Largest Derangement of a Sequence

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 <bits/stdc++.h>
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<Integer> 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

<script>

// JavaScript program to find the largest derangement


function printLargest(seq, N) {
    let res = new Array(N); // Stores result

    // Insert all elements into a priority queue
    let pq = new Array();
    for (let i = 0; i < N; i++)
        pq.push(seq[i]);

    // Fill Up res[] from left to right
    for (let i = 0; i < N; i++) {
        pq.sort((a, b) => a - b);
        pq.reverse();
        let d = pq[0];
        pq.shift();
        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.shift();
            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];
    }
    document.write("Largest Derangement<br>");
    for (let i = 0; i < N; i++)
        document.write(res[i] + " ");
}

// Driver code
let seq = [92, 3, 52, 13, 2, 31, 1];
let n = seq.length;
printLargest(seq, n);

// This code is contributed by gfgking

</script>

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