查找数组中对的数量，以使它们的XOR为0

给定一个数组$A[ ]$的大小为N。找到对(i，j)的数量，使得$A_i$异或$A_j$ = 0，并且1 <= i 例子 ：

Input : A[] = {1, 3, 4, 1, 4}
Output : 2
Explanation : Index (0, 3) and (2, 4)

Input : A[] = {2, 2, 2}
Output : 3

第一种方法：排序
$A_i$异或$A_j$ = 0仅在以下情况下满足$A_i = A_j$ 。因此，我们将首先对数组进行排序，然后计算每个元素的频率。通过组合，我们可以观察到如果某个元素的频率是$count$然后，它将有助于$count*(count-1)/2$答案。
下面是上述方法的实现：

C++

// C++ program to find number
// of pairs in an array such
// that their XOR is 0
#include 
using namespace std;
 
// Function to calculate the
// count
int calculate(int a[], int n)
{
    // Sorting the list using
    // built in function
    sort(a, a + n);
 
    int count = 1;
    int answer = 0;
 
    // Traversing through the
    // elements
    for (int i = 1; i < n; i++) {
     
        if (a[i] == a[i - 1]){
 
            // Counting frequency of each
            // elements
            count += 1;
             
        }
        else
        {
            // Adding the contribution of
            // the frequency to the answer
            answer = answer + (count * (count - 1)) / 2;
            count = 1;
        }
    }
 
    answer = answer + (count * (count - 1)) / 2;
 
    return answer;
}
 
// Driver Code
int main()
{
 
    int a[] = { 1, 2, 1, 2, 4 };
    int n = sizeof(a) / sizeof(a[0]);
 
    // Print the count
    cout << calculate(a, n);
    return 0;
}
 
// This article is contributed by Sahil_Bansall.

Java

// Java program to find number
// of pairs in an array such
// that their XOR is 0
import java.util.*;
 
class GFG
{
    // Function to calculate
    // the count
    static int calculate(int a[], int n)
    {
        // Sorting the list using
        // built in function
        Arrays.sort(a);
     
        int count = 1;
        int answer = 0;
     
        // Traversing through the
        // elements
        for (int i = 1; i < n; i++)
        {
         
            if (a[i] == a[i - 1])
            {
                // Counting frequency of each
                // elements
                count += 1;
                 
            }
            else
            {
                // Adding the contribution of
                // the frequency to the answer
                answer = answer + (count * (count - 1)) / 2;
                count = 1;
            }
        }
     
        answer = answer + (count * (count - 1)) / 2;
     
        return answer;
    }
     
    // Driver Code
    public static void main (String[] args)
    {
        int a[] = { 1, 2, 1, 2, 4 };
        int n = a.length;
     
        // Print the count
        System.out.println(calculate(a, n));
    }
}
 
// This code is contributed by Ansu Kumari.

Python3

# Python3 program to find number of pairs
# in an array such that their XOR is 0
 
# Function to calculate the count
def calculate(a) :
 
    # Sorting the list using
    # built in function
    a.sort()
     
    count = 1
    answer = 0
     
    # Traversing through the elements
    for i in range(1, len(a)) :
 
        if a[i] == a[i - 1] :
             
            # Counting frequncy of each elements
            count += 1
 
        else :
 
            # Adding the contribution of
            # the frequency to the answer
            answer = answer + count * (count - 1) // 2
            count = 1
 
    answer = answer + count * (count - 1) // 2
     
    return answer
 
 
# Driver Code
if __name__ == '__main__':
     
    a = [1, 2, 1, 2, 4]
 
    # Print the count
    print(calculate(a))

C#

// C# program to find number
// of pairs in an array such
// that their XOR is 0
using System;
 
class GFG
{
    // Function to calculate
    // the count
    static int calculate(int []a, int n)
    {
        // Sorting the list using
        // built in function
        Array.Sort(a);
     
        int count = 1;
        int answer = 0;
     
        // Traversing through the
        // elements
        for (int i = 1; i < n; i++)
        {
         
            if (a[i] == a[i - 1])
            {
                // Counting frequency of each
                // elements
                count += 1;
                 
            }
            else
            {
                // Adding the contribution of
                // the frequency to the answer
                answer = answer + (count * (count - 1)) / 2;
                count = 1;
            }
        }
     
        answer = answer + (count * (count - 1)) / 2;
     
        return answer;
    }
     
    // Driver Code
    public static void Main ()
    {
        int []a = { 1, 2, 1, 2, 4 };
        int n = a.Length;
     
        // Print the count
        Console.WriteLine(calculate(a, n));
    }
}
 
// This code is contributed by vt_m.

Javascript

 
// JavaScript program to find number
// of pairs in an array such
// that their XOR is 0
 
// Function to calculate the
// count
function calculate(a, n)
{
    // Sorting the list using
    // built in function
    a.sort();
 
    let count = 1;
    let answer = 0;
 
    // Traversing through the
    // elements
    for (let i = 1; i < n; i++) {
     
        if (a[i] == a[i - 1]){
 
            // Counting frequency of each
            // elements
            count += 1;
             
        }
        else
        {
            // Adding the contribution of
            // the frequency to the answer
            answer = answer + Math.floor((count * (count - 1)) / 2);
            count = 1;
        }
    }
 
    answer = answer + Math.floor((count * (count - 1)) / 2);
 
    return answer;
}
 
// Driver Code
 
    let a = [ 1, 2, 1, 2, 4 ];
    let n = a.length;
 
    // Print the count
    document.write(calculate(a, n));
 
// This code is contributed by Surbhi Tyagi.
 

