Given an array containing N positive integers, the task is to find the sum of XOR of all sub-arrays of the array.
Examples:
Input : arr[] = {1, 3, 7, 9, 8, 7}
Output : 128
Input : arr[] = {3, 8, 13}
Output : 46
Explanation for second test-case:
XOR of {3} = 3
XOR of {3, 8} = 11
XOR of {3, 8, 13} = 6
XOR of {8} = 8
XOR of {8, 13} = 5
XOR of {13} = 13
Sum = 3 + 11 + 6 + 8 + 5 + 13 = 46
Simple solution: A simple solution will be to generate all the sub-arrays and then iterate through them all to find the required XOR values and then sum them up.
C++
// C++ program to find the sum of XOR of// all subarray of the array#include <iostream>#include <vector>using namespace std;// Function to calculate the sum of XOR// of all subarraysint findXorSum(int arr[], int n){ // variable to store // the final sum int sum = 0; for (int i = 0; i < n; i++) { int xorr = 0; for (int j = i; j < n; j++) { xorr = xorr ^ arr[j]; sum += xorr; } } return sum;}// Driver Codeint main(){ int arr[] = { 3, 8, 13 }; int n = sizeof(arr) / sizeof(arr[0]); cout << findXorSum(arr, n); return 0;} |
Java
import java.util.Arrays;public class Main { // Function to calculate the sum of XOR // of all subarrays public static int findXorSum(int[] arr, int n) { // variable to store // the final sum int sum = 0; for (int i = 0; i < n; i++) { int xorr = 0; for (int j = i; j < n; j++) { xorr = xorr ^ arr[j]; sum += xorr; } } return sum; } // Driver Code public static void main(String[] args) { int[] arr = { 3, 8, 13 }; int n = arr.length; System.out.println(findXorSum(arr, n)); }} |
Python3
# Python code for the above approachdef findXorSum(arr, n): # variable to store the final sum sum = 0 for i in range(n): xorr = 0 for j in range(i, n): xorr = xorr ^ arr[j] sum += xorr return sum# Driver Codearr = [3, 8, 13]n = len(arr)print(findXorSum(arr, n))# This code is contributed by lokeshpotta20. |
C#
using System;class Gfg { // Function to calculate the sum of XOR // of all subarrays public static int findXorSum(int[] arr, int n) { // variable to store // the final sum int sum = 0; for (int i = 0; i < n; i++) { int xorr = 0; for (int j = i; j < n; j++) { xorr = xorr ^ arr[j]; sum += xorr; } } return sum; } // Driver Code public static void Main(string[] args) { int[] arr = { 3, 8, 13 }; int n = arr.Length; Console.WriteLine(findXorSum(arr, n)); }} |
Javascript
// JavaScript code for the above approach// Function to calculate the sum of XOR// of all subarraysfunction findXorSum(arr, n){ // variable to store // the final sum let sum = 0; for(let i = 0; i < n; i++){ let xorr = 0; for(let j = i; j < n; j++){ xorr = xorr ^ arr[j]; sum += xorr; } } return sum;}let arr = [ 3, 8, 13 ];let n = arr.length;console.log(findXorSum(arr, n));// This code is contributed by lokeshmvs21. |
46
Time Complexity: O(n*n)
Auxiliary Space: O(1)
Better solution: A better solution will be using a prefix array i.e. , for every index ‘i’ of the array ‘arr[]’ create a prefix array to store the XOR of all the elements from the left end of the array ‘arr[]’ up to the ith element of ‘arr[]’ Creating a prefix array will take a time of O(N).
Now, using this prefix array, we can find the XOR value of any sub-array in O(1) time.
We can find the XOR from index l to r using the formula:
if l is not zero
XOR = prefix[r] ^ prefix[l-1]
else
XOR = prefix[r].
After this, all we have to do is, to sum up, the XOR values of all the sub-arrays.
Since a total number of sub-arrays is of the order (N2), the time complexity of this approach will be O(N2).
