Given an integer, N, the task is to count the number of ways to generate an array, arr[] of consisting of N integers such that for every index i(1-based indexing), arr[i] is either a factor or a multiple of i, or both. The arr[] must be the permutations of all the numbers from the range [1, N].
Examples:
Input: N=2
Output: 2
Explanation:
Two possible arrangements are {1, 2} and {2, 1}Input: N=3
Output: 3
Explanation:
The 6 possible arrangements are {1, 2, 3}, {2, 1, 3}, {3, 2, 1}, {3, 1, 2}, {2, 3, 1} and {1, 3, 2}.
Among them, the valid arrangements are {1, 2, 3}, {2, 1, 3} and {3, 2, 1}.
Approach: The problem can be solved using Backtracking technique and the concept of print all permutations using recursion. Follow the steps below to find the recurrence relation:
- Traverse the range [1, N].
- For the current index pos, if i % pos == 0 and i % pos == 0, then insert i into the arrangement and use the concept of Backtracking to find valid permutations.
- Remove i.
- Repeat the above steps for all values in the range [1, N], and finally, print the count of valid permutations.
Below is the implementation of the above approach:
C++
// C++ Program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to find the count of// desired permutationsint findPermutation(unordered_set<int>& arr, int N){ int pos = arr.size() + 1; // Base case if (pos > N) return 1; int res = 0; for (int i = 1; i <= N; i++) { // If i has not been inserted if (arr.find(i) == arr.end()) { // Backtrack if (i % pos == 0 or pos % i == 0) { // Insert i arr.insert(i); // Recur to find valid permutations res += findPermutation(arr, N); // Remove i arr.erase(arr.find(i)); } } } // Return the final count return res;}// Driver Codeint main(){ int N = 5; unordered_set<int> arr; cout << findPermutation(arr, N); return 0;} |
Java
// Java program to implement// the above approachimport java.util.*;class GFG{ // Function to find the count of// desired permutationsstatic int findPermutation(Set<Integer>arr, int N){ int pos = arr.size() + 1; // Base case if (pos > N) return 1; int res = 0; for(int i = 1; i <= N; i++) { // If i has not been inserted if (! arr.contains(i)) { // Backtrack if (i % pos == 0 || pos % i == 0) { // Insert i arr.add(i); // Recur to find valid permutations res += findPermutation(arr, N); // Remove i arr.remove(i); } } } // Return the final count return res;}// Driver Codepublic static void main(String []args){ int N = 5; Set<Integer> arr = new HashSet<Integer>(); System.out.print(findPermutation(arr, N));}}// This code is contributed by chitranayal |
Python3
# Python3 program to implement# the above approach# Function to find the count of# desired permutationsdef findPermutation(arr, N): pos = len(arr) + 1 # Base case if(pos > N): return 1 res = 0 for i in range(1, N + 1): # If i has not been inserted if(i not in arr): # Backtrack if(i % pos == 0 or pos % i == 0): # Insert i arr.add(i) # Recur to find valid permutations res += findPermutation(arr, N) # Remove i arr.remove(i) # Return the final count return res# Driver CodeN = 5arr = set()# Function callprint(findPermutation(arr, N))# This code is contributed by Shivam Singh |
C#
// C# program to implement// the above approachusing System;using System.Collections.Generic;class GFG{ // Function to find the count of// desired permutationsstatic int findPermutation(HashSet<int>arr, int N){ int pos = arr.Count + 1; // Base case if (pos > N) return 1; int res = 0; for(int i = 1; i <= N; i++) { // If i has not been inserted if (! arr.Contains(i)) { // Backtrack if (i % pos == 0 || pos % i == 0) { // Insert i arr.Add(i); // Recur to find valid permutations res += findPermutation(arr, N); // Remove i arr.Remove(i); } } } // Return the readonly count return res;}// Driver Codepublic static void Main(String []args){ int N = 5; HashSet<int> arr = new HashSet<int>(); Console.Write(findPermutation(arr, N));}}// This code is contributed by gauravrajput1 |
Javascript
<script>// Javascript program to implement// the above approach// Function to find the count of// desired permutationsfunction findPermutation(arr, N){ var pos = arr.size + 1; // Base case if (pos > N) return 1; var res = 0; for(var i = 1; i <= N; i++) { // If i has not been inserted if (!arr.has(i)) { // Backtrack if (i % pos == 0 || pos % i == 0) { // Insert i arr.add(i); // Recur to find valid permutations res += findPermutation(arr, N); // Remove i arr.delete(i); } } } // Return the final count return res;}// Driver Codevar N = 5;var arr = new Set();document.write(findPermutation(arr, N));// This code is contributed by importantly</script> |
Output:
10
Time Complexity: O(N×N!)
Auxiliary Space: O(N)
Using Brute Force in python:
Approach:
The first approach is to generate all possible permutations of the array and check if each element is a multiple or a factor of its index. We can do this using a recursive function to generate permutations and then checking each permutation.
- Define a function is_valid(arr) that takes an array arr and checks if each element is a multiple or a factor of its index. It returns True if all elements satisfy this condition, otherwise False.
- Define a recursive function generate_permutations(arr, l, r, valid_permutations) that generates all possible permutations of the array arr from index l to index r. For each permutation, it checks if it is valid using the function is_valid(arr). If a valid permutation is found, it is added to the list valid_permutations.
- Define a function count_permutations(n) that creates an array arr of size n, initializes it with values 1 to n, and calls the function generate_permutations() with arr, 0, n-1, and an empty list valid_permutations. It then returns the length of valid_permutations.
Call the function count_permutations() with the input n and print the result.
Python3
def is_valid(arr): for i in range(len(arr)): if arr[i] % (i+1) != 0 and (i+1) % arr[i] != 0: return False return Truedef generate_permutations(arr, l, r, valid_permutations): if l == r: if is_valid(arr): valid_permutations.append(arr[:]) else: for i in range(l, r+1): arr[l], arr[i] = arr[i], arr[l] generate_permutations(arr, l+1, r, valid_permutations) arr[l], arr[i] = arr[i], arr[l]def count_permutations(n): arr = [i+1 for i in range(n)] valid_permutations = [] generate_permutations(arr, 0, n-1, valid_permutations) return len(valid_permutations)print(count_permutations(2)) # Output: 2print(count_permutations(3)) # Output: 3 |
Output
2 3
Time Complexity: O(n!)
Space Complexity: O(n!)
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