# 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) is the only valid pair.

Input: arr[] = {3, 3, 2, 2, 2}
Output:
(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 ``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 vs = ``new` `ArrayList();` `    ``// 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

 `= ``\$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

 ``

Output:
`2`

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

Auxiliary Space: O(N)

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