Best solution: For the sake of better understanding, let’s assume any bit of an element is represented by the variable ‘i’ and the variable ‘sum’ is used to store the final sum.
The idea here is, we will try to find the number of XOR values with ith bit set. Let us suppose, there is ‘Si‘ number of sub-arrays with ith bit set. For, ith bit, the sum can be updated as sum += (2i * S)
So, the question is how to implement the above idea?
We will break the task into multiple steps. At each step, we will try to find the number of XOR values with ith bit set.
Now, we will break each step into sub-steps. In each sub-step, we will try to find the number of sub-arrays starting from an index ‘j'(where j varies between 0 to n – 1) with ith bit set in their XOR value. For, ith bit is to be set, an odd number of elements of the sub-array should have their ith bit set.
For all the bits, in a variable c_odd, we will store the count of the number of sub-arrays starting from j = 0 with ith bit set in an odd number of elements. Then, we will iterate through all the elements of the array updating the value of c_odd when needed. If we reach an element ‘j’ with ith bit set, we will update c_odd as c_odd = (n – j – c_odd). Its because, since we encountered a set bit, the number of sub-arrays with an even the number of elements with ith bit set will switch to a number of sub-arrays with an odd number of elements with ith bit set.
Below is the implementation of this approach:
C++
// C++ program to find the sum of XOR of// all subarray of the array#include <iostream>#include <vector>using namespace std;// Function to calculate the sum of XOR// of all subarraysint findXorSum(int arr[], int n){ // variable to store // the final sum int sum = 0; // multiplier int mul = 1; for (int i = 0; i < 30; i++) { // variable to store number of // sub-arrays with odd number of elements // with ith bits starting from the first // element to the end of the array int c_odd = 0; // variable to check the status // of the odd-even count while // calculating c_odd bool odd = 0; // loop to calculate initial // value of c_odd for (int j = 0; j < n; j++) { if ((arr[j] & (1 << i)) > 0) odd = (!odd); if (odd) c_odd++; } // loop to iterate through // all the elements of the // array and update sum for (int j = 0; j < n; j++) { sum += (mul * c_odd); if ((arr[j] & (1 << i)) > 0) c_odd = (n - j - c_odd); } // updating the multiplier mul *= 2; } // returning the sum return sum;}// Driver Codeint main(){ int arr[] = { 3, 8, 13 }; int n = sizeof(arr) / sizeof(arr[0]); cout << findXorSum(arr, n); return 0;} |
Java
// Java program to find the sum of XOR // of all subarray of the array import java.util.*; class GFG { // Function to calculate the sum of XOR // of all subarrays static int findXorSum(int arr[], int n) { // variable to store // the final sum int sum = 0; // multiplier int mul = 1; for (int i = 0; i < 30; i++) { // variable to store number of // sub-arrays with odd number of elements // with ith bits starting from the first // element to the end of the array int c_odd = 0; // variable to check the status // of the odd-even count while // calculating c_odd boolean odd = false; // loop to calculate initial // value of c_odd for (int j = 0; j < n; j++) { if ((arr[j] & (1 << i)) > 0) odd = (!odd); if (odd) c_odd++; } // loop to iterate through // all the elements of the // array and update sum for (int j = 0; j < n; j++) { sum += (mul * c_odd); if ((arr[j] & (1 << i)) > 0) c_odd = (n - j - c_odd); } // updating the multiplier mul *= 2; } // returning the sum return sum; } // Driver code public static void main(String[] args) { int arr[] = { 3, 8, 13 }; int n = arr.length; System.out.println(findXorSum(arr, n)); } } // This code is contributed by Rituraj Jain. |
Python3
