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 :
- make a copy of both array.
- sort copy_A[] in increasing, copy_B[] in decreasing order.
- 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.
- Return Count Value. That will be our answer.
Implementation:
// 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 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. |
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# 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 |
// 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)); |
2
Time Complexity: O(n logn).
Auxiliary Space: O(n)