Tutorialspoint.dev

Count Derangements (Permutation such that no element appears in its original position)

A Derangement is a permutation of n elements, such that no element appears in its original position. For example, a derangement of {0, 1, 2, 3} is {2, 3, 1, 0}.

Given a number n, find total number of Derangements of a set of n elements.

Examples :

Input: n = 2
Output: 1
For two elements say {0, 1}, there is only one 
possible derangement {1, 0}

Input: n = 3
Output: 2
For three elements say {0, 1, 2}, there are two 
possible derangements {2, 0, 1} and {1, 2, 0}

Input: n = 4
Output: 9
For four elements say {0, 1, 2, 3}, there are 9
possible derangements {1, 0, 3, 2} {1, 2, 3, 0}
{1, 3, 0, 2}, {2, 3, 0, 1}, {2, 0, 3, 1}, {2, 3,
1, 0}, {3, 0, 1, 2}, {3, 2, 0, 1} and {3, 2, 1, 0}



Let countDer(n) be count of derangements for n elements. Below is recursive relation for it.

countDer(n) = (n - 1) * [countDer(n - 1) + countDer(n - 2)]

How does above recursive relation work?
There are n – 1 ways for element 0 (this explains multiplication with n – 1).

Let 0 be placed at index i. There are now two possibilities, depending on whether or not element i is placed at 0 in return.

  1. i is placed at 0: This case is equivalent to solving the problem for n-2 elements as two elements have just swapped their positions.
  2. i is not placed at 0: This case is equivalent to solving the problem for n-1 elements as now there are n-1 elements, n-1 positions and every element has n-2 choices

Below is Simple Solution based on above recursive formula.

div class="responsive-tabs">

C++

// A Naive Recursive C++ program 
// to count derangements
#include <bits/stdc++.h>
using namespace std;
  
int countDer(int n)
{
// Base cases
if (n == 1) return 0;
if (n == 0) return 1;
if (n == 2) return 1;
  
// countDer(n) = (n-1)[countDer(n-1) + der(n-2)]
return (n - 1) * (countDer(n - 1) + 
                  countDer(n - 2));
}
  
// Driver Code
int main()
{
    int n = 4;
    cout << "Count of Derangements is " 
         << countDer(n);
    return 0;
}

Java

// A Naive Recursive java 
// program to count derangements
import java.io.*;
  
class GFG 
{
      
    // Function to count
    // derangements
    static int countDer(int n)
    {
        // Base cases
        if (n == 1) return 0;
        if (n == 0) return 1;
        if (n == 2) return 1;
          
        // countDer(n) = (n-1)[countDer(n-1) + der(n-2)]
        return (n - 1) * (countDer(n - 1) + 
                          countDer(n - 2));
    }
      
    // Driver Code
    public static void main (String[] args)
    {
        int n = 4;
        System.out.println( "Count of Derangements is "
                             +countDer(n));
  
    }
}
  
// This code is contributed by vt_m

Python3

# A Naive Recursive Python3 
# program to count derangements
  
def countDer(n):
      
    # Base cases
    if (n == 1): return 0
    if (n == 0): return 1
    if (n == 2): return 1
      
    # countDer(n) = (n-1)[countDer(n-1) + der(n-2)]
    return (n - 1) * (countDer(n - 1) + 
                      countDer(n - 2))
  
# Driver Code
n = 4
print("Count of Derangements is ", countDer(n))
  
  
# This code is contributed by Azkia Anam.

C#

// A Naive Recursive C#
// program to count derangements
using System;
  
class GFG 
{
      
    // Function to count
    // derangements
    static int countDer(int n)
    {
        // Base cases
        if (n == 1) return 0;
        if (n == 0) return 1;
        if (n == 2) return 1;
          
        // countDer(n) = (n-1)[countDer(n-1) + der(n-2)]
        return (n - 1) * (countDer(n - 1) + 
                          countDer(n - 2));
    }
      
    // Driver Code
    public static void Main ()
    {
        int n = 4;
        Console.Write( "Count of Derangements is " +
                        countDer(n));
  
    }
}
  
// This code is contributed by nitin mittal.

PHP

<?php
// A Naive Recursive PHP program 
// to count derangements
  
function countDer($n)
{
      
    // Base cases
    if ($n == 1) 
        return 0;
    if ($n == 0)
        return 1;
    if ($n == 2) 
        return 1;
      
    // countDer(n) = (n-1)[countDer(n-1) +
    // der(n-2)]
    return ($n - 1) * (countDer($n - 1) + 
                       countDer($n - 2));
}
  
    // Driver Code
    $n = 4;
    echo "Count of Derangements is ", countDer($n);
  
// This code is contributed by nitin mittal.
?>


Output:

Count of Derangements is 9

Time Complexity: T(n) = T(n-1) + T(n-2) which is exponential.
We can observe that this implementation does repeated work. For example see recursion tree for countDer(5), countDer(3) is being being evaluated twice.

cdr() ==> countDer()

                    cdr(5)   
                 /              
             cdr(4)          cdr(3)   
           /               /     
       cdr(3)     cdr(2)  cdr(2)   cdr(1)

An Efficient Solution is to use Dynamic Programming to store results of subproblems in an array and build the array in bottom up manner.