# Python3 program to find the Sum of# XOR of all subarray of the array# Function to calculate the Sum of XOR# of all subarraysdef findXorSum(arr, n): # variable to store the final Sum Sum = 0 # multiplier mul = 1 for i in range(30): # variable to store number of sub-arrays # with odd number of elements with ith # bits starting from the first element # to the end of the array c_odd = 0 # variable to check the status of the # odd-even count while calculating c_odd odd = 0 # loop to calculate initial # value of c_odd for j in range(n): if ((arr[j] & (1 << i)) > 0): odd = (~odd) if (odd): c_odd += 1 # loop to iterate through all the # elements of the array and update Sum for j in range(n): Sum += (mul * c_odd) if ((arr[j] & (1 << i)) > 0): c_odd = (n - j - c_odd) # updating the multiplier mul *= 2 # returning the Sum return Sum# Driver Codearr = [3, 8, 13]n = len(arr)print(findXorSum(arr, n))# This code is contributed by Mohit Kumar |
C#
// C# program to find the sum of XOR of// all subarray of the arrayusing System;class GFG{ // Function to calculate the sum // of XOR of all subarraysstatic int findXorSum(int []arr, int n){ // variable to store // the final sum int sum = 0; // multiplier int mul = 1; for (int i = 0; i < 30; i++) { // variable to store number of sub-arrays // with odd number of elements with ith // bits starting from the first element // to the end of the array int c_odd = 0; // variable to check the status // of the odd-even count while // calculating c_odd bool odd = false; // loop to calculate initial // value of c_odd for (int j = 0; j < n; j++) { if ((arr[j] & (1 << i)) > 0) odd = (!odd); if (odd) c_odd++; } // loop to iterate through // all the elements of the // array and update sum for (int j = 0; j < n; j++) { sum += (mul * c_odd); if ((arr[j] & (1 << i)) > 0) c_odd = (n - j - c_odd); } // updating the multiplier mul *= 2; } // returning the sum return sum;}// Driver Codestatic void Main(){ int []arr = { 3, 8, 13 }; int n = arr.Length; Console.WriteLine(findXorSum(arr, n));}}// This code is contributed by mits |
PHP
<?php// PHP program to find the sum of XOR // of all subarray of the array // Function to calculate the sum of// XOR of all subarrays function findXorSum($arr, $n) { // variable to store // the final sum $sum = 0; // multiplier $mul = 1; for ($i = 0; $i < 30; $i++) { // variable to store number of // sub-arrays with odd number of // elements with ith bits starting // from the first element to the // end of the array $c_odd = 0; // variable to check the status // of the odd-even count while // calculating c_odd $odd = 0; // loop to calculate initial // value of c_odd for ($j = 0; $j < $n; $j++) { if (($arr[$j] & (1 << $i)) > 0) $odd = (!$odd); if ($odd) $c_odd++; } // loop to iterate through // all the elements of the // array and update sum for ($j = 0; $j < $n; $j++) { $sum += ($mul * $c_odd); if (($arr[$j] & (1 << $i)) > 0) $c_odd = ($n - $j - $c_odd); } // updating the multiplier $mul *= 2; } // returning the sum return $sum; } // Driver Code $arr = array(3, 8, 13); $n = sizeof($arr); echo findXorSum($arr, $n);// This code is contributed by Ryuga?> |
Javascript
<script>// Javascript program to find // the sum of XOR of// all subarray of the array// Function to calculate the sum of XOR// of all subarraysfunction findXorSum(arr, n){ // variable to store // the final sum let sum = 0; // multiplier let mul = 1; for (let i = 0; i < 30; i++) { // variable to store number of // sub-arrays with odd number of elements // with ith bits starting from the first // element to the end of the array let c_odd = 0; // variable to check the status // of the odd-even count while // calculating c_odd let odd = 0; // loop to calculate initial // value of c_odd for (let j = 0; j < n; j++) { if ((arr[j] & (1 << i)) > 0) odd = (!odd); if (odd) c_odd++; } // loop to iterate through // all the elements of the // array and update sum for (let j = 0; j < n; j++) { sum += (mul * c_odd); if ((arr[j] & (1 << i)) > 0) c_odd = (n - j - c_odd); } // updating the multiplier mul *= 2; } // returning the sum return sum;}// Driver Code let arr = [ 3, 8, 13 ]; let n = arr.length; document.write(findXorSum(arr, n));</script> |
46
Time Complexity: O(N)
Auxiliary Space: O(1)
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