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📜  XOR为零的唯一三元组数

📅  最后修改于: 2021-05-06 23:36:13             🧑  作者: Mango

给定N个没有重复的数字,请计算唯一三元组(a i ,a j ,a k )的数量,以使它们的XOR为0。如果三元组中的所有三个数字都是唯一的,则认为三元组是唯一的。
例子:

Input : a[] = {1, 3, 5, 10, 14, 15};
Output : 2 
Explanation : {1, 14, 15} and {5, 10, 15} are the 
              unique triplets whose XOR is 0. 
              {1, 14, 15} and all other combinations of 
              1, 14, 15 are considered as 1 only.

Input : a[] = {4, 7, 5, 8, 3, 9};
Output : 1
Explanation : {4, 7, 3} is the only triplet whose XOR is 0 

天真的方法:天真的方法是运行三个嵌套循环,第一个从0到n,第二个从i + 1到n,最后一个从j + 1到n,以获得唯一的三元组。计算a i ,a jk的XOR,检查其是否等于0,如果等于0,则增加计数。
时间复杂度:O(n 3 )
高效的方法:一种有效的方法是使用两个相同数字的XOR等于0的XOR属性之一。因此,我们只需要计算唯一对的XOR,如果计算出的XOR是数组元素之一,则我们得到XOR为0的三元组。下面是计算唯一三元组数的步骤:
以下是此方法的完整算法:

  1. 使用map标记所有数组元素。
  2. 运行两个嵌套循环,一个来自in-1,另一个来自i + 1-n,以获取所有对。
  3. 获取该对的XOR。
  4. 检查XOR是否为数组元素,而不是ij之一
  5. 如果条件成立,请增加计数。
  6. 返回count / 3,因为我们只需要唯一的三元组。由于in和j + 1-n给我们唯一的对而不是三胞胎,所以我们做一个count / 3来删除其他两个可能的组合。

下面是上述想法的实现:

C++
// CPP program to count the number of
// unique triplets whose XOR is 0
#include 
using namespace std;
 
// function to count the number of
// unique triplets whose xor is 0
int countTriplets(int a[], int n)
{
    // To store values that are present
    unordered_set s;
    for (int i = 0; i < n; i++)
        s.insert(a[i]);
     
    // stores the count of unique triplets
    int count = 0;
     
    // traverse for all i, j pairs such that j>i
    for (int i = 0; i < n-1; i++) {
        for (int j = i + 1; j < n; j++) {
 
          // xor of a[i] and a[j]
          int xr = a[i] ^ a[j];
     
          // if xr of two numbers is present,
          // then increase the count
          if (s.find(xr) != s.end() && xr != a[i] &&
                                       xr != a[j])
            count++;
        }
    }
     
    // returns answer
    return count / 3;
}
 
// Driver code to test above function
int main()
{
    int a[] = {1, 3, 5, 10, 14, 15};
    int n = sizeof(a) / sizeof(a[0]);  
    cout << countTriplets(a, n);   
    return 0;
}


Java
// Java program to count
// the number of unique
// triplets whose XOR is 0
import java.io.*;
import java.util.*;
 
class GFG
{
    // function to count the
    // number of unique triplets
    // whose xor is 0
    static int countTriplets(int []a,
                             int n)
    {
        // To store values
        // that are present
        ArrayList s =
                  new ArrayList();
        for (int i = 0; i < n; i++)
            s.add(a[i]);
         
        // stores the count
        // of unique triplets
        int count = 0;
         
        // traverse for all i,
        // j pairs such that j>i
        for (int i = 0; i < n; i++)
        {
            for (int j = i + 1;
                     j < n; j++)
            {
     
            // xor of a[i] and a[j]
            int xr = a[i] ^ a[j];
         
            // if xr of two numbers
            // is present, then
            // increase the count
            if (s.contains(xr) &&
                xr != a[i] && xr != a[j])
                count++;
            }
        }
         
        // returns answer
        return count / 3;
    }
     
    // Driver code
    public static void main(String srgs[])
    {
        int []a = {1, 3, 5,
                   10, 14, 15};
        int n = a.length;
        System.out.print(countTriplets(a, n));
    }
}
 
// This code is contributed by
// Manish Shaw(manishshaw1)


Python3
# Python 3 program to count the number of
# unique triplets whose XOR is 0
 
# function to count the number of
# unique triplets whose xor is 0
def countTriplets(a, n):
     
    # To store values that are present
    s = set()
    for i in range(n):
        s.add(a[i])
     
    # stores the count of unique triplets
    count = 0
     
    # traverse for all i, j pairs such that j>i
    for i in range(n):
        for j in range(i + 1, n, 1):
             
            # xor of a[i] and a[j]
            xr = a[i] ^ a[j]
             
            # if xr of two numbers is present,
            # then increase the count
            if (xr in s and xr != a[i] and
                            xr != a[j]):
                count += 1;
         
    # returns answer
    return int(count / 3)
 
# Driver code
if __name__ == '__main__':
    a = [1, 3, 5, 10, 14, 15]
    n = len(a)
    print(countTriplets(a, n))
     
# This code is contributed by
# Surendra_Gangwar


C#
// C# program to count
// the number of unique
// triplets whose XOR is 0
using System;
using System.Collections.Generic;
 
class GFG
{
    // function to count the
    // number of unique triplets
    // whose xor is 0
    static int countTriplets(int []a,
                             int n)
    {
        // To store values
        // that are present
        List s = new List();
        for (int i = 0; i < n; i++)
            s.Add(a[i]);
         
        // stores the count
        // of unique triplets
        int count = 0;
         
        // traverse for all i,
        // j pairs such that j>i
        for (int i = 0; i < n; i++)
        {
            for (int j = i + 1;
                     j < n; j++)
            {
     
            // xor of a[i] and a[j]
            int xr = a[i] ^ a[j];
         
            // if xr of two numbers
            // is present, then
            // increase the count
            if (s.Exists(item => item == xr) &&
                   xr != a[i] && xr != a[j])
                count++;
            }
        }
         
        // returns answer
        return count / 3;
    }
     
    // Driver code
    static void Main()
    {
        int []a = new int[]{1, 3, 5,
                            10, 14, 15};
        int n = a.Length;
        Console.Write(countTriplets(a, n));
    }
}
 
// This code is contributed by
// Manish Shaw(manishshaw1)


输出:

2

时间复杂度: O(n 2 )