C++

// A Dynamic programming based C++ 
// program to count derangements
#include <bits/stdc++.h>
using namespace std;
  
int countDer(int n)
{
    // Create an array to store 
    // counts for subproblems
    int der[n + 1];
  
    // Base cases
    der[0] = 1;
    der[1] = 0;
    der[2] = 1;
  
    // Fill der[0..n] in bottom up manner 
    // using above recursive formula
    for (int i = 3; i <= n; ++i)
        der[i] = (i - 1) * (der[i - 1] + 
                            der[i - 2]);
  
    // Return result for n
    return der[n];
}
  
// Driver code
int main()
{
    int n = 4;
    cout << "Count of Derangements is " 
         << countDer(n);
    return 0;
}

Java

// A Dynamic programming based 
// java program to count derangements
import java.io.*;
  
class GFG 
{
      
    // Function to count
    // derangements 
    static int countDer(int n)
    {
        // Create an array to store
        // counts for subproblems
        int der[] = new int[n + 1];
      
        // Base cases
        der[0] = 1;
        der[1] = 0;
        der[2] = 1;
      
        // Fill der[0..n] in bottom up 
        // manner using above recursive
        // formula
        for (int i = 3; i <= n; ++i)
            der[i] = (i - 1) * (der[i - 1] + 
                                der[i - 2]);
      
        // Return result for n
        return der[n];
    }
      
    // Driver program
    public static void main (String[] args) 
    {
        int n = 4;
        System.out.println("Count of Derangements is "
                            countDer(n));
      
    }
}
  
// This code is contributed by vt_m

Python3

# A Dynamic programming based Python3
# program to count derangements
  
def countDer(n):
      
    # Create an array to store
    # counts for subproblems
    der = [0 for i in range(n + 1)]
      
    # Base cases
    der[0] = 1
    der[1] = 0
    der[2] = 1
      
    # Fill der[0..n] in bottom up manner 
    # using above recursive formula
    for i in range(3, n + 1):
        der[i] = (i - 1) * (der[i - 1] + 
                            der[i - 2])
          
    # Return result for n
    return der[n]
  
# Driver Code
n = 4
print("Count of Derangements is ", countDer(n))
  
# This code is contributed by Azkia Anam.

C#

   
// A Dynamic programming based 
// C# program to count derangements
using System;
  
class GFG 
{
      
    // Function to count
    // derangements 
    static int countDer(int n)
    {
        // Create an array to store
        // counts for subproblems
        int []der = new int[n + 1];
      
        // Base cases
        der[0] = 1;
        der[1] = 0;
        der[2] = 1;
      
        // Fill der[0..n] in bottom up 
        // manner using above recursive
        // formula
        for (int i = 3; i <= n; ++i)
            der[i] = (i - 1) * (der[i - 1] + 
                                der[i - 2]);
      
        // Return result for n
        return der[n];
    }
      
    // Driver code
    public static void Main () 
    {
        int n = 4;
        Console.Write("Count of Derangements is "
                       countDer(n));
      
    }
}
  
// This code is contributed by nitin mittal

PHP

<?php
// A Dynamic programming based PHP
// program to count derangements
  
function countDer($n)
{
    // Create an array to store 
    // counts for subproblems
  
    // Base cases
    $der[0] = 1;
    $der[1] = 0;
    $der[2] = 1;
  
    // Fill der[0..n] in bottom up manner 
    // using above recursive formula
    for ($i = 3; $i <= $n; ++$i)
        $der[$i] = ($i - 1) * ($der[$i - 1] + 
                               $der[$i - 2]);
  
    // Return result for n
    return $der[$n];
}
  
// Driver code
$n = 4;
echo "Count of Derangements is ",
                    countDer($n);
  
// This code is contributed by aj_36
?>



Output :

Count of Derangements is 9

Time Complexity : O(n)
Auxiliary Space : O(n)

Thanks to Utkarsh Trivedi for suggesting above solution.

References:
https://en.wikipedia.org/wiki/Derangement

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above



This article is attributed to GeeksforGeeks.org

You Might Also Like

leave a comment

code

0 Comments

load comments

Subscribe to Our Newsletter