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Largest Derangement of a Sequence
  • Difficulty Level : Medium
  • Last Updated : 22 Jun, 2021

Given any sequence S = \{a_1, a_2 \dots a_n \}              , find the largest derangement of S              .
A derangement D              is any permutation of S              , such that no two elements at the same position in S              and D              are equal. 
The Largest Derangement is such that D_i > D_{i+1}, \forall i              .

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 largest derangement, we start putting larger elements in more significant positions.
Start from left, at any position i              place the next largest element among the values of the sequence which have not yet been placed in positions before i              .
To scan all positions takes N iteration. In each iteration we are required to find a maximum numbers, so a trivial implementation would be O(N^2)              complexity,
However if we use a data structure like max-heap to find the maximum element, then the complexity reduces to O(N * log{N})
 

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 poped 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 poped 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 poped 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 poped 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 poped 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)

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

This article is contributed by Sayan Mahapatra. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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