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:
# 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)) |
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 |
11
Note: For the above recurrence you can check oeis.org.
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