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}

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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.

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.
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.

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

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Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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Java

Python3

C#

PHP

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