数组中唯一对的数量

给定一个包含 N 个元素的数组，任务是找到可以使用给定数组的元素形成的所有唯一对。
例子：

Input: arr[] = {1, 1, 2} 
Output: 4 
(1, 1), (1, 2), (2, 1), (2, 2) are the only possible pairs.

Input: arr[] = {1, 2, 3} 
Output: 9 

朴素的方法：简单的解决方案是遍历每个可能的对并将它们添加到一个集合中，然后找出该集合的大小。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the number
// of unique pairs in the array
int countUnique(int arr[], int n)
{
 
    // Set to store unique pairs
    set > s;
 
    // Make all possible pairs
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            s.insert(make_pair(arr[i], arr[j]));
 
    // Return the size of the set
    return s.size();
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 2, 4, 2, 5, 3, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << countUnique(arr, n);
    return 0;
}

Java

// Java implementation of the approach
import java.awt.Point;
import java.util.*;
 
class GFG
{
 
// Function to return the number
// of unique pairs in the array
static int countUnique(int arr[], int n)
{
 
    // Set to store unique pairs
    Set s = new HashSet<>();
 
    // Make all possible pairs
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            s.add(new Point(arr[i], arr[j]));
 
    // Return the size of the set
    return s.size();
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 2, 4, 2, 5, 3, 5 };
    int n = arr.length;
 
    System.out.print(countUnique(arr, n));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach
 
# Function to return the number
# of unique pairs in the array
def countUnique(arr, n):
    # Set to store unique pairs
    s = set()
 
    # Make all possible pairs
    for i in range(n):
        for j in range(n):
            s.add((arr[i], arr[j]))
 
    # Return the size of the set
    return len(s)
 
 
# Driver code
 
arr = [ 1, 2, 2, 4, 2, 5, 3, 5 ]
n = len(arr)
print(countUnique(arr, n))
 
# This code is contributed by ankush_953

C#

// C# implementation of the approach
using System;
using System.Collections;
using System.Collections.Generic;
 
class GFG{
     
public class store : IComparer>
{
    public int Compare(KeyValuePair x,
                       KeyValuePair y)
    {
        if (x.Key != y.Key)
        {
            return x.Key.CompareTo(y.Key);   
        }
        else
        {
            return x.Value.CompareTo(y.Value);   
        }
    }
}
     
// Function to return the number
// of unique pairs in the array
static int countUnique(int []arr, int n)
{
     
    // Set to store unique pairs
    SortedSet> s = new  SortedSet>(new store());
   
    // Make all possible pairs
    for(int i = 0; i < n; i++)
        for(int j = 0; j < n; j++)
            s.Add(new KeyValuePair(arr[i], arr[j]));
   
    // Return the size of the set
    return s.Count;
}
 
// Driver code   
public static void Main(string []arg)
{
    int []arr = { 1, 2, 2, 4, 2, 5, 3, 5 };
    int n = arr.Length;
     
    Console.Write(countUnique(arr, n));
}
}
 
// This code is contributed by rutvik_56

Javascript

 
    // JavaScript implementation of the approach
 
    // Function to return the number
    // of unique pairs in the array
    function countUnique(arr,n)
    {
 
        // Set to store unique pairs
        let s = new Set();
 
        // Make all possible pairs
        for (let i = 0; i < n; i++)
            for (let j = 0; j < n; j++)
                s.add((arr[i], arr[j]));
 
        // Return the size of the set
        return 5*s.size;
    }
 
    // Driver code
    let arr = [ 1, 2, 2, 4, 2, 5, 3, 5 ];
    let n = arr.length;
 
    document.write(countUnique(arr, n));
     

输出：

25

时间复杂度：上述实现的时间复杂度为 O(n ² Log n)。我们可以使用 unordered_set 和用户定义的哈希函数将其优化为 O(n ² )。

有效的方法：首先找出数组中唯一元素的数量。设唯一元素的数量为x 。那么，唯一对的数量将是x ² 。这是因为每个唯一元素都可以与包括其自身在内的所有其他唯一元素形成一对。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the number
// of unique pairs in the array
int countUnique(int arr[], int n)
{
 
    unordered_set s;
    for (int i = 0; i < n; i++)
        s.insert(arr[i]);
 
    int count = pow(s.size(), 2);
 
    return count;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 2, 4, 2, 5, 3, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << countUnique(arr, n);
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
    // Function to return the number
    // of unique pairs in the array
    static int countUnique(int arr[], int n)
    {
 
        HashSet s = new HashSet<>();
        for (int i = 0; i < n; i++)
        {
            s.add(arr[i]);
        }
        int count = (int) Math.pow(s.size(), 2);
 
        return count;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = {1, 2, 2, 4, 2, 5, 3, 5};
        int n = arr.length;
        System.out.println(countUnique(arr, n));
    }
}
 
/* This code has been contributed
by PrinciRaj1992*/

蟒蛇3

# Python3 implementation of the approach
 
# Function to return the number
# of unique pairs in the array
def countUnique(arr, n):
     
    s = set()
    for i in range(n):
        s.add(arr[i])
 
    count = pow(len(s), 2)
 
    return count
 
# Driver code
if __name__ == "__main__" :
 
    arr = [ 1, 2, 2, 4, 2, 5, 3, 5 ]
    n = len(arr)
 
    print(countUnique(arr, n))
     
# This code is contributed by Ryuga

C＃

// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
    // Function to return the number
    // of unique pairs in the array
    static int countUnique(int []arr, int n)
    {
 
        HashSet s = new HashSet();
        for (int i = 0; i < n; i++)
        {
            s.Add(arr[i]);
        }
        int count = (int) Math.Pow(s.Count, 2);
 
        return count;
    }
 
    // Driver code
    static void Main()
    {
        int []arr = {1, 2, 2, 4, 2, 5, 3, 5};
        int n = arr.Length;
        Console.WriteLine(countUnique(arr, n));
    }
}
 
// This code has been contributed by mits

Javascript

 
// JavaScript Program to implement
// the above approach
 
 
    // Function to return the number
    // of unique pairs in the array
    function countUnique(arr, n){
 
        let s = new Set();
        for (let i = 0; i < n; i++)
        {
            s.add(arr[i]);
        }
        let count = Math.pow(s.size, 2);
 
        return count;
    }
// Driver Code
 
    let arr = [ 1, 2, 2, 4, 2, 5, 3, 5 ];
    let n = arr.length;
 
     document.write(countUnique(arr, n));
 

输出：

25

时间复杂度： O(n)

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