# Count number of permutation of an Array having no SubArray of size two or more from original Array

Given an array of distinct integer A, the task is to count the number of possible permutations of the given array A[] such that the permutations do not contain any subarray of size 2 or more from the original array.

Examples:

Input: A = [ 1, 3, 9 ]
Output: 3
All the permutation of [ 1, 3, 9 ] are : [ 1, 3, 9 ], [ 1, 9, 3 ], [ 3, 9, 1 ], [ 3, 1, 9 ], [ 9, 1, 3 ], [ 9, 3, 1 ]

Here [ 1, 3, 9 ], [ 9, 1, 3 ] are removed as they contain sub-array [ 1, 3 ] from original list
and [ 3, 9, 1 ] removed as it contains sub-array [3, 9] from original list so,
Following are the 3 arrays that satisfy the condition : [1, 9, 3], [3, 1, 9], [9, 3, 1]

Input : A = [1, 3, 9, 12]
Output :11

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach: Iterate through list of all permutations and remove those arrays which contains any sub-array [ i, i+1 ] from A.

Below is the implementation of the above approach:

 `# Python implementation of the approach ` ` `  `# Importing the itertools  ` `from` `itertools ``import` `permutations ` ` `  `# Function that return count of all the permutation ` `# having no sub-array of [ i, i + 1 ] ` `def` `count(arr): ` `    ``z ``=``[] ` `    ``perm ``=` `permutations(arr) ` `    ``for` `i ``in` `list``(perm): ` `        ``z.append(``list``(i)) ` ` `  `    ``q ``=``[] ` `    ``for` `i ``in` `range``(``len``(arr)``-``1``): ` `        ``x, y ``=` `arr[i], arr[i ``+` `1``] ` `        ``for` `j ``in` `range``(``len``(z)): ` ` `  `            ``# Finding the indexes where x is present ` `            ``if` `z[j].index(x)!``=` `len``(z[j])``-``1``: ` ` `  `                ``# If y is present at position of x + 1 ` `                ``# append into a temp list q ` `                ``if` `z[j][z[j].index(x)``+``1``]``=``=` `y: ` `                    ``q.append(z[j]) ` ` `  `    ``# Removing all the lists that are present ` `    ``# in z ( list of all premutations ) ` `    ``for` `i ``in` `range``(``len``(q)): ` `         ``if` `q[i] ``in` `z: ` `             ``z.remove(q[i]) ` `    ``return` `len``(z) ` ` `  `# Driver Code ` `A ``=``[``1``, ``3``, ``9``] ` `print``(count(A)) `

Output:

```3
```

Efficient Solution : After making the solution for smaller size of array, we can observe a pattern:

The following pattern generates a recurrence:
Suppose the length of array A is n, then:

```count(0) = 1
count(1) = 1
count(n) = n * count(n-1) + (n-1) * count(n-2)
```

Below is the implementation of the approach:

## C++

 `// C++ implementation of the approach  ` `#include ` `using` `namespace` `std; ` ` `  `// Recursive function that returns  ` `// the count of permutation-based  ` `// on the length of the array.  ` `int` `count(``int` `n) ` `{  ` `    ``if``(n == 0) ` `        ``return` `1; ` `    ``if``(n == 1) ` `        ``return` `1; ` `    ``else` `        ``return` `(n * count(n - 1)) +  ` `              ``((n - 1) * count(n - 2)); ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `A[] = {1, 2, 3, 9}; ` `     `  `    ``// length of array ` `    ``int` `n = 4; ` `         `  `    ``// Output required answer ` `    ``cout << count(n - 1);  ` `         `  `    ``return` `0; ` `} ` ` `  `// This code is contributed by Sanjit Prasad `

## Java

 `// Java implementation of the approach  ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `// Recursive function that returns  ` `// the count of permutation-based  ` `// on the length of the array.  ` `static` `int` `count(``int` `n) ` `{  ` `    ``if``(n == ``0``) ` `        ``return` `1``; ` `    ``if``(n == ``1``) ` `        ``return` `1``; ` `    ``else` `        ``return` `(n * count(n - ``1``)) +  ` `              ``((n - ``1``) * count(n - ``2``)); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `A[] = {``1``, ``2``, ``3``, ``9``}; ` `     `  `    ``// length of array ` `    ``int` `n = ``4``; ` `         `  `    ``// Output required answer ` `    ``System.out.println(count(n - ``1``));  ` `} ` `}  ` ` `  `// This code is contributed by PrinciRaj1992 `

## Python3

 `# Python implementation of the approach ` ` `  `# Recursive function that returns ` `# the count of permutation-based ` `# on the length of the array. ` ` `  `def` `count(n): ` `    ``if` `n ``=``=` `0``: ` `        ``return` `1` `    ``if` `n ``=``=` `1``: ` `        ``return` `1` `    ``else``: ` `        ``return` `(n ``*` `count(n``-``1``)) ``+` `((n``-``1``) ``*` `count(n``-``2``)) ` ` `  `# Driver Code ` `A ``=``[``1``, ``2``, ``3``, ``9``] ` `print``(count(``len``(A)``-``1``)) `

## C#

 `// C# implementation of the above approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `// Recursive function that returns  ` `// the count of permutation-based  ` `// on the length of the array.  ` `static` `int` `count(``int` `n) ` `{  ` `    ``if``(n == 0) ` `        ``return` `1; ` `    ``if``(n == 1) ` `        ``return` `1; ` `    ``else` `        ``return` `(n * count(n - 1)) +  ` `              ``((n - 1) * count(n - 2)); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `[]A = {1, 2, 3, 9}; ` `     `  `    ``// length of array ` `    ``int` `n = 4; ` `         `  `    ``// Output required answer ` `    ``Console.WriteLine(count(n - 1));  ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

Output:

```11
```

Note: For the above recurrence you can check oeis.org.

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