通过将三元组替换为其按位异或，使所有数组元素相等

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📜 通过将三元组替换为其按位异或，使所有数组元素相等

📅 最后修改于: 2021-05-17 01:47:48 🧑 作者: Mango

给定大小为N的数组arr [] ，任务是找到所有三元组(i，j，k) ，以便用其按位XOR值替换三元组的元素，即替换arr [i]，arr [j] ，带有(arr [i] ^ arr [j] ^ arr [k])的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

为什么编程需要懂一点英语

方法：可以通过以下观察来解决问题：

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.

为什么编程需要懂一点英语

请按照以下步骤解决问题：

选择形式为{ (0，1，2) ， (2，3，4) …}的三元组将使{ (arr [0]，arr [1]) ， (arr [2]，arr [3]) …}相等。
从上述观察中，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，打印三胞胎。

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

输出：

时间复杂度： O(N)
辅助空间： O(1)