数组中唯一对的数量

给定N个元素的数组，任务是查找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 

天真的方法：简单的解决方案是遍历每个可能的对，然后将它们添加到集合中，然后找出集合的大小。

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

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 )。

高效的方法：首先找出数组中唯一元素的数量。令唯一元素的数量为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其他每个唯一元素(包括自身)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*/

Python3

# 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

输出：

25

时间复杂度： O(n)