Count pairs (p, q) such that p occurs in array at least q times and q occurs at least p times

Given an array arr[], the task is to count unordered pairs of positive integers (p, q) such that p occurs in the array at least q times and q occurs at least p times.

Examples:

Input: arr[] = {1, 2, 3, 4, 5} 
Output: 1 
(1, 1) is the only valid pair.

Input: arr[] = {3, 3, 2, 2, 2} 
Output: 2 
(2, 3) and (2, 2) are the only possible pairs.  

Approach:

1. The idea is to hash every element of an array with its count and create a vector of unique elements from the array elements. Initialize the number of pairs to 0.
2. Loop through the vector of unique elements to count the number of pairs.
3. Inside the loop, if the count of the element is less than the element itself, then continue, else if the count of the element is equal to the element, then increment the number of pairs by 1 (Pair of the form (x, x)), else go to Step 4.
4. Increment the number of pairs by 1 and loop from j = (element + 1) to the count of the element updating j by 1 if the count of j is greater than or equal to an element, then increment the number of pairs by 1 as the pairs are unordered.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the count of required pairs
int get_unordered_pairs(int a[], int n)
{
    // To store unique elements
    vector<int> vs;

    // To hash elements with their frequency
    unordered_map<int, int> m;

    // Store frequencies in m and all distinct
    // items in vs
    for (int i = 0; i < n; i++) {
        m[a[i]]++;
        if (m[a[i]] == 1)
            vs.push_back(a[i]);
    }

    // Traverse through distinct elements
    int number_of_pairs = 0;
    for (int i = 0; i < vs.size(); i++) {

        // If current element is greater than
        // its frequency in the array
        if (m[vs[i]] < vs[i])
            continue;

        // If element is equal to its frequency,
        // a pair of the form (x, x) is formed.
        else if (m[vs[i]] == vs[i])
            number_of_pairs += 1;

        // If element is less than its frequency
        else {
            number_of_pairs += 1;
            for (int j = vs[i] + 1; j <= m[vs[i]]; j++) {
                if (m[j] >= vs[i])
                    number_of_pairs += 1;
            }
        }
    }
    return number_of_pairs;
}

// Driver code
int main()
{
    int arr[] = { 3, 3, 2, 2, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << get_unordered_pairs(arr, n);
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;

class GFG
{

// Function to return the count of required pairs
static int get_unordered_pairs(int []a, int n)
{
    // To store unique elements
    ArrayList<Integer> vs = new ArrayList<Integer>();

    // To hash elements with their frequency
    int[] m = new int[maximum(a)+1];

    // Store frequencies in m and all distinct
    // items in vs
    for (int i = 0; i < n; i++)
    {
        m[(int)a[i]]++;
        if (m[a[i]] == 1)
            vs.add(a[i]);
    }

    // Traverse through distinct elements
    int number_of_pairs = 0;
    for (int i = 0; i < vs.size(); i++)
    {

        // If current element is greater than
        // its frequency in the array
        if (m[(int)vs.get(i)] < (int)vs.get(i))
            continue;

        // If element is equal to its frequency,
        // a pair of the form (x, x) is formed.
        else if (m[(int)vs.get(i)] == (int)vs.get(i))
            number_of_pairs += 1;

        // If element is less than its frequency
        else
        {
            number_of_pairs += 1;
            for (int j = (int)vs.get(i) + 1; j <= m[(int)vs.get(i)]; j++)
            {
                if (m[j] >= (int)vs.get(i))
                    number_of_pairs += 1;
            }
        }
    }
    return number_of_pairs;
}
static int maximum(int []arr)
{
    int max = Integer.MIN_VALUE;
    for(int i = 0; i < arr.length; i++)
    {
        if(arr[i] > max)
        {
            max = arr[i];
        }
    }
    return max;
}

// Driver code
public static void main (String[] args)
{

    int []arr = { 3, 3, 2, 2, 2 };
    int n = arr.length;
    System.out.println(get_unordered_pairs(arr, n));
}
}

// This code is contributed by mits

Python3

# Python3 implementation of the approach
from collections import defaultdict

# Function to return the count of
# required pairs
def get_unordered_pairs(a, n):

    # To store unique elements
    vs = []

    # To hash elements with their frequency
    m = defaultdict(lambda:0)

    # Store frequencies in m and
    # all distinct items in vs
    for i in range(0, n):
        m[a[i]] += 1
        if m[a[i]] == 1:
            vs.append(a[i])

    # Traverse through distinct elements
    number_of_pairs = 0
    for i in range(0, len(vs)):

        # If current element is greater
        # than its frequency in the array
        if m[vs[i]] < vs[i]:
            continue

        # If element is equal to its frequency,
        # a pair of the form (x, x) is formed.
        elif m[vs[i]] == vs[i]:
            number_of_pairs += 1

