通过用它们的按位异或替换三元组使所有数组元素相等

给定一个大小为N的数组arr[] ，任务是找到所有的三元组(i, j, k) ，以便用它们的按位异或值替换三元组的元素，即替换 arr[i], arr[j] , arr[k] 与(arr[i] ^ arr[j] ^ arr[k])使所有数组元素相等。如果存在多个解决方案，请打印其中任何一个。否则，打印-1 。

例子：

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

方法：该问题可以基于以下观察来解决：

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. 
 

请按照以下步骤解决问题：

• 选择形式为 { (0, 1, 2) , (2, 3, 4) …} 的三元组使 { (arr[0], arr[1]) , (arr[2], arr [3]) … } 相等。
• 根据以上观察，选择 { (0, 1, N – 1) , (2, 3, N -1) , … } 形式的三元组使所有数组元素等于数组的最后一个元素。
• 最后，打印三元组。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
 
#include 
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

// 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

Python3

# 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#

// 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

Javascript

 
// 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);
 

输出：

0 1 2
2 3 4
0 1 4
2 3 4

时间复杂度： O(N)
辅助空间： O(1)

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