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Count of distinct possible pairs such that the element from A is greater than the element from B
• Difficulty Level : Medium
• Last Updated : 05 Jun, 2020

Given two given arrays A and B of equal length, the task is to find the maximum number of distinct pairs of elements that can be chosen such that the element from A is strictly greater than the element from B.

Examples:

Input:
A[]={20, 30, 50} , B[]={25, 60, 40}
Output: 2
Explanation:
(30, 25) and (50, 40) are the two possible pairs.

Input:
A[]={20, 25, 60} , B[]={25, 60, 40}
Output: 1
Explanation:
(60, 25) or (60, 40) can be the required pair.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: In order to solve this problem we need to adopt the following approach:

• Sort both the arrays.
• Traverse the entire length of array A and check if the current element in A is greater than the element in B currently pointed at B.
• If so, point to the next element in both the arrays. Otherwise, move to the next element in A and check if it is greater than the element currently pointed at B.
• The number of elements traversed in array B after the entire traversal of array A gives the required answer.

Illustration:
Let us consider the two following arrays:
A[] = { 30, 28, 45, 22 } , B[] = { 35, 25, 22, 48 }
After sorting, the arrays appear to be
A[] = { 22, 28, 30, 45 } , B[] = { 22, 25, 35, 48}

After the first iteration, since A is not greater than B, we move to A.
After the second iteration, we move to B as A is greater than B.
After the third iteration, we move to B as A is greater than B.
Similarly, A is greater than B and we move to B.
Hence the number of elements traversed in B,i.e. 3 is the final answer.
The possible pairs are (28,22), (30,25) are (45, 35).

Below is the implementation of the above approach:

## C++

 `// C++ Program to count number of distinct ``// pairs possible from the two arrays``// such that element selected from one array is ``// always greater than the one selected from ``// the other array`` ` `#include ``using` `namespace` `std;`` ` `// Function to return the count ``// of pairs``int` `countPairs(vector<``int``> A,``                    ``vector<``int``> B)``{``    ``int` `n = A.size();``     ` `    ``sort(A.begin(),A.end());``    ``sort(B.begin(),B.end());``     ` `    ``int` `ans = 0, i;``    ``for` `(``int` `i = 0; i < n; i++) {``        ``if` `(A[i] > B[ans]) {``            ``ans++;``        ``}``    ``}``     ` `    ``return` `ans;``}`` ` `// Driver code``int` `main()``{``    ``vector<``int``> A = { 30, 28, 45, 22 };``    ``vector<``int``> B = { 35, 25, 22, 48 };``     ` `    ``cout << countPairs(A,B);``     ` `    ``return` `0;``}`

## Java

 `// Java program to count number of distinct ``// pairs possible from the two arrays such ``// that element selected from one array is ``// always greater than the one selected from ``// the other array``import` `java.util.*;`` ` `class` `GFG{`` ` `// Function to return the count ``// of pairs``static` `int` `countPairs(``int` `[] A,``                      ``int` `[] B)``{``    ``int` `n = A.length;``    ``int` `ans = ``0``;``     ` `    ``Arrays.sort(A);``    ``Arrays.sort(B);``     ` `    ``for``(``int` `i = ``0``; i < n; i++)``    ``{``       ``if` `(A[i] > B[ans])``       ``{``           ``ans++;``       ``}``    ``}``    ``return` `ans;``}`` ` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `[] A = { ``30``, ``28``, ``45``, ``22` `};``    ``int` `[] B = { ``35``, ``25``, ``22``, ``48` `};``     ` `    ``System.out.print(countPairs(A, B));``}``}`` ` `// This code is contributed by sapnasingh4991`

## Python3

 `# Python3 program to count number of distinct``# pairs possible from the two arrays``# such that element selected from one array is``# always greater than the one selected from``# the other array`` ` `# Function to return the count``# of pairs``def` `countPairs(A, B):`` ` `    ``n ``=` `len``(A)`` ` `    ``A.sort()``    ``B.sort()`` ` `    ``ans ``=` `0``    ``for` `i ``in` `range``(n):``        ``if``(A[i] > B[ans]):``            ``ans ``+``=` `1`` ` `    ``return` `ans`` ` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``A ``=` `[``30``, ``28``, ``45``, ``22``]``    ``B ``=` `[``35``, ``25``, ``22``, ``48``]`` ` `    ``print``(countPairs(A, B))`` ` `# This code is contributed by Shivam Singh`

## C#

 `// C# program to count number of distinct ``// pairs possible from the two arrays such ``// that element selected from one array is ``// always greater than the one selected from ``// the other array``using` `System;``class` `GFG{`` ` `// Function to return the count ``// of pairs``static` `int` `countPairs(``int` `[] A,``                      ``int` `[] B)``{``    ``int` `n = A.Length;``    ``int` `ans = 0;``     ` `    ``Array.Sort(A);``    ``Array.Sort(B);``     ` `    ``for``(``int` `i = 0; i < n; i++)``    ``{``        ``if` `(A[i] > B[ans])``        ``{``            ``ans++;``        ``}``    ``}``    ``return` `ans;``}`` ` `// Driver code``public` `static` `void` `Main()``{``    ``int` `[]A = { 30, 28, 45, 22 };``    ``int` `[]B = { 35, 25, 22, 48 };``     ` `    ``Console.Write(countPairs(A, B));``}``}`` ` `// This code is contributed by Code_Mech`
Output:
```3
```

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