XOR of all subarray XORs | Set 1
Given an array of integers, we need to get the total XOR of all subarray XORs where subarray XOR can be obtained by XORing all elements of it.
Examples :
Input : arr[] = [3, 5, 2, 4, 6] Output : 7 Total XOR of all subarray XORs is, (3) ^ (5) ^ (2) ^ (4) ^ (6) (3^5) ^ (5^2) ^ (2^4) ^ (4^6) (3^5^2) ^ (5^2^4) ^ (2^4^6) (3^5^2^4) ^ (5^2^4^6) ^ (3^5^2^4^6) = 7 Input : arr[] = {1, 2, 3} Output : 2 Input : arr[] = {1, 2, 3, 4} Output : 0
A simple solution is to generate all subarrays and compute XOR of all of them. Below is the implementation of the above idea :
Implementation:
C++
// C++ program to get total xor of all subarray xors #include <bits/stdc++.h> using namespace std; // Returns XOR of all subarray xors int getTotalXorOfSubarrayXors( int arr[], int N) { // initialize result by 0 as (a xor 0 = a) int res = 0; // select the starting element for ( int i=0; i<N; i++) // select the eNding element for ( int j=i; j<N; j++) // Do XOR of elements in current subarray for ( int k=i; k<=j; k++) res = res ^ arr[k]; return res; } // Driver code to test above methods int main() { int arr[] = {3, 5, 2, 4, 6}; int N = sizeof (arr) / sizeof (arr[0]); cout << getTotalXorOfSubarrayXors(arr, N); return 0; } |
Java
// java program to get total XOR // of all subarray xors public class GFG { // Returns XOR of all subarray xors static int getTotalXorOfSubarrayXors( int arr[], int N) { // initialize result by // 0 as (a xor 0 = a) int res = 0 ; // select the starting element for ( int i = 0 ; i < N; i++) // select the eNding element for ( int j = i; j < N; j++) // Do XOR of elements // in current subarray for ( int k = i; k <= j; k++) res = res ^ arr[k]; return res; } // Driver code public static void main(String args[]) { int arr[] = { 3 , 5 , 2 , 4 , 6 }; int N = arr.length; System.out.println( getTotalXorOfSubarrayXors(arr, N)); } } // This code is contributed by Sam007. |
Python 3
# python program to get total xor # of all subarray xors # Returns XOR of all subarray xors def getTotalXorOfSubarrayXors(arr, N): # initialize result by 0 as # (a xor 0 = a) res = 0 # select the starting element for i in range ( 0 , N): # select the eNding element for j in range (i, N): # Do XOR of elements in # current subarray for k in range (i, j + 1 ): res = res ^ arr[k] return res # Driver code to test above methods arr = [ 3 , 5 , 2 , 4 , 6 ] N = len (arr) print (getTotalXorOfSubarrayXors(arr, N)) # This code is contributed by Sam007. |
C#
// C# program to get total XOR // of all subarray xors using System; class GFG { // Returns XOR of all subarray xors static int getTotalXorOfSubarrayXors( int []arr, int N) { // initialize result by // 0 as (a xor 0 = a) int res = 0; // select the starting element for ( int i = 0; i < N; i++) // select the eNding element for ( int j = i; j < N; j++) // Do XOR of elements // in current subarray for ( int k = i; k <= j; k++) res = res ^ arr[k]; return res; } // Driver Code static void Main() { int []arr = {3, 5, 2, 4, 6}; int N = arr.Length; Console.Write(getTotalXorOfSubarrayXors(arr, N)); } } // This code is contributed by Sam007 |
PHP
<?php // PHP program to get total // xor of all subarray xors // Returns XOR of all subarray xors function getTotalXorOfSubarrayXors( $arr , $N ) { // initialize result by // 0 as (a xor 0 = a) $res = 0; // select the starting element for ( $i = 0; $i < $N ; $i ++) // select the eNding element for ( $j = $i ; $j < $N ; $j ++) // Do XOR of elements in // current subarray for ( $k = $i ; $k <= $j ; $k ++) $res = $res ^ $arr [ $k ]; return $res ; } // Driver code $arr = array (3, 5, 2, 4, 6); $N = sizeof( $arr ); echo getTotalXorOfSubarrayXors( $arr , $N ); // This code is contributed by nitin mittal. ?> |
Javascript
<script> // JavaScript program for the above approach // Returns XOR of all subarray xors function getTotalXorOfSubarrayXors( arr, N) { // initialize result by // 0 as (a xor 0 = a) let res = 0; // select the starting element for (let i = 0; i < N; i++) // select the eNding element for (let j = i; j < N; j++) // Do XOR of elements // in current subarray for (let k = i; k <= j; k++) res = res ^ arr[k]; return res; } // Driver Code // Both a[] and b[] must be of same size. let arr = [3, 5, 2, 4, 6]; let N = arr.length; document.write(getTotalXorOfSubarrayXors(arr, N)); // This code is contributed by code_hunt. </script> |
7
Time Complexity: O(N3)
Auxiliary Space: O(1)
An efficient solution is based on the idea to enumerate all subarrays, we can count the frequency of each element that occurred totally in all subarrays, if the frequency of an element is odd then it will be included in the final result otherwise not.
