# Largest Derangement of a Sequence

Last Updated : 22 Dec, 2022

Given any sequence, find the largest derangement of .
A derangement is any permutation of, such that no two elements at the same position in and are equal.
The Largest Derangement is such that.

Examples:

Input : seq[] = {5, 4, 3, 2, 1}
Output : 4 5 2 1 3

Input : seq[] = {56, 21, 42, 67, 23, 74}
Output : 74, 67, 56, 42, 21, 23

Since we are interested in generating the largest derangement, we start putting larger elements in more significant positions.
Start from left, at any position place the next largest element among the values of the sequence which have not yet been placed in positions before.
To scan all positions takes N iteration. In each iteration we are required to find a maximum number, so a trivial implementation would be complexity,
However, if we use a data structure like max-heap to find the maximum element, then the complexity reduces to

Below is the implementation.

## C++

 `// C++ program to find the largest derangement``#include ``using` `namespace` `std;` `void` `printLargest(``int` `seq[], ``int` `N)``{``    ``int` `res[N]; ``// Stores result` `    ``// Insert all elements into a priority queue``    ``std::priority_queue<``int``> pq; ``    ``for` `(``int` `i = 0; i < N; i++) ``        ``pq.push(seq[i]);    ` `    ``// Fill Up res[] from left to right``    ``for` `(``int` `i = 0; i < N; i++) {``        ``int` `d = pq.top();``        ``pq.pop();``        ``if` `(d != seq[i] || i == N - 1) {``            ``res[i] = d;``        ``} ``else` `{` `            ``// New Element popped equals the element ``            ``// in original sequence. Get the next``            ``// largest element``            ``res[i] = pq.top();``            ``pq.pop();``            ``pq.push(d);``        ``}``    ``}` `    ``// If given sequence is in descending order then ``    ``// we need to swap last two elements again``    ``if` `(res[N - 1] == seq[N - 1]) {``        ``res[N - 1] = res[N - 2];``        ``res[N - 2] = seq[N - 1];``    ``}` `    ``printf``(``"\nLargest Derangement \n"``);``    ``for` `(``int` `i = 0; i < N; i++) ``        ``printf``(``"%d "``, res[i]);``}` `// Driver code``int` `main()``{``    ``int` `seq[] = { 92, 3, 52, 13, 2, 31, 1 }; ``    ``int` `n = ``sizeof``(seq)/``sizeof``(seq[0]);``    ``printLargest(seq, n);``    ``return` `0;``}`

## Java

 `// Java program to find the largest derangement``import` `java.io.*;``import` `java.util.Collections;``import` `java.util.PriorityQueue;` `class` `GFG{``    ` `public` `static` `void` `printLargest(``int` `a[],``int` `n)``{``     ``PriorityQueue pq = ``new` `PriorityQueue<>(``         ``Collections.reverseOrder());``      ` `      ``// Insert all elements into a priority queue``      ``for``(``int` `i = ``0``; i < n; i++)``    ``{``        ``pq.add(a[i]);``    ``}``    ` `    ``// Stores result``      ``int` `res[] = ``new` `int``[n]; ``      ` `      ``// Fill Up res[] from left to right``    ``for``(``int` `i = ``0``; i < n; i++)``    ``{``        ``int` `p = pq.peek();``        ``pq.remove();``        ` `        ``if` `(p != a[i] || i == n - ``1``)``        ``{``            ``res[i] = p;``        ``}``        ``else``        ``{``            ` `            ``// New Element popped equals the element ``            ``// in original sequence. Get the next``            ``// largest element``            ``res[i] = pq.peek();``            ``pq.remove();``            ``pq.add(p);``        ``}``    ``}``    ` `      ``// If given sequence is in descending ``      ``// order then we need to swap last two``      ``// elements again``      ``if` `(res[n - ``1``] == a[n - ``1``])``    ``{``        ``res[n - ``1``] = res[n - ``2``];``        ``res[n - ``2``] = a[n - ``1``];``    ``}``  ` `      ``System.out.println(``"Largest Derangement"``);``    ``for``(``int` `i = ``0``; i < n; i++)``    ``{``        ``System.out.print(res[i] + ``" "``);``    ``}``}` `// Driver code``public` `static` `void` `main(String[] args) ``{``      ``int` `n = ``7``;``    ``int` `seq[] = { ``92``, ``3``, ``52``, ``13``, ``2``, ``31``, ``1` `};``    ` `      ``printLargest(seq, n);``}``}` `// This code is contributed by aditya7409`

