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++
// 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 equalvoid 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 Codeint 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
// 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 equalstatic 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 codepublic 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 |
Python3
# Python program to implement# the above approach# Function to find triplets such that# replacing them with their XOR make# all array elements equaldef 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 codeif __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#
// 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 equalstatic 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 codepublic 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 |
Javascript
<script>// Javascript program to implement// the above approach// Function to find triplets such that// replacing them with their XOR make // all array elements equalfunction 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> |
Output:
0 1 2 2 3 4 0 1 4 2 3 4
Time Complexity: O(N)
Auxiliary Space: O(1)
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