Find bitwise XOR of all triplets formed from given three Arrays
Last Updated :
12 Dec, 2022
Given three arrays arr1[], arr2[], and arr3[] consisting of non-negative integers. The task is to find the bitwise XOR of the XOR of all possible triplets that are formed by taking one element from each array.
Examples:
Input: arr1[] = {2, 3, 1}, arr2[] = {2, 4}, arr3[] = {3, 5}
Output: 0
Explanation: All possible triplets are (2, 2, 3), (2, 2, 5), (2, 4, 3), (2, 4, 5), (3, 2, 3), (3, 2, 5), (3, 4, 3), (3, 4, 5), (1, 2, 3), (1, 2, 5), (1, 4, 3), (1, 4, 5). Bitwise XOR of all possible triplets are =(2^2^3) ^ (2^2^5) ^ (2^4^3) ^ (2^4^5) ^ (3^2^3) ^ (3^2^5) ^ (3^4^3) ^ (3^4^5) ^ (1^2^3) ^ (1^2^5) ^ (1^4^3) ^ (1^4^5) = 0
Input: arr1[] = {1}, arr2[] = {3}, arr3[] = {4, 2, 3}
Output:7
Explanation: All possible triplets are (1, 3, 4), (1, 3, 2), (1, 3, 3)
Bitwise XOR of all possible triplets are = (1^3^4) ^ (1^3^2) ^ (1^3^3) =7
Naive Approach:
The basic idea is to traverse the arrays and generate all possible triplets from the given arrays. Finally, print the Bitwise XOR of each of these triplets from the given arrays.
Follow the steps below to solve the problem:
- Initialize a variable, say totalXOR, to store the bitwise XOR of each pair from these array sets.
- Traverse the given arrays and generate all possible triplets (arr[i], arr[j], arr[k]) from the given arrays.
- For each triplet arr[i], arr[j] and arr[k] update the value:
- totalXOR = (totalXOR ^ arr[i] ^ arr[j] ^ arr[k]).
- Finally, return the value of totalXOR.
Below is the Implementation of the above approach.
C++
#include <bits/stdc++.h>
using namespace std;
int XorOfAllTriplets(int arr1[], int arr2[], int arr3[],
int n1, int n2, int n3)
{
int result = 0;
for (int i = 0; i < n1; i++) {
for (int j = 0; j < n2; j++) {
for (int k = 0; k < n3; k++) {
result = result
^ (arr1[i] ^ arr2[j] ^ arr3[k]);
}
}
}
return result;
}
int main()
{
int arr1[] = { 2, 3, 1 };
int arr2[] = { 2, 4 };
int arr3[] = { 3, 5 };
int n1 = sizeof(arr1) / sizeof(arr1[0]);
int n2 = sizeof(arr2) / sizeof(arr2[0]);
int n3 = sizeof(arr3) / sizeof(arr3[0]);
cout << XorOfAllTriplets(arr1, arr2, arr3, n1, n2, n3);
return 0;
}
|
Java
import java.io.*;
class GFG
{
public static int XorOfAllTriplets(int arr1[],
int arr2[],
int arr3[], int n1,
int n2, int n3)
{
int result = 0;
for (int i = 0; i < n1; i++) {
for (int j = 0; j < n2; j++) {
for (int k = 0; k < n3; k++) {
result
= result
^ (arr1[i] ^ arr2[j] ^ arr3[k]);
}
}
}
return result;
}
public static void main(String[] args)
{
int arr1[] = { 2, 3, 1 };
int arr2[] = { 2, 4 };
int arr3[] = { 3, 5 };
int n1 = arr1.length;
int n2 = arr2.length;
int n3 = arr3.length;
System.out.print(
XorOfAllTriplets(arr1, arr2, arr3, n1, n2, n3));
}
}
|
Python3
def XorOfAllTriplets(arr1, arr2, arr3, n1, n2, n3) :
result = 0;
for i in range(n1) :
for j in range(n2) :
for k in range(n3) :
result = result ^ (arr1[i] ^ arr2[j] ^ arr3[k]);
return result;
if __name__ == "__main__" :
arr1 = [ 2, 3, 1 ];
arr2 = [ 2, 4 ];
arr3 = [ 3, 5 ];
n1 = len(arr1);
n2 = len(arr2);
n3 = len(arr3);
print(XorOfAllTriplets(arr1, arr2, arr3, n1, n2, n3));
|
C#
using System;
public class GFG {
public static int XorOfAllTriplets(int[] arr1,
int[] arr2,
int[] arr3, int n1,
int n2, int n3)
{
int result = 0;
for (int i = 0; i < n1; i++) {
for (int j = 0; j < n2; j++) {
for (int k = 0; k < n3; k++) {
result
= result
^ (arr1[i] ^ arr2[j] ^ arr3[k]);
}
}
}
return result;
}
static public void Main()
{
int[] arr1 = { 2, 3, 1 };
int[] arr2 = { 2, 4 };
int[] arr3 = { 3, 5 };
int n1 = arr1.Length;
int n2 = arr2.Length;
int n3 = arr3.Length;
Console.Write(
XorOfAllTriplets(arr1, arr2, arr3, n1, n2, n3));
}
}
|
Javascript
function XorOfAllTriplets( arr1, arr2, arr3,
n1,n2, in3)
{
let result = 0;
for (let i = 0; i < n1; i++) {
for (let j = 0; j < n2; j++) {
for (let k = 0; k < n3; k++) {
result = result ^ (arr1[i] ^ arr2[j] ^ arr3[k]);
}
}
}
return result;
}
let arr1 = [ 2, 3, 1 ];
let arr2 = [ 2, 4 ];
let arr3 = [ 3, 5 ];
let n1 = arr1.length;
let n2 = arr2.length;
let n3 = arr3.length;
console.log(XorOfAllTriplets(arr1, arr2, arr3, n1, n2, n3));
|
Output:
0
Time Complexity: O(N * M * P) where N, M and P are sizes of the three arrays
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, follow the observations below
Property of Bitwise XOR:
a ^ a ^ . . .( Even times ) = 0
a ^ a ^ a ^ . . .( Odd times ) = a
Say the sizes of arr1[], arr2[] and arr3[] are N1, N2 and N3 respectively. Element of arr1[] occurs exactly N2 * N3 times. Similarly, arr2[] occurs exactly N3 * N1 and element of arr3[] occur exactly N1 * N2 times.
