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
// 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-arrangementdef 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 functionA = [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-arrangementfunction 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 functionconst 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)
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