# De-arrangements for minimum product sum of two arrays

• Difficulty Level : Medium
• Last Updated : 12 Dec, 2017

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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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.
 // CPP program to count de-arrangements for // minimum product.#includeusing namespace std;  // function for finding de-arrangementint findDearrange (int A[], int B[], int n){    // create copy of array    vector copy_A (A, A+n);    vector 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());      // count no. of de arrangements    int count = 0;    for (int i=0; i::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 functionint 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;}

Output:

2

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