Follow the steps below to solve the problem:
- For each array check if the number of occurrences of its elements is odd or above using the above mentioned formula for occurrences:
- If the number of occurrences is odd, then XOR all the elements of that array with the final result.
- Otherwise, move to the next array.
- Return the final value of XOR.
Below is the implementation of the above approach.
C++
#include <bits/stdc++.h>
using namespace std;
int XorOfAllTriplets(int arr1[], int arr2[], int arr3[],
int n1, int n2, int n3)
{
int result = 0;
if ((n2 * n3) % 2)
for (int i = 0; i < n1; i++)
result ^= arr1[i];
if ((n3 * n1) % 2)
for (int j = 0; j < n2; j++)
result ^= arr2[j];
if ((n1 * n2) % 2)
for (int k = 0; k < n3; k++)
result ^= arr3[k];
return result;
}
int main()
{
int arr1[] = { 1 };
int arr2[] = { 3 };
int arr3[] = { 4, 2, 3 };
int N1 = sizeof(arr1) / sizeof(arr1[0]);
int N2 = sizeof(arr2) / sizeof(arr2[0]);
int N3 = sizeof(arr3) / sizeof(arr3[0]);
cout << XorOfAllTriplets(arr1, arr2, arr3, N1, N2, N3);
return 0;
}
|
Java
import java.io.*;
class GFG
{
public static int XorOfAllTriplets(int arr1[], int arr2[], int arr3[],int n1, int n2, int n3)
{
int result = 0;
if ((n2 * n3) % 2==1)
for (int i = 0; i < n1; i++)
result ^= arr1[i];
if ((n3 * n1) % 2==1)
for (int j = 0; j < n2; j++)
result ^= arr2[j];
if ((n1 * n2) % 2==1)
for (int k = 0; k < n3; k++)
result ^= arr3[k];
return result;
}
public static void main(String[] args)
{
int arr1[] = { 1 };
int arr2[] = { 3 };
int arr3[] = { 4, 2, 3 };
int N1 = arr1.length;
int N2 = arr2.length;
int N3 = arr3.length;
System.out.println(XorOfAllTriplets(arr1, arr2, arr3, N1, N2, N3));
}
}
|
Python3
def XorOfAllTriplets(arr1, arr2, arr3,n1, n2, n3) :
result = 0;
if ((n2 * n3) % 2) :
for i in range(n1) :
result ^= arr1[i];
if ((n3 * n1) % 2) :
for j in range(n2) :
result ^= arr2[j];
if ((n1 * n2) % 2) :
for k in range(n3) :
result ^= arr3[k];
return result;
if __name__ == "__main__" :
arr1 = [ 1 ];
arr2 = [ 3 ];
arr3 = [ 4, 2, 3 ];
N1 = len(arr1);
N2 = len(arr2);
N3 = len(arr3);
print(XorOfAllTriplets(arr1, arr2, arr3, N1, N2, N3));
|
C#
using System;
public class GFG
{
public static int XorOfAllTriplets(int []arr1, int []arr2,
int []arr3,int n1,
int n2, int n3)
{
int result = 0;
if ((n2 * n3) % 2==1)
for (int i = 0; i < n1; i++)
result ^= arr1[i];
if ((n3 * n1) % 2==1)
for (int j = 0; j < n2; j++)
result ^= arr2[j];
if ((n1 * n2) % 2==1)
for (int k = 0; k < n3; k++)
result ^= arr3[k];
return result;
}
public static void Main(string[] args)
{
int []arr1 = { 1 };
int []arr2 = { 3 };
int []arr3 = { 4, 2, 3 };
int N1 = arr1.Length;
int N2 = arr2.Length;
int N3 = arr3.Length;
Console.WriteLine(XorOfAllTriplets(arr1, arr2, arr3, N1, N2, N3));
}
}
|
Javascript
function XorOfAllTriplets(arr1, arr2, arr3, n1, n2, n3){
var result = 0;
if ((n2 * n3) % 2==1)
for (let i = 0; i < n1; i++)
result ^= arr1[i];
if ((n3 * n1) % 2==1)
for (let j = 0; j < n2; j++)
result ^= arr2[j];
if ((n1 * n2) % 2==1)
for (let k = 0; k < n3; k++)
result ^= arr3[k];
return result;
}
var arr1 = [ 1 ];
var arr2 = [ 3 ];
var arr3 = [ 4, 2, 3 ];
var N1 = arr1.length;
var N2 = arr2.length;
var N3 = arr3.length;
console.log(XorOfAllTriplets(arr1, arr2, arr3, N1, N2, N3));
|
Time Complexity: O(N1 + N2 + N3)
Auxiliary Space: O(1)
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