De-arrangements for minimum product sum of two arrays

Given two arrays A[] and B[] of same size n. We need to first permute any of arrays such that the sum of product of pairs( 1 element from each) is minimum. That is SUM ( Ai*Bi) for all i is minimum. We also need to count number of de-arrangements present in original array as compared to permuted array.

Examples:

Input : A[] = {4, 3, 2},  
        B[] = {7, 12, 5}
Output : 3
Explanation : A[] = {4, 3, 2} and B[] = {5, 7, 12}
results in minimum product sum. B[] = {7, 12, 5} 
is 3-position different from new B[].

Input : A[] = {4, 3, 2},  
        B[] = { 1, 2, 3}
Output : 0
Explanation : A[] = {4, 3, 2} and B[] = {1, 2, 3}
results in minimum product sum. B[] = {1, 2, 3} 
is exactly same as new one.

Idea behind finding the minimum sum of product from two array is to sort both array one in increasing and other in decreasing manner. These type of arrays will always produce minimum sum of pair product. Sorting both array will give the pair value i.e. which element from A is paired to which element from B[]. After that count the de-arrangement from original arrays.

Algorithm :

1. make a copy of both array.
2. sort copy_A[] in increasing, copy_B[] in decreasing order.
3. Iterate for all Ai, find Ai in copy_A[] as copy_A[j] and check whether copy_B[j] == B[i] or not. Increment count if not equal.
4. Return Count Value. That will be our answer.

Implementation:

CPP

// CPP program to count de-arrangements for
// minimum product.
#include<bits/stdc++.h>
using namespace std;

// function for finding de-arrangement
int findDearrange (int A[], int B[], int n)
{
 // create copy of array
 vector <int> copy_A (A, A+n);
 vector <int> copy_B (B, B+n);

 // sort array in inc & dec way
 sort(copy_A.begin(), copy_A.end());
 sort(copy_B.begin(), copy_B.end(),greater<int>());

 // count no. of de arrangements
 int count = 0;
 for (int i=0; i<n;i++)
 {
  vector<int>::iterator itA;

  // find position of A[i] in sorted array
  itA = lower_bound(copy_A.begin(),
      copy_A.end(), A[i]);

  // check whether B[i] is same as required or not
  if (B[i] != copy_B[itA-copy_A.begin()])
   count++;
 }

 // return count
 return count;
}

// driver function
int main()
{
 int A[] = {1, 2, 3, 4};
 int B[] = {6, 3, 4, 5};
 int n = sizeof(A) /sizeof(A[0]);;
 cout << findDearrange(A,B,n);
 return 0;
}

Java

// Java program to count de-arrangements for
// minimum product.
import java.io.*;
import java.util.*;

public class GFG {

  // function for finding de-arrangement
  static Integer findDearrange (Integer A[], Integer B[], Integer n)
  {

    // create copy of array
    Integer copy_A[]=A.clone();
    Integer copy_B[]=B.clone();

    // sort array in inc & dec way
    Arrays.sort(copy_A);
    Arrays.sort(copy_B, Collections.reverseOrder());

    // count no. of de arrangements
    Integer count = 0;
    for (Integer i=0; i<n;i++)
    {
      Integer itA;

      // find position of A[i] in sorted array
      itA = Arrays.binarySearch(copy_A, A[i]);

      // check whether B[i] is same as required or not
      if (B[i] != copy_B[itA])
        count++;
    }

    // return count
    return count;
  }

  // driver function
  public static void main (String[] args)
  {
    Integer A[] = {1, 2, 3, 4};
    Integer B[] = {6, 3, 4, 5};
    Integer n = A.length;
    System.out.println(findDearrange(A,B,n));
  }
}

// This code is contributed by Pushpesh Raj.

Python3

import copy

# function for finding de-arrangement
def findDearrange(A, B, n):

    # create copy of array
    copy_A = copy.deepcopy(A)
    copy_B = copy.deepcopy(B)

    # sort array in inc & dec way
    copy_A.sort()
    copy_B.sort(reverse=True)

    # count no. of de arrangements
    count = 0
    for i in range(n):
        itA = None

        # find position of A[i] in sorted array
        itA = copy_A.index(A[i])

        # check whether B[i] is same as required or not
        if B[i] != copy_B[itA]:
            count += 1

    # return count
    return count

# driver function
A = [1, 2, 3, 4]
B = [6, 3, 4, 5]
n = len(A)
print(findDearrange(A, B, n))

C#

// C# program to count de-arrangements for
// minimum product.
using System;

class GFG
{
    // function for finding de-arrangement
    static int FindDearrange(int[] A, int[] B, int n)
    {
        // create copy of array
        int[] copy_A = (int[])A.Clone();
        int[] copy_B = (int[])B.Clone();

        // sort array in inc & dec way
        Array.Sort(copy_A);
        Array.Sort(copy_B, new Comparison<int>((a, b) => b.CompareTo(a)));

        // count no. of de arrangements
        int count = 0;
        for (int i = 0; i < n; i++)
        {
            // find position of A[i] in sorted array
            int itA = Array.BinarySearch(copy_A, A[i]);

            // check whether B[i] is same as required or not
            if (B[i] != copy_B[itA])
                count++;
        }

        // return count
        return count;
    }

    // Driver function
    static void Main(string[] args)
    {
        int[] A = { 1, 2, 3, 4 };
        int[] B = { 6, 3, 4, 5 };
        int n = A.Length;
        Console.WriteLine(FindDearrange(A, B, n));
    }
}

// This code is contributed by Aman Kumar

Javascript

// function for finding de-arrangement
function findDearrange(A, B, n) {
    // create copy of array
    const copy_A = [...A];
    const copy_B = [...B];

    // sort array in inc & dec way
    copy_A.sort((a, b) => a - b);
    copy_B.sort((a, b) => b - a);

    // count no. of de arrangements
    let count = 0;
    for (let i = 0; i < n; i++) {
        let itA = null;

        // find position of A[i] in sorted array
        itA = copy_A.indexOf(A[i]);

        // check whether B[i] is same as required or not
        if (B[i] !== copy_B[itA]) {
            count += 1;
        }
    }

    // return count
    return count;
}

// driver function
const A = [1, 2, 3, 4];
const B = [6, 3, 4, 5];
const n = A.length;
console.log(findDearrange(A, B, n));

Output

2

Time Complexity: O(n logn).
Auxiliary Space: O(n)