Given an array of integer A[] with no duplicate elements, write a function that returns the count of all the permutations of an array having no subarray of [i, i+1] for every i in A[].
Examples:
Input : 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 subarray [1, 3] from original list and [3, 9, 1] removed as it contains subarray [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 : 1 3 9 12 Output : 11
Naive Solution :Iterate through list of all permutations and remove those arrays which contains any subarray [i, i+1] from A[ ].
Code: Python code for finding out the permutation in the array
# Python implementation of the approachfrom itertools import permutations # Function that returns the count of all the permutation# having no subarray 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)): if z[j].index(x)!=len(z[j])-1: if z[j][z[j].index(x)+1]==y: q.append(z[j]) for i in range(len(q)): if q[i] in z: z.remove(q[i]) return len(z) # Driver CodeA = [1,3, 8, 9]print(count(A)) |
Output :
11
Efficient Solution : Following is a recursive solution based on fact that length of the array decides the number of all permutations having no subarray [i, i+1] for every i in A[ ]
Suppose the length of A[ ] is n, then
n = n-1 count(0) = 1 count(1) = 1 count(n) = n * count(n-1) + (n-1) * count(n-2)
Code : Below is the code for implementing the recursive function that return count of permutations
C++
// C++ implementation of the approach// Recursive function that return count of // permutation based on the length of array.#include <bits/stdc++.h>using namespace std; 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 codeint main(){ int A[] = {1, 2, 3, 9}; int len = sizeof(A) / sizeof(A[0]); cout << count(len - 1);} // This code is contributed by 29AjayKumar |
Java
// Java implementation of the approach// Recursive function that return count of // permutation based on the length of array.import java.util.*; class GFG{ 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 Codestatic public void main(String[] arg) { int []A = {1, 2, 3, 9}; System.out.print(count(A.length - 1));}} // This code is contributed by PrinciRaj1992 |
Python3
# Python implementation of the approach# Recursive function that return count of # permutation based on the length of 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 CodeA = [1, 2, 3, 9]print(count(len(A)-1)) |
C#
// C# implementation of the approach// Recursive function that return count of // permutation based on the length of array.using System; class GFG{ 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 Codestatic public void Main(String[] arg) { int []A = {1, 2, 3, 9}; Console.Write(count(A.Length - 1));}} // This code is contributed by Princi Singh |
Output :
11
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