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 

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: 

C++

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int indexOf(vector<int> arr, int target) {
    for (int i = 0; i < arr.size(); i++) {
        if (arr[i] == target) {
            return i;
        }
    }
    return -1;
}
void permute(int arr[], int l, int r, vector<vector<int>>& result) {
    if (l == r) {
        vector<int> v(arr, arr + r + 1);
        result.push_back(v);
    } else {
        for (int i = l; i <= r; i++) {
            swap(arr[l], arr[i]);
            permute(arr, l + 1, r, result);
            swap(arr[l], arr[i]);
        }
    }
}
// Function that returns count of all the permutation
// having no sub-array of [ i, i + 1 ]
int count(int arr[], int n) {
    vector<vector<int>> z;
    int perm[n];
    for (int i = 0; i < n; i++) {
        perm[i] = arr[i];
    }
    permute(perm, 0, n - 1, z);

    vector<vector<int>> q;
    for (int i = 0; i < n - 1; i++) {
        int x = arr[i];
        int y = arr[i + 1];
        for (int j = 0; j < z.size(); j++) {
            vector<int> curr = z[j];
            int idx = indexOf(curr, x);
            if (idx != n - 1 && curr[idx + 1] == y) {
                q.push_back(curr);
            }
        }
    }

    for (int i = 0; i < q.size(); i++) {
        z.erase(remove(z.begin(), z.end(), q[i]), z.end());
    }
    return z.size();
}



void swap(int& a, int& b) {
    int temp = a;
    a = b;
    b = temp;
}


// Driver code
int main() {
    int A[] = {1, 3, 9};
    int n = sizeof(A) / sizeof(A[0]);
    cout << count(A, n) << endl;
    return 0;
}

Java

import java.util.ArrayList;
import java.util.List;
import java.util.Arrays;

public class Main {
// Function that return count of all the permutation
// having no sub-array of [ i, i + 1 ]
public static int count(int[] arr) {
    List<int[]> z = new ArrayList<int[]>();
    int n = arr.length;
    int[] perm = new int[n];
    for (int i = 0; i < n; i++) {
        perm[i] = arr[i];
    }
    permute(perm, 0, n - 1, z);

    List<int[]> q = new ArrayList<int[]>();
    for (int i = 0; i < n - 1; i++) {
        int x = arr[i];
        int y = arr[i + 1];
        for (int j = 0; j < z.size(); j++) {
            int[] curr = z.get(j);
            int idx = indexOf(curr, x);
            if (idx != n - 1 && curr[idx + 1] == y) {
                q.add(curr);
            }
        }
    }

    z.removeAll(q);
    return z.size();
}

public static void permute(int[] arr, int l, int r, List<int[]> result) {
    if (l == r) {
        result.add(Arrays.copyOf(arr, arr.length));
    } else {
        for (int i = l; i <= r; i++) {
            swap(arr, l, i);
            permute(arr, l + 1, r, result);
            swap(arr, l, i);
        }
    }
}

public static void swap(int[] arr, int i, int j) {
    int temp = arr[i];
    arr[i] = arr[j];
    arr[j] = temp;
}

public static int indexOf(int[] arr, int target) {
    for (int i = 0; i < arr.length; i++) {
        if (arr[i] == target) {
            return i;
        }
    }
    return -1;
}

// Driver code
public static void main(String[] args) {
    int[] A = {1, 3, 9};
    System.out.println(count(A));
}
}

Python3

# 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 permutations )
    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))

C#

using System;
using System.Collections.Generic;
using System.Linq;

public class MainClass
{

  // Function that return count of all the permutation
  // having no sub-array of [ i, i + 1 ]
  public static int Count(int[] arr)
  {
    List<int[]> z = new List<int[]>();
    int n = arr.Length;
    int[] perm = new int[n];
    for (int i = 0; i < n; i++)
    {
      perm[i] = arr[i];
    }
    Permute(perm, 0, n - 1, z);
    List<int[]> q = new List<int[]>();
    for (int i = 0; i < n - 1; i++)
    {
      int x = arr[i];
      int y = arr[i + 1];
      for (int j = 0; j < z.Count(); j++)
      {
        int[] curr = z.ElementAt(j);
        int idx = IndexOf(curr, x);
        if (idx != n - 1 && curr[idx + 1] == y)
        {
          q.Add(curr);
        }
      }
    }

    z = z.Except(q).ToList();
    return z.Count();
  }

  public static void Permute(int[] arr, int l, int r, List<int[]> result)
  {
    if (l == r)
    {
      result.Add((int[])arr.Clone());
    }
    else
    {
      for (int i = l; i <= r; i++)
      {
        Swap(arr, l, i);
        Permute(arr, l + 1, r, result);
        Swap(arr, l, i);
      }
    }
  }

  public static void Swap(int[] arr, int i, int j)
  {
    int temp = arr[i];
    arr[i] = arr[j];
    arr[j] = temp;
  }

  public static int IndexOf(int[] arr, int target)
  {
    for (int i = 0; i < arr.Length; i++)
    {
      if (arr[i] == target)
      {
        return i;
      }
    }
    return -1;
  }

  // Driver code
  public static void Main()
  {
    int[] A = { 1, 3, 9 };
    Console.WriteLine(Count(A));
  }
}

Javascript

// JavaScript implementation of the approach

// Function that return count of all the permutation
// having no sub-array of [ i, i + 1 ]
function count(arr) {
    let z = [];
    let perm = permutations(arr);
    for (let i of perm) {
        z.push(Array.from(i));
    }

    let q = [];
    for (let i = 0; i < arr.length - 1; i++) {
        let x = arr[i];
        let y = arr[i + 1];
        for (let j = 0; j < z.length; j++) {

            // Finding the indexes where x is present
            if (z[j].indexOf(x) !== z[j].length - 1) {

                // If y is present at position of x + 1
                // push into a temp array q
                if (z[j][z[j].indexOf(x) + 1] === y) {
                    q.push(z[j]);
                }
            }
        }

        // Removing all the arrays that are present
        // in z ( array of all permutations )
        for (let i = 0; i < q.length; i++) {
            if (z.includes(q[i])) {
                let index = z.indexOf(q[i]);
                z.splice(index, 1);
            }
        }
    }
    return z.length;
}
function permutations(inputArr) {
    let result = [];

    function permute(arr, m = []) {
        if (arr.length === 0) {
            result.push(m)
        } else {
            for (let i = 0; i < arr.length; i++) {
                let curr = arr.slice();
                let next = curr.splice(i, 1);
                permute(curr.slice(), m.concat(next))
             }
        }
    }

    permute(inputArr)

    return result;
}

// Driver Code
let A = [1, 3, 9];
console.log(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<bits/stdc++.h>
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

Javascript

<script>

// JavaScript implementation of the approach

// Recursive function that returns
// the count of permutation-based
// on the length of the array.
function 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

let A = [1, 2, 3, 9];

// length of array
let n = 4;

// Output required answer
document.write(count(n - 1));


// This code is contributed by _saurabh_jaiswal

</script>

Output:

11

Time Complexity: O(2^n)
Auxiliary Space: O(2^n)