Given an array arr[] of size N, the task is to find all the triplets (i, j, k) such that replacing the elements of the triplets with their Bitwise XOR values, i.e. replacing arr[i], arr[j], arr[k] with (arr[i] ^ arr[j] ^ arr[k]) makes all array elements equal. If more than one solution exists, print any of them. Otherwise, print -1.
Examples:
Input: arr[] = { 4, 2, 1, 7, 2 }
Output: { (0, 1, 2), (2, 3, 4), (0, 1, 4) }
Explanation:
Selecting a triplet (0, 1, 2) and replacing them with arr[0] ^ arr[1] ^ arr[2] modifies arr[] to { 7, 7, 7, 7, 2 }
Selecting a triplet (2, 3, 4) and replacing them with arr[2] ^ arr[3] ^ arr[4] modifies arr[] to { 7, 7, 2, 2, 2 }
Selecting a triplet (0, 1, 4) and replacing them with arr[0] ^ arr[1] ^ arr[2] modifies arr[] to { 2, 2, 2, 2, 2 }Input: arr[] = { 1, 3, 2, 2 }
Output: -1
Approach: The problem can be solved based on the following observation:
x ^ X ^ Y = Y
X ^ Y ^ Y = X
If any two elements of a triplet are equal, then replacing all the elements of the triplet with their Bitwise XOR makes all elements of the triplet equal to the third element of the triplet.
Follow the steps below to solve the problem:
- Selecting the triplets of the form { (0, 1, 2), (2, 3, 4) …} makes the elements of the pairs { (arr[0], arr[1]), (arr[2], arr[3])… } equal.
- From the above observations, selecting the triplets of the form { (0, 1, N – 1), (2, 3, N -1), … } make all the array elements equal to the last element of the array.
- Finally, print the triplets.
Below is the implementation of the above approach:
// C++ program to implement // the above approach #include <bits/stdc++.h> using namespace std;
// Function to find triplets such that // replacing them with their XOR make // all array elements equal void checkXOR( int arr[], int N)
{ // If N is even
if (N % 2 == 0) {
// Calculate xor of
// array elements
int xro = 0;
// Traverse the array
for ( int i = 0; i < N; i++) {
// Update xor
xro ^= arr[i];
}
// If xor is not equal to 0
if (xro != 0) {
cout << -1 << endl;
return ;
}
// Selecting the triplets such that
// elements of the pairs (arr[0], arr[1]),
// (arr[2], arr[3])... can be made equal
for ( int i = 0; i < N - 3; i += 2) {
cout << i << " " << i + 1
<< " " << i + 2 << endl;
}
// Selecting the triplets such that
// all array elements can be made
// equal to arr[N - 1]
for ( int i = 0; i < N - 3; i += 2) {
cout << i << " " << i + 1
<< " " << N - 1 << endl;
}
}
else {
// Selecting the triplets such that
// elements of the pairs (arr[0], arr[1]),
// (arr[2], arr[3])... can be made equal
for ( int i = 0; i < N - 2; i += 2) {
cout << i << " " << i + 1 << " "
<< i + 2 << endl;
}
// Selecting the triplets such that
// all array elements can be made
// equal to arr[N - 1]
for ( int i = 0; i < N - 2; i += 2) {
cout << i << " " << i + 1
<< " " << N - 1 << endl;
}
}
} // Driver Code int main()
{ // Given array
int arr[] = { 4, 2, 1, 7, 2 };
// Size of array
int N = sizeof (arr) / sizeof (arr[0]);
// Function call
checkXOR(arr, N);
} |
// Java program to implement // the above approach import java.util.*;
class GFG{
// Function to find triplets such that // replacing them with their XOR make // all array elements equal static void checkXOR( int arr[], int N)
{ // If N is even
if (N % 2 == 0 )
{
// Calculate xor of
// array elements
int xro = 0 ;
// Traverse the array
for ( int i = 0 ; i < N; i++)
{
// Update xor
xro ^= arr[i];
}
// If xor is not equal to 0
if (xro != 0 )
{
System.out.println(- 1 );
return ;
}
// Selecting the triplets such that
// elements of the pairs (arr[0], arr[1]),
// (arr[2], arr[3])... can be made equal
for ( int i = 0 ; i < N - 3 ; i += 2 )
{
System.out.println(i + " " + (i + 1 ) +
" " + (i + 2 ));
}
// Selecting the triplets such that
// all array elements can be made
// equal to arr[N - 1]
for ( int i = 0 ; i < N - 3 ; i += 2 )
{
System.out.println(i + " " + (i + 1 ) +
" " + (N - 1 ));
}
}
else
{
// Selecting the triplets such that
// elements of the pairs (arr[0], arr[1]),
// (arr[2], arr[3])... can be made equal
for ( int i = 0 ; i < N - 2 ; i += 2 )
{
System.out.println(i + " " + (i + 1 ) +
" " + (i + 2 ));
}
// Selecting the triplets such that
// all array elements can be made
// equal to arr[N - 1]
for ( int i = 0 ; i < N - 2 ; i += 2 )
{
System.out.println(i + " " + (i + 1 ) +
" " + (N - 1 ));
}
}
} // Driver code public static void main(String[] args)
{ // Given array
int arr[] = { 4 , 2 , 1 , 7 , 2 };
// Size of array