输出 ：

2

时间复杂度：O(N Log N)
第二种方法：散列(索引映射)
如果我们可以计算数组中每个元素的频率，则解决方案很方便。索引映射技术可用于计算每个元素的频率。
下面是上述方法的实现：

C++

// C++ program to find number of pairs
// in an array such that their XOR is 0
#include 
using namespace std;
 
// Function to calculate the answer
int calculate(int a[], int n){
     
    // Finding the maximum of the array
    int *maximum = max_element(a, a + n);
 
    // Creating frequency array
    // With initial value 0
    int frequency[*maximum + 1] = {0};
     
    // Traversing through the array
    for(int i = 0; i < n; i++)
    {
        // Counting frequency
        frequency[a[i]] += 1;
    }
    int answer = 0;
     
    // Traversing through the frequency array
    for(int i = 0; i < (*maximum)+1; i++)
    {
        // Calculating answer
        answer = answer + frequency[i] * (frequency[i] - 1) ;
    }
    return answer/2;
}
 
// Driver Code
int main()
{
   int a[] = {1, 2, 1, 2, 4};
   int n = sizeof(a) / sizeof(a[0]);
    
   // Function calling
   cout << (calculate(a,n));
}
 
// This code is contributed by Smitha

Java

// Java program to find number of pairs
// in an array such that their XOR is 0
import java.util.*;
 
class GFG
{
 
    // Function to calculate the answer
    static int calculate(int a[], int n)
    {
 
        // Finding the maximum of the array
        int maximum = Arrays.stream(a).max().getAsInt();
 
        // Creating frequency array
        // With initial value 0
        int frequency[] = new int[maximum + 1];
 
        // Traversing through the array
        for (int i = 0; i < n; i++)
        {
             
            // Counting frequency
            frequency[a[i]] += 1;
        }
        int answer = 0;
 
        // Traversing through the frequency array
        for (int i = 0; i < (maximum) + 1; i++)
        {
             
            // Calculating answer
            answer = answer + frequency[i] * (frequency[i] - 1);
        }
        return answer / 2;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int a[] = {1, 2, 1, 2, 4};
        int n = a.length;
 
        // Function calling
        System.out.println(calculate(a, n));
    }
}
 
// This code is contributed by 29AjayKumar

的Python 3

# Python3 program to find number of pairs
# in an array such that their XOR is 0
 
# Function to calculate the answer
def calculate(a) :
     
    # Finding the maximum of the array
    maximum = max(a)
     
    # Creating frequency array
    # With initial value 0
    frequency = [0 for x in range(maximum + 1)]
     
    # Traversing through the array
    for i in a :
          
        # Counting frequency
        frequency[i] += 1
     
    answer = 0
     
    # Traversing through the frequency array
    for i in frequency :
         
        # Calculating answer
        answer = answer + i * (i - 1) // 2
     
    return answer
 
# Driver Code
a = [1, 2, 1, 2, 4]
print(calculate(a))

C＃

// C# program to find number of pairs
// in an array such that their XOR is 0
using System;
using System.Linq;
class GFG
{
 
    // Function to calculate the answer
    static int calculate(int []a, int n)
    {
 
        // Finding the maximum of the array
        int maximum = a.Max();
 
        // Creating frequency array
        // With initial value 0
        int []frequency = new int[maximum + 1];
 
        // Traversing through the array
        for (int i = 0; i < n; i++)
        {
             
            // Counting frequency
            frequency[a[i]] += 1;
        }
        int answer = 0;
 
        // Traversing through the frequency array
        for (int i = 0; i < (maximum) + 1; i++)
        {
             
            // Calculating answer
            answer = answer + frequency[i] *
                             (frequency[i] - 1);
        }
        return answer / 2;
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        int []a = {1, 2, 1, 2, 4};
        int n = a.Length;
 
        // Function calling
        Console.WriteLine(calculate(a, n));
    }
}
 
// This code is contributed by PrinciRaj1992

Java脚本

// Javascript program to find number of pairs
// in an array such that their XOR is 0
 
// Function to calculate the answer
function calculate(a, n){
     
    // Finding the maximum of the array
    let maximum = Math.max(...a);
 
    // Creating frequency array
    // With initial value 0
    let frequency = new Array(maximum + 1).fill(0);
     
    // Traversing through the array
    for(let i = 0; i < n; i++)
    {
        // Counting frequency
        frequency[a[i]] += 1;
    }
    let answer = 0;
     
    // Traversing through the frequency array
    for(let i = 0; i < maximum+1; i++)
    {
        // Calculating answer
        answer = answer + frequency[i] * (frequency[i] - 1) ;
    }
    return parseInt(answer/2);
}
 
// Driver Code
   let a = [1, 2, 1, 2, 4];
   let n = a.length;
    
   // Function calling
   document.write(calculate(a,n));
 

输出 ：

2

时间复杂度：O(N)
注意：仅当数组中的数字不大时才可以使用索引映射方法。在这种情况下，可以使用分类方法。