计数紊乱（排列使得没有元素出现在其原始位置）

Derangement 是 n 个元素的排列，因此没有元素出现在其原始位置。例如，{0, 1, 2, 3} 的错位是 {2, 3, 1, 0}。
给定一个数 n，找出一组 n 个元素的乱序总数。

例子 ：

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}

令 countDer(n) 为 n 个元素的紊乱计数。下面是它的递归关系。

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

上述递归关系如何工作?

元素 0 有 n – 1 种方式(这解释了与 n – 1 的乘法)。
将0 置于索引 i 处。现在有两种可能性，这取决于元素 i 是否被放置在 0 作为回报。

1. i 被放置在 0：这种情况相当于解决 n-2 个元素的问题，因为两个元素刚刚交换了它们的位置。
2. i 未放置在 0：这种情况相当于解决 n-1 个元素的问题，因为现在有 n-1 个元素，n-1 个位置，每个元素都有 n-2 个选择

以下是基于上述递归公式的简单解决方案：

C++

// A Naive Recursive C++ program
// to count derangements
#include 
using namespace std;
 
int countDer(int n)
{
  // Base cases
  if (n == 1) return 0;
  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 == 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 == 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 == 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.

Javascript

 
    // A Naive Recursive Javascript
    // program to count derangements
     
    // Function to count
    // derangements
    function countDer(n)
    {
        // Base cases
        if (n == 1) return 0;
        if (n == 2) return 1;
          
        // countDer(n) = (n-1)[countDer(n-1) + der(n-2)]
        return (n - 1) * (countDer(n - 1) + countDer(n - 2));
    }
     
    let n = 4;
    document.write("Count of Derangements is " + countDer(n));
     

输出

Count of Derangements is 9

时间复杂度：T(n) = T(n-1) + T(n-2) 这是指数。

我们可以观察到这个实现确实重复工作。例如，请参阅 countDer(5) 的递归树，countDer(3) 被评估两次。

cdr() ==> countDer()

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

一个有效的解决方案是使用动态规划将子问题的结果存储在数组中，并以自底向上的方式构建数组。

C++

// A Dynamic programming based C++
// program to count derangements
#include 
using namespace std;
 
int countDer(int n)
{
    // Create an array to store
    // counts for subproblems
    int der[n + 1] = {0};
 
    // Base cases
    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[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

蟒蛇3

# 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[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[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

Javascript

 
    // A Dynamic programming based
    // javascript program to count
    // derangements
     
    // Function to count
    // derangements
    function countDer(n)
    {
        // Create an array to store
        // counts for subproblems
        let der = new Array(n + 1);
      
        // Base cases
        der[1] = 0;
        der[2] = 1;
      
        // Fill der[0..n] in bottom up
        // manner using above recursive
        // formula
        for (let i = 3; i <= n; ++i)
            der[i] = (i - 1) * (der[i - 1] +
                                der[i - 2]);
      
        // Return result for n
        return der[n];
    }
     
    let n = 4;
    document.write(
    "Count of Derangements is " + countDer(n)
    );
     

输出

Count of Derangements is 9

时间复杂度： O(n)
辅助空间： O(n)
感谢 Utkarsh Trivedi 提出上述解决方案。

不使用额外空间的更有效的解决方案。

因为我们只需要记住两个先前的值所以，不是将值存储在数组中，而是可以使用两个变量来仅存储所需的先前值。

下面是上述方法的实现：

C++

// C++ implementation of the above
// approach
 
#include 
using namespace std;
 
int countDer(int n)
{
 
    // base case
    if (n == 1 or n == 2) {
        return n - 1;
    }
 
    // Variable for just storing
    // previous values
    int a = 0;
    int b = 1;
 
    // using above recursive formula
    for (int i = 3; i <= n; ++i) {
        int cur = (i - 1) * (a + b);
        a = b;
        b = cur;
    }
 
    // Return result for n
    return b;
}
 
// Driver Code
int main()
{
 
    cout << "Count of Dearrangements is " << countDer(4);
    return 0;
}
 
 // Code contributed by skagnihotri

Java

// Java implementation of the
// above approach
 
import java.io.*;
 
class GFG {
   
    // Function to count derangements 
    static int countDer(int n) {
        // Base case
          if(n == 1 || n == 2) {
            return n-1;
        }
       
        // Variable for storing prev values
        int a = 0;
          int b = 1;
       
        // manner using above recursive formula
        for (int i = 3; i <= n; ++i) {
            int cur = (i-1)*(a+b);
            a = b;
              b = cur;
        }
             
        // Return result for n
        return b;
    }
       
    // Driver Code
    public static void main (String[] args) 
    {
        int n = 4;
        System.out.println("Count of Dearrangements is " + 
                            countDer(n));
       
    }
   
  // Code contributed by skagnihotri
}

Python

# Python program to count derangements
   
def countDer(n):
       
    # Base Case
    if n == 1 or n == 2:
      return n-1;
         
    # Variables for storing prevoius values
    a = 0
    b = 1
     
    # using above recursive formula
    for i in range(3, n + 1):
        cur = (i-1)*(a+b)
        a = b
        b = cur
         
    # Return result for n
    return b
   
# Driver Code
n = 4
print("Count of Dearrangements is ", countDer(n))
# Code contributed by skagnihotri

C＃

// C# implementation of the above
// approach
using System;
 
class GFG{
     
// Function to count
// derangements
static int countDer(int n)
{
     
    // Base case
    if (n == 1 || n == 2)
    {
        return n - 1;
    }
 
    // Variable for just storing
    // previous values
    int a = 0;
    int b = 1;
 
    // Using above recursive formula
    for(int i = 3; i <= n; ++i)
    {
        int cur = (i - 1) * (a + b);
        a = b;
        b = cur;
    }
 
    // Return result for n
    return b;
}
 
// Driver code
public static void Main()
{
    Console.Write("Count of Dearrangements is " +
                   countDer(4));
}
}
 
// This code is contributed by koulick_sadhu

Javascript

 
    // Javascript implementation
    // of the above approach
     
    // Function to count
    // derangements
    function countDer(n)
    {
 
        // Base case
        if (n == 1 || n == 2)
        {
            return n - 1;
        }
 
        // Variable for just storing
        // previous values
        let a = 0;
        let b = 1;
 
        // Using above recursive formula
        for(let i = 3; i <= n; ++i)
        {
            let cur = (i - 1) * (a + b);
            a = b;
            b = cur;
        }
 
        // Return result for n
        return b;
    }
     
    document.write("Count of Derangements is "
    + countDer(4));
 

输出

Count of Derangements is 9

时间复杂度： O(n)  
辅助空间： O(1)

感谢 Shubham Kumar 提出上述解决方案。

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

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。