int N = arr.length;
// Function call
checkXOR(arr, N);
} } // This code is contributed by susmitakundugoaldanga |
# Python program to implement # the above approach # Function to find triplets such that # replacing them with their XOR make # all array elements equal def checkXOR(arr, N):
# If N is even
if (N % 2 = = 0 ):
# Calculate xor of
# array elements
xro = 0 ;
# Traverse the array
for i in range (N):
# Update xor
xro ^ = arr[i];
# If xor is not equal to 0
if (xro ! = 0 ):
print ( - 1 );
return ;
# Selecting the triplets such that
# elements of the pairs (arr[0], arr[1]),
# (arr[2], arr[3])... can be made equal
for i in range ( 0 , N - 3 , 2 ):
print (i, " " , (i + 1 ), " " , (i + 2 ), end = " " );
# Selecting the triplets such that
# all array elements can be made
# equal to arr[N - 1]
for i in range ( 0 , N - 3 , 2 ):
print (i, " " , (i + 1 ), " " , (N - 1 ), end = " " );
else :
# Selecting the triplets such that
# elements of the pairs (arr[0], arr[1]),
# (arr[2], arr[3])... can be made equal
for i in range ( 0 , N - 2 , 2 ):
print (i, " " , (i + 1 ), " " , (i + 2 ));
# Selecting the triplets such that
# all array elements can be made
# equal to arr[N - 1]
for i in range ( 0 , N - 2 , 2 ):
print (i, " " , (i + 1 ), " " , (N - 1 ));
# Driver code if __name__ = = '__main__' :
# Given array
arr = [ 4 , 2 , 1 , 7 , 2 ];
# Size of array
N = len (arr);
# Function call
checkXOR(arr, N);
# This code is contributed by 29AjayKumar |
// C# program to implement // the above approach using System;
class GFG{
// Function to find triplets such that // replacing them with their XOR make // all array elements equal static void checkXOR( int [] arr, int N)
{ // If N is even
if (N % 2 == 0)
{
// Calculate xor of
// array elements
int xro = 0;
// Traverse the array
for ( int i = 0; i < N; i++)
{
// Update xor
xro ^= arr[i];
}
// If xor is not equal to 0
if (xro != 0)
{
Console.WriteLine(-1);
return ;
}
// Selecting the triplets such that
// elements of the pairs (arr[0], arr[1]),
// (arr[2], arr[3])... can be made equal
for ( int i = 0; i < N - 3; i += 2)
{
Console.WriteLine(i + " " + (i + 1) +
" " + (i + 2));
}
// Selecting the triplets such that
// all array elements can be made
// equal to arr[N - 1]
for ( int i = 0; i < N - 3; i += 2)
{
Console.WriteLine(i + " " + (i + 1) +
" " + (N - 1));
}
}
else
{
// Selecting the triplets such that
// elements of the pairs (arr[0], arr[1]),
// (arr[2], arr[3])... can be made equal
for ( int i = 0; i < N - 2; i += 2)
{
Console.WriteLine(i + " " + (i + 1) +
" " + (i + 2));
}
// Selecting the triplets such that
// all array elements can be made
// equal to arr[N - 1]
for ( int i = 0; i < N - 2; i += 2)
{
Console.WriteLine(i + " " + (i + 1) +
" " + (N - 1));
}
}
} // Driver code public static void Main()
{ // Given array
int [] arr = { 4, 2, 1, 7, 2 };
// Size of array
int N = arr.Length;
// Function call
checkXOR(arr, N);
} } // This code is contributed by sanjoy_62 |
<script> // Javascript program to implement // the above approach // Function to find triplets such that // replacing them with their XOR make // all array elements equal function checkXOR(arr, N)
{ // If N is even
if (N % 2 == 0) {
// Calculate xor of
// array elements
let xro = 0;
// Traverse the array
for (let i = 0; i < N; i++) {
// Update xor
xro ^= arr[i];
}
// If xor is not equal to 0
if (xro != 0) {
document.write(-1 + "<br>" );
return ;
}
// Selecting the triplets such that
// elements of the pairs (arr[0], arr[1]),
// (arr[2], arr[3])... can be made equal
for (let i = 0; i < N - 3; i += 2) {
document.write(i + " " + (i + 1)
+ " " + (i + 2) + "<br>" );
}
// Selecting the triplets such that
// all array elements can be made
// equal to arr[N - 1]
for (let i = 0; i < N - 3; i += 2) {
document.write(i + " " + (i + 1)
+ " " + (N - 1) + "<br>" );
}
}
else {
// Selecting the triplets such that
// elements of the pairs (arr[0], arr[1]),
// (arr[2], arr[3])... can be made equal
for (let i = 0; i < N - 2; i += 2) {
document.write(i + " " + (i + 1) + " "
+ (i + 2) + "<br>" );
}
// Selecting the triplets such that
// all array elements can be made
// equal to arr[N - 1]
for (let i = 0; i < N - 2; i += 2) {
document.write(i + " " + (i + 1)
+ " " + (N - 1) + "<br>" );
}
}
} // Driver Code // Given array
let arr = [ 4, 2, 1, 7, 2 ];
// Size of array
let N = arr.length;
// Function call
checkXOR(arr, N);
</script> |
0 1 2 2 3 4 0 1 4 2 3 4
Time Complexity: O(N)
Auxiliary Space: O(1)