# Find the compatibility difference between two arrays

• Difficulty Level : Hard
• Last Updated : 30 Apr, 2021

Suppose there are two friends and now they want to test their friendship that how much compatible they are. Given the numbers n numbered from 1 …n and they are asked to rank the numbers. The task is find the compatibility difference between them. Compatibility difference is the number of mis-matches in the relative ranking of the same movie given by them.

Examples :

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```Input : a1[] = {3, 1, 2, 4, 5}
a2[] = {3, 2, 4, 1, 5}
Output : 2
Explanation : Compatibility difference is two
because first ranks movie 1 before 2 and 4 but
other ranks it after.

Input : a1[] = {5, 3, 1, 2, 4}
a2[] = {3, 1, 2, 4, 5}
Output : 5
Total difference is four due to mis-match in
position of 5```

The idea is traverse both arrays.
1) If current elements are same, do nothing.
2) Find next position of a1[i] in a2[]. Let this position be j. One by one move a2[j] to a2[i] (Similar to bubble step of bubble sort)

Below is the implementation of above steps.

## C++

 `// C++ program to count of misplacements``#include ``using` `namespace` `std;``int` `findDifference(``int` `a1[], ``int` `a2[], ``int` `n)``{``    ``int` `res = 0;` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``// If elements at current position``        ``// are not same``        ``if` `(a1[i] != a2[i]) {` `            ``// Find position of a1[i] in a2[]``            ``int` `j = i + 1;``            ``while` `(a1[i] != a2[j])``                ``j++;``            ` `            ``// Insert the element a2[j] at``            ``// a2[i] by moving all intermediate``            ``// elements one position ahead.``            ``while` `(j != i) {``                ``swap(a2[j], a2[j - 1]);``                ``j--;``                ``res++;``            ``}``        ``}``    ``}``    ``return` `res;``}` `// Driver code``int` `main()``{``    ``int` `a1[] = { 3, 1, 2, 4, 5 };``    ``int` `a2[] = { 3, 2, 4, 1, 5 };``    ``int` `n = ``sizeof``(a1)/``sizeof``(a1);``    ``cout << findDifference(a1, a2, n);``    ``return` `0;``}`

## Java

 `// Java program to count of misplacements``public` `class` `Compatability_difference {` `    ``static` `int` `findDifference(``int` `a1[], ``int` `a2[], ``int` `n)``    ``{``        ``int` `res = ``0``;``     ` `        ``for` `(``int` `i = ``0``; i < n; i++) {``     ` `            ``// If elements at current position``            ``// are not same``            ``if` `(a1[i] != a2[i]) {``     ` `                ``// Find position of a1[i] in a2[]``                ``int` `j = i + ``1``;``                ``while` `(a1[i] != a2[j])``                    ``j++;``                 ` `                ``// Insert the element a2[j] at``                ``// a2[i] by moving all intermediate``                ``// elements one position ahead.``                ``while` `(j != i) {``                    ` `                    ``//swap``                    ``int` `temp = a2[j - ``1``];``                    ``a2[j - ``1``] = a2[j];``                    ``a2[j] = temp;``                    ``j--;``                    ``res++;``                ``}``            ``}``        ``}``        ``return` `res;``    ``}``     ` `    ``// Driver code``    ``public` `static` `void` `main(String args[])``    ``{``        ``int` `a1[] = { ``3``, ``1``, ``2``, ``4``, ``5` `};``        ``int` `a2[] = { ``3``, ``2``, ``4``, ``1``, ``5` `};``        ``int` `n = a1.length;``        ` `        ``System.out.println(findDifference(a1, a2, n));``    ``}``}``// This code is contributed by Sumit Ghosh`

## Python3

 `# Python3 program to count misplacements` `def` `findDifference(a1, a2, n):` `    ``res ``=` `0` `    ``for` `i ``in` `range``(``0``, n):` `        ``# If elements at current``        ``# position are not same``        ``if` `a1[i] !``=` `a2[i]:` `            ``# Find position of a1[i] in a2[]``            ``j ``=` `i ``+` `1``            ``while` `(a1[i] !``=` `a2[j]):``                ``j ``+``=` `1``                ``if` `i >``=` `n ``or` `j >``=` `n:``                    ``break``            ` `            ``# Insert the element a2[j] at``            ``# a2[i] by moving all intermediate``            ``# elements one position ahead.``            ``while` `(j !``=` `i):``                ``a2[j],a2[j``-``1``] ``=` `a2[j``-``1``],a2[j]``                ``res ``+``=` `1``                ``j ``-``=` `1``                ``if` `i >``=` `n ``or` `j >``=` `n:``                    ``break``        ` `    ``return` `res` `# Driver code``a1 ``=` `[ ``3``, ``1``, ``2``, ``4``, ``5` `]``a2 ``=` `[ ``3``, ``2``, ``4``, ``1``, ``5` `]``n ``=` `len``(a1)``print``(findDifference(a1, a2, n))` `# This code is contributed by Smitha Dinesh Semwal`

## C#

 `// C# program to count of misplacements``using` `System;` `public` `class` `Compatability_difference``{``    ``static` `int` `findDifference(``int` `[]a1, ``int` `[]a2, ``int` `n)``    ``{``        ``int` `res = 0;``    ` `        ``for` `(``int` `i = 0; i < n; i++) {``    ` `            ``// If elements at current``            ``// position are not same``            ``if` `(a1[i] != a2[i]) {``    ` `                ``// Find position of a1[i] in a2[]``                ``int` `j = i + 1;``                ``while` `(a1[i] != a2[j])``                    ``j++;``                ` `                ``// Insert the element a2[j] at``                ``// a2[i] by moving all intermediate``                ``// elements one position ahead.``                ``while` `(j != i) {``                    ` `                    ``//swap``                    ``int` `temp = a2[j - 1];``                    ``a2[j - 1] = a2[j];``                    ``a2[j] = temp;``                    ``j--;``                    ``res++;``                ``}``            ``}``        ``}``        ``return` `res;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int` `[]a1 = {3, 1, 2, 4, 5};``        ``int` `[]a2 = {3, 2, 4, 1, 5};``        ``int` `n = a1.Length;``        ` `        ``// Function calling``        ``Console.WriteLine(findDifference(a1, a2, n));``    ``}``}` `// This code is contributed by vt_m.`

## Javascript

 ``

Output:

`2`

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