Permutation Sequence in C++

Suppose the set is like [1,2,3,...,n], contains a total of n! unique permutations. By listing and labeling all of the permutations in order, we get these sequence for n = 3: ["123","132","213","231","312","321"] So if n and k are given, then return the kth permutation sequence. The n will be between 1 to 9 (inclusive) and k will be between 1 to n! (inclusive). For example if n = 3.

Let us see the steps −

• ans := empty string, define array called candidates of size n
• for i in range 0 to n – 1
  • candidates[i] := ((i + 1) + character ‘0’)
• create one array called fact of size n + 1, set fact[0] := 1
• for i in range 1 to n
  • fact[i] := fact[i – 1] * i
• decrease k by 1
• for i := n – 1 down to 0
  • idx := k / fact[i]
  • ans := ans + candidates[idx]
  • for j := idx, j + 1 < size of candidates
    • candidates[j] := candidates[j + 1]
  • k := k mod fact[i]
• return ans

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
typedef long long int lli;
class Solution {
   public:
   string getPermutation(int n, int k) {
      string ans = "";
      vector <char> candidates(n);
      for(lli i = 0; i < n; i++)
         candidates[i] = ((i + 1) + '0');
      vector <lli> fact(n + 1);
      fact[0] = 1;
      for(lli i = 1; i <= n; i++)
         fact[i] = fact[i - 1] * i;
      k--;
      for(lli i = n - 1; i >= 0; i--){
         lli idx = k / fact[i];
         ans += candidates[idx];
         for(lli j = idx; j + 1< candidates.size(); j++)
            candidates[j] = candidates[j + 1];
         k = k % fact[i];
      }
      return ans;
   }
};
main(){
   Solution ob;
   cout << ob.getPermutation(4, 9);
}

Input

4
9

Output

2314