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Find the compatibility difference between two arrays
  • Difficulty Level : Hard
  • Last Updated : 16 Apr, 2018

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 :

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

Asked in Walmart Labs

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 <bits/stdc++.h>
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[0]);
    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.



Output:

2

This article is contributed by Rakesh Kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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