictionary< int , int >(); // Initialize first element of prefix array xorArr[0] = arr[0]; // Computing the prefix array. for ( int i = 1; i < n; i++) xorArr[i] = xorArr[i - 1] ^ arr[i]; // Calculate the answer for ( int i = 0; i < n; i++) { // Find XOR of current prefix with m. int tmp = m ^ xorArr[i]; // If above XOR exists in map, then there // is another previous prefix with same // XOR, i.e., there is a subarray ending // at i with XOR equal to m. ans = ans + (mp.ContainsKey(tmp) == false ? 0 : (( long )mp[tmp])); // If this subarray has XOR equal to m itself. if (xorArr[i] == m) ans++; // Add the XOR of this subarray to the map if (mp.ContainsKey(xorArr[i])) mp[xorArr[i]] = mp[xorArr[i]] + 1; else mp.Add(xorArr[i], 1); } // Return total count of subarrays having XOR of // elements as given value m return ans; } // Driver code public static void Main(String[] args) { int [] arr = { 4, 2, 2, 6, 4 }; int n = arr.Length; int m = 6; Console.Write( "Number of subarrays having given XOR is " + subarrayXor(arr, n, m)); } } // This code is contributed by Rajput-Ji |
Output:
Number of subarrays having given XOR is 4
Time Complexity: O(n)
Alternate Approach: Using Python Dictionary to store Prefix XOR
Python3
from collections import defaultdict def subarrayXor(arr, n, m): HashTable = defaultdict( bool ) HashTable[ 0 ] = 1 count = 0 curSum = 0 for i in arr: curSum^ = i if HashTable[curSum^m]: count + = HashTable[curSum^m] HashTable[curSum] + = 1 return (count) # Driver program to test above function def main(): arr = [ 5 , 6 , 7 , 8 , 9 ] n = len (arr) m = 5 print ( "Number of subarrays having given XOR is " , subarrayXor(arr, n, m)) if __name__ = = '__main__' : main() # This code is contributed by mrmechanical26052000 |
Output:
Number of subarrays having given XOR is 4
Time Complexity: O(n)
Space Complexity: O(n)
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