Related Articles

# Permutation of an array that has smaller values from another array

• Difficulty Level : Medium
• Last Updated : 25 Mar, 2021

Given two arrays A and B of equal size. The task is to print any permutation of array A such that the number of indices i for which A[i] > B[i] is maximized.

Examples:

```Input: A = [12, 24, 8, 32],
B = [13, 25, 32, 11]
Output: 24 32 8 12

Input: A = [2, 7, 11, 15],
B = [1, 10, 4, 11]
Output: 2 11 7 15```

If the smallest element in A beats the smallest element in B, we should pair them. Otherwise, it is useless for our score, as it can’t beat any other element of B.
With above strategy we make two vector of pairs, Ap for A and Bp for B with their element and respective index. Then sort both vectors and simulate them. Whenever we found any element in vector Ap such that Ap[i].first > Bp[j].first for some (i, j) we pair them i:e we update our answer array to ans[Bp[j].second] = Ap[i].first. However if Ap[i].first < Bp[j].first for some (i, j) then we store them in vector remain and finally pair them with any one.

Below is the implementation of above approach:

## C++

 `// C++ program to find permutation of an array that``// has smaller values from another array``#include ``using` `namespace` `std;` `// Function to print required permutation``void` `anyPermutation(``int` `A[], ``int` `B[], ``int` `n)``{``    ``// Storing elements and indexes``    ``vector > Ap, Bp;``    ``for` `(``int` `i = 0; i < n; i++)``        ``Ap.push_back(make_pair(A[i], i));``    ``for` `(``int` `i = 0; i < n; i++)``        ``Bp.push_back(make_pair(B[i], i));` `    ``sort(Ap.begin(), Ap.end());``    ``sort(Bp.begin(), Bp.end());` `    ``int` `i = 0, j = 0, ans[n] = { 0 };` `    ``// Filling the answer array``    ``vector<``int``> remain;``    ``while` `(i < n && j < n) {` `        ``// pair element of A and B``        ``if` `(Ap[i].first > Bp[j].first) {``            ``ans[Bp[j].second] = Ap[i].first;``            ``i++;``            ``j++;``        ``}``        ``else` `{``            ``remain.push_back(i);``            ``i++;``        ``}``    ``}` `    ``// Fill the remaining elements of answer``    ``j = 0;``    ``for` `(``int` `i = 0; i < n; ++i)``        ``if` `(ans[i] == 0) {``            ``ans[i] = Ap[remain[j]].first;``            ``j++;``        ``}` `    ``// Output required permutation``    ``for` `(``int` `i = 0; i < n; ++i)``        ``cout << ans[i] << ``" "``;``}` `// Driver program``int` `main()``{``    ``int` `A[] = { 12, 24, 8, 32 };``    ``int` `B[] = { 13, 25, 32, 11 };``    ``int` `n = ``sizeof``(A) / ``sizeof``(A);``    ``anyPermutation(A, B, n);``    ``return` `0;``}` `// This code is written by Sanjit_Prasad`

## Java

 `// Java program to find permutation of an``// array that has smaller values from``// another array``import` `java.io.*;``import` `java.lang.*;``import` `java.util.*;` `class` `GFG{` `// Function to print required permutation``static` `void` `anyPermutation(``int` `A[], ``int` `B[], ``int` `n)``{``    ` `    ``// Storing elements and indexes``    ``ArrayList<``int``[]> Ap = ``new` `ArrayList<>();``    ``ArrayList<``int``[]> Bp = ``new` `ArrayList<>();` `    ``for``(``int` `i = ``0``; i < n; i++)``        ``Ap.add(``new` `int``[] { A[i], i });` `    ``for``(``int` `i = ``0``; i < n; i++)``        ``Bp.add(``new` `int``[] { B[i], i });` `    ``// Sorting the Both Ap and Bp``    ``Collections.sort(Ap, (x, y) -> {``        ``if` `(x[``0``] != y[``0``])``            ``return` `x[``0``] - y[``0``];``            ` `        ``return` `y[``1``] - y[``1``];``    ``});` `    ``Collections.sort(Bp, (x, y) -> {``        ``if` `(x[``0``] != y[``0``])``            ``return` `x[``0``] - y[``0``];``            ` `        ``return` `y[``1``] - y[``1``];``    ``});` `    ``int` `i = ``0``, j = ``0``;``    ``int` `ans[] = ``new` `int``[n];` `    ``// Filling the answer array``    ``ArrayList remain = ``new` `ArrayList<>();``    ``while` `(i < n && j < n)``    ``{``        ` `        ``// Pair element of A and B``        ``if` `(Ap.get(i)[``0``] > Bp.get(j)[``0``])``        ``{``            ``ans[Bp.get(j)[``1``]] = Ap.get(i)[``0``];``            ``i++;``            ``j++;``        ``}``        ``else``        ``{``            ``remain.add(i);``            ``i++;``        ``}``    ``}` `    ``// Fill the remaining elements of answer``    ``j = ``0``;``    ``for``(i = ``0``; i < n; ++i)``        ``if` `(ans[i] == ``0``)``        ``{``            ``ans[i] = Ap.get(remain.get(j))[``0``];``            ``j++;``        ``}` `    ``// Output required permutation``    ``for``(i = ``0``; i < n; ++i)``        ``System.out.print(ans[i] + ``" "``);``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `A[] = { ``12``, ``24``, ``8``, ``32` `};``    ``int` `B[] = { ``13``, ``25``, ``32``, ``11` `};``    ``int` `n = A.length;` `    ``anyPermutation(A, B, n);``}``}` `// This code is contributed by Kingash`

## Python3

 `# Python3 program to find permutation of``# an array that has smaller values from``# another array` `# Function to print required permutation``def` `anyPermutation(A, B, n):` `    ``# Storing elements and indexes``    ``Ap, Bp ``=` `[], []``    ` `    ``for` `i ``in` `range``(``0``, n):``        ``Ap.append([A[i], i])``    ``for` `i ``in` `range``(``0``, n):``        ``Bp.append([B[i], i])``    ` `    ``Ap.sort()``    ``Bp.sort()``    ` `    ``i, j ``=` `0``, ``0``,``    ``ans ``=` `[``0``] ``*` `n` `    ``# Filling the answer array``    ``remain ``=` `[]``    ``while` `i < n ``and` `j < n:` `        ``# pair element of A and B``        ``if` `Ap[i][``0``] > Bp[j][``0``]:``            ``ans[Bp[j][``1``]] ``=` `Ap[i][``0``]``            ``i ``+``=` `1``            ``j ``+``=` `1``        ` `        ``else``:``            ``remain.append(i)``            ``i ``+``=` `1``        ` `    ``# Fill the remaining elements``    ``# of answer``    ``j ``=` `0``    ``for` `i ``in` `range``(``0``, n):``        ``if` `ans[i] ``=``=` `0``:``            ``ans[i] ``=` `Ap[remain[j]][``0``]``            ``j ``+``=` `1``        ` `    ``# Output required permutation``    ``for` `i ``in` `range``(``0``, n):``        ``print``(ans[i], end ``=` `" "``)` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``    ` `    ``A ``=` `[ ``12``, ``24``, ``8``, ``32` `]``    ``B ``=` `[ ``13``, ``25``, ``32``, ``11` `]``    ``n ``=` `len``(A)``    ``anyPermutation(A, B, n)``    ` `# This code is contributed``# by Rituraj Jain`
Output:

`24 32 8 12`

Time Complexity: O(N*log(N)), where N is the length of array.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up