As in above example, 3 occurred 5 times, 5 occurred 8 times, 2 occurred 9 times, 4 occurred 8 times, 6 occurred 5 times So our final result will be xor of all elements which occurred odd number of times i.e. 3^2^6 = 7 From above occurrence pattern we can observe that number at i-th index will have (i + 1) * (N - i) frequency.
So we can iterate over all elements once and calculate their frequencies and if it is odd then we can include that in our final result by XORing it with the result.
Total time complexity of the solution will be O(N)
Implementation:
C++
// C++ program to get total // xor of all subarray xors #include <bits/stdc++.h> using namespace std; // Returns XOR of all subarray xors int getTotalXorOfSubarrayXors( int arr[], int N) { // initialize result by 0 // as (a XOR 0 = a) int res = 0; // loop over all elements once for ( int i = 0; i < N; i++) { // get the frequency of // current element int freq = (i + 1) * (N - i); // Uncomment below line to print // the frequency of arr[i] // cout << arr[i] << " " << freq << endl; // if frequency is odd, then // include it in the result if (freq % 2 == 1) res = res ^ arr[i]; } // return the result return res; } // Driver Code int main() { int arr[] = {3, 5, 2, 4, 6}; int N = sizeof (arr) / sizeof (arr[0]); cout << getTotalXorOfSubarrayXors(arr, N); return 0; } |
Java
// java program to get total xor // of all subarray xors import java.io.*; public class GFG { // Returns XOR of all subarray // xors static int getTotalXorOfSubarrayXors( int arr[], int N) { // initialize result by 0 // as (a XOR 0 = a) int res = 0 ; // loop over all elements once for ( int i = 0 ; i < N; i++) { // get the frequency of // current element int freq = (i + 1 ) * (N - i); // Uncomment below line to print // the frequency of arr[i] // if frequency is odd, then // include it in the result if (freq % 2 == 1 ) res = res ^ arr[i]; } // return the result return res; } public static void main(String[] args) { int arr[] = { 3 , 5 , 2 , 4 , 6 }; int N = arr.length; System.out.println( getTotalXorOfSubarrayXors(arr, N)); } } // This code is contributed by Sam007. |
Python 3
# Python3 program to get total # xor of all subarray xors # Returns XOR of all # subarray xors def getTotalXorOfSubarrayXors(arr, N): # initialize result by 0 # as (a XOR 0 = a) res = 0 # loop over all elements once for i in range ( 0 , N): # get the frequency of # current element freq = (i + 1 ) * (N - i) # Uncomment below line to print # the frequency of arr[i] # if frequency is odd, then # include it in the result if (freq % 2 = = 1 ): res = res ^ arr[i] # return the result return res # Driver Code arr = [ 3 , 5 , 2 , 4 , 6 ] N = len (arr) print (getTotalXorOfSubarrayXors(arr, N)) # This code is contributed # by Smitha |
C#
// C# program to get total xor // of all subarray xors using System; class GFG { // Returns XOR of all subarray xors static int getTotalXorOfSubarrayXors( int []arr, int N) { // initialize result by 0 // as (a XOR 0 = a) int res = 0; // loop over all elements once for ( int i = 0; i < N; i++) { // get the frequency of // current element int freq = (i + 1) * (N - i); // Uncomment below line to print // the frequency of arr[i] // if frequency is odd, then // include it in the result if (freq % 2 == 1) res = res ^ arr[i]; } // return the result return res; } // Driver Code public static void Main() { int []arr = {3, 5, 2, 4, 6}; int N = arr.Length; Console.Write(getTotalXorOfSubarrayXors(arr, N)); } } // This code is contributed by Sam007 |
PHP
<?php // PHP program to get total // xor of all subarray xors // Returns XOR of all subarray xors function getTotalXorOfSubarrayXors( $arr , $N ) { // initialize result by 0 // as (a XOR 0 = a) $res = 0; // loop over all elements once for ( $i = 0; $i < $N ; $i ++) { // get the frequency of // current element $freq = ( $i + 1) * ( $N - $i ); // if frequency is odd, then // include it in the result if ( $freq % 2 == 1) $res = $res ^ $arr [ $i ]; } // return the result return $res ; } // Driver Code $arr = array (3, 5, 2, 4, 6); $N = count ( $arr ); echo getTotalXorOfSubarrayXors( $arr , $N ); // This code is contributed by anuj_67. ?> |
Javascript
<script> // Javascript program to get total // xor of all subarray xors // Returns XOR of all subarray xors function getTotalXorOfSubarrayXors(arr, N) { // initialize result by 0 // as (a XOR 0 = a) let res = 0; // loop over all elements once for (let i = 0; i < N; i++) { // get the frequency of // current element let freq = (i + 1) * (N - i); // Uncomment below line to print // the frequency of arr[i] // cout << arr[i] << " " << freq << endl; // if frequency is odd, then // include it in the result if (freq % 2 == 1) res = res ^ arr[i]; } // return the result return res; } // Driver Code let arr = [3, 5, 2, 4, 6]; let N = arr.length; document.write(getTotalXorOfSubarrayXors(arr, N)); </script> |
7
Time Complexity : O(N)
Auxiliary Space: O(1)
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