Beautiful Arrangement in C++

C++Server Side ProgrammingProgramming

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
raja
Published on 02-May-2020 17:07:41
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