# Count of distinct possible pairs such that the element from A is greater than the element from B

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:
Explanation:
(30, 25) and (50, 40) are the two possible pairs.

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

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[0] is not greater than B[0], we move to A[1].
After the second iteration, we move to B[1] as A[1] is greater than B[0].
After the third iteration, we move to B[2] as A[2] is greater than B[1].
Similarly, A[3] is greater than B[2] and we move to B[3].
Hence, the number of elements traversed in B, i.e. 3, is the final answer.
The possible pairs are (28,22), (30,25), (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`

## Javascript

 ``

Output:

`3`

Time Complexity: O(n * log n)

Auxiliary Space: O(1)

Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!

Previous
Next