## Python3

 `# Python3 program to find the largest derangement``def` `printLargest(seq, N) :` `    ``res ``=` `[``0``]``*``N ``# Stores result``  ` `    ``# Insert all elements into a priority queue``    ``pq ``=` `[]``    ``for` `i ``in` `range``(N) :``        ``pq.append(seq[i])   ``  ` `    ``# Fill Up res[] from left to right``    ``for` `i ``in` `range``(N) :    ``        ``pq.sort()``        ``pq.reverse()``        ``d ``=` `pq[``0``]``        ``del` `pq[``0``]``        ``if` `(d !``=` `seq[i] ``or` `i ``=``=` `N ``-` `1``) :``            ``res[i] ``=` `d        ``        ``else` `:        ``  ` `            ``# New Element popped equals the element ``            ``# in original sequence. Get the next``            ``# largest element``            ``res[i] ``=` `pq[``0``]``            ``del` `pq[``0``]``            ``pq.append(d)``  ` `    ``# If given sequence is in descending order then ``    ``# we need to swap last two elements again``    ``if` `(res[N ``-` `1``] ``=``=` `seq[N ``-` `1``]) :    ``        ``res[N ``-` `1``] ``=` `res[N ``-` `2``]``        ``res[N ``-` `2``] ``=` `seq[N ``-` `1``]``         ` `    ``print``(``"Largest Derangement"``)``    ``for` `i ``in` `range``(N) :``        ``print``(res[i], end ``=` `" "``)` `# Driver code``seq ``=` `[ ``92``, ``3``, ``52``, ``13``, ``2``, ``31``, ``1` `] ``n ``=` `len``(seq)``printLargest(seq, n)` `# This code is contributed by divyesh072019.`

## C#

 `// C# program to find the largest derangement``using` `System;``using` `System.Collections.Generic;``class` `GFG ``{``    ` `    ``static` `void` `printLargest(``int``[] seq, ``int` `N)``    ``{``        ``int``[] res = ``new` `int``[N]; ``// Stores result``     ` `        ``// Insert all elements into a priority queue``        ``List<``int``> pq = ``new` `List<``int``>();``        ``for` `(``int` `i = 0; i < N; i++) ``            ``pq.Add(seq[i]);    ``     ` `        ``// Fill Up res[] from left to right``        ``for` `(``int` `i = 0; i < N; i++) ``        ``{``            ``pq.Sort();``            ``pq.Reverse();``            ``int` `d = pq[0];``            ``pq.RemoveAt(0);``            ``if` `(d != seq[i] || i == N - 1) ``            ``{``                ``res[i] = d;``            ``} ``          ``else``            ``{``     ` `                ``// New Element popped equals the element ``                ``// in original sequence. Get the next``                ``// largest element``                ``res[i] = pq[0];``                ``pq.RemoveAt(0);``                ``pq.Add(d);``            ``}``        ``}``     ` `        ``// If given sequence is in descending order then ``        ``// we need to swap last two elements again``        ``if` `(res[N - 1] == seq[N - 1]) ``        ``{``            ``res[N - 1] = res[N - 2];``            ``res[N - 2] = seq[N - 1];``        ``}     ``        ``Console.WriteLine(``"Largest Derangement"``);``        ``for` `(``int` `i = 0; i < N; i++) ``            ``Console.Write(res[i] + ``" "``);``    ``}` `  ``// Driver code``  ``static` `void` `Main()``  ``{``    ``int``[] seq = { 92, 3, 52, 13, 2, 31, 1 }; ``    ``int` `n = seq.Length;``    ``printLargest(seq, n);``  ``}``}` `// This code is contributed by divyeshrabadiya07`

## Javascript

 ``

Output
```Largest Derangement
52 92 31 3 13 1 2 ```

Time Complexity: O(n log n)
Auxiliary Space: O(N), because, we use an N size array to store results.

Note:

The method can be easily modified to obtain the smallest derangement as well.
Instead of a Max Heap, we should use a Min Heap to consecutively get minimum elements

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