        # If element is less than its
        # frequency
        else:
            number_of_pairs += 1
            for j in range(vs[i] + 1, m[vs[i]] + 1):
                if m[j] >= vs[i]:
                    number_of_pairs += 1

    return number_of_pairs

# Driver code
if __name__ == "__main__":

    arr = [3, 3, 2, 2, 2]
    n = len(arr)
    print(get_unordered_pairs(arr, n))

# This code is contributed
# by Rituraj Jain

C#

// C# implementation of the approach
using System;
using System.Collections;
using System.Linq;

class GFG
{
// Function to return the count of required pairs
static int get_unordered_pairs(int []a, int n)
{
    // To store unique elements
    ArrayList vs = new ArrayList();

    // To hash elements with their frequency
    int[] m = new int[a.Max()+1];

    // Store frequencies in m and all distinct
    // items in vs
    for (int i = 0; i < n; i++)
    {
        m[(int)a[i]]++;
        if (m[a[i]] == 1)
            vs.Add(a[i]);
    }

    // Traverse through distinct elements
    int number_of_pairs = 0;
    for (int i = 0; i < vs.Count; i++)
    {

        // If current element is greater than
        // its frequency in the array
        if (m[(int)vs[i]] < (int)vs[i])
            continue;

        // If element is equal to its frequency,
        // a pair of the form (x, x) is formed.
        else if (m[(int)vs[i]] == (int)vs[i])
            number_of_pairs += 1;

        // If element is less than its frequency
        else
        {
            number_of_pairs += 1;
            for (int j = (int)vs[i] + 1; j <= m[(int)vs[i]]; j++)
            {
                if (m[j] >= (int)vs[i])
                    number_of_pairs += 1;
            }
        }
    }
    return number_of_pairs;
}

// Driver code
static void Main()
{
    int []arr = { 3, 3, 2, 2, 2 };
    int n = arr.Length;
    Console.WriteLine(get_unordered_pairs(arr, n));
}
}

// This code is contributed by mits

PHP

<?php
// PHP implementation of the approach

// Function to return the count of
// required pairs
function get_unordered_pairs($a, $n)
{
    // To store unique elements
    $vs = array();

    // To hash elements with
    // their frequency
    $m = array();

    for($i = 0; $i < $n; $i++)
        $m[$a[$i]] = 0 ;


    // Store frequencies in m and
    // all distinct items in vs
    for ($i = 0; $i < $n; $i++)
    {
        $m[$a[$i]]++;
        if ($m[$a[$i]] == 1)
            array_push($vs, $a[$i]);
    }

    // Traverse through distinct elements
    $number_of_pairs = 0;
    for ($i = 0; $i < sizeof($vs); $i++)
    {

        // If current element is greater
        // than its frequency in the array
        if ($m[$vs[$i]] < $vs[$i])
            continue;

        // If element is equal to its frequency,
        // a pair of the form (x, x) is formed.
        else if ($m[$vs[$i]] == $vs[$i])
            $number_of_pairs += 1;

        // If element is less than its
        // frequency
        else
        {
            $number_of_pairs += 1;
            for ($j = $vs[$i] + 1;
                 $j <= $m[$vs[$i]]; $j++)
            {
                if ($m[$j] >= $vs[$i])
                    $number_of_pairs += 1;
            }
        }
    }
    return $number_of_pairs;
}

// Driver code
$arr = array(3, 3, 2, 2, 2);
$n = sizeof($arr);
echo get_unordered_pairs($arr, $n);

// This code is contributed by Ryuga
?>

Javascript

<script>

// Javascript implementation of the approach

// Function to return the count of required pairs
function get_unordered_pairs(a, n)
{
    // To store unique elements
    let vs = [];

    // To hash elements with their frequency
    let m = Array.from({length: Math.max(...a)+1}, (_, i) => 0);

    // Store frequencies in m and all distinct
    // items in vs
    for (let i = 0; i < n; i++)
    {
        m[a[i]]++;
        if (m[a[i]] == 1)
            vs.push(a[i]);
    }

    // Traverse through distinct elements
    let number_of_pairs = 0;
    for (let i = 0; i < vs.length; i++)
    {

        // If current element is greater than
        // its frequency in the array
        if (m[vs[i]] < vs[i])
            continue;

        // If element is equal to its frequency,
        // a pair of the form (x, x) is formed.
        else if (m[vs[i]] == vs[i])
            number_of_pairs += 1;

        // If element is less than its frequency
        else
        {
            number_of_pairs += 1;
            for (let j = vs[i] + 1; j <= m[vs[i]]; j++)
            {
                if (m[j] >= vs[i])
                    number_of_pairs += 1;
            }
        }
    }
    return number_of_pairs;
}

    // Driver code

    let arr = [ 3, 3, 2, 2, 2 ];
    let n = arr.length;
    document.write(get_unordered_pairs(arr, n));

</script>

Output:

2

Time Complexity: O(N* log(N))

Auxiliary Space: O(N)