Beautiful Arrangement in C++

Suppose we have N integers from 1 to N. We will define a beautiful arrangement as an array that is constructed by these N numbers completely if one of the following is true for the ith position (1 <= i <= N) in this array −

• The number at the ith position is can be divided by i.
• i is divisible by the number at the ith position.

So if the input is 2, then the result will be also 2, as the first beautiful arrangement is [1,2]. Here number at the 1st position (i=1) is 1, and 1 is divisible by i (i=1). Then number at the 2nd position (i=2) is 2, and 2 is divisible by i (i=2). The second beautiful arrangement is [2, 1]: Here number at the 1st position (i=1) is 2, and 2 is divisible by i (i=1). Then number at the 2nd position (i=2) is 1, and i (i=2) is divisible by 1.

To solve this, we will follow these steps −

• Define a recursive method solve(), that will take visited array, end and pos. The pos is initially 1.
• if pos = end + 1, then increase ans by 1 and return
• for i in range 1 to end
  • if i is not visited and pos is divisible by 0 or i is divisible by 0, then
    • mark i as visited
    • solve(visited, end, pos + 1)
    • mark i as unvisited
• from the main method, do the following −
• ans := 0, make an array called visited
• call solve(visited, N, 1)r
• return ans.

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h> using namespace std; class Solution {    public:    int ans;    void solve(vector <bool>& visited, int end, int pos = 1){       if(pos == end + 1){          ans++;          return;       }       for(int i = 1; i <= end; i++){          if(!visited[i] && (pos % i == 0 || i % pos == 0)){             visited[i] = true;             solve(visited, end, pos + 1);             visited[i] = false;          }       }    }    int countArrangement(int N) {       ans = 0;       vector <bool> visited(N);       solve(visited, N);       return ans;    } }; main(){    Solution ob;    cout << (ob.countArrangement(2)); }

Input

2

Output

2