Find Permutation in C++

Suppose we have a secret signature consisting of character 'D' and 'I'. 'D' denotes the decreasing relationship between two numbers, 'I' denotes increasing relationship between two numbers. And the secret signature was constructed by a special integer array, which contains uniquely all the different number from 1 to n.

For example, the secret signature "DI" can be constructed from an array like [2,1,3] or [3,1,2], but not be constructed using array like [3,2,4] or [2,1,3,4], which are both illegal constructing special string that can't represent the "DI" secret signature.

Now we have to find the lexicographically smallest permutation of [1, 2, ... n] that could refer to the given secret signature in the input.

So, if the input is like "DI", then the output will be [2,1,3], As we know [3,1,2] can also construct the secret signature "DI", but since we want to find the one with the smallest lexicographical permutation, we need to output [2,1,3]

To solve this, we will follow these steps −

Define one stack st
cnt := 2
Define an array ret
for initialize i := 1, when i <= size of s, update (increase i by 1), do −
- if s[i - 1] is same as 'D', then −
  - insert i into st
- Otherwise
  - insert i at the end of ret
  - while (not st is empty), do −
    - insert top element of st at the end of ret
    - delete element from st
insert size of s into st
while (not st is empty), do −
- insert top element of st at the end of ret
- delete element from st
return ret

Example

Let us see the following implementation to get a better understanding −

Live Demo

#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<auto< v){
   cout << "[";
   for(int i = 0; i<v.size(); i++){
      cout << v[i] << ", ";
   }
   cout << "]"<<endl;
}
class Solution {
public:
   vector<int< findPermutation(string s) {
      stack <int< st;
      int cnt = 2;
      vector <int< ret;
      for(int i = 1; i <= s.size(); i++){
         if(s[i - 1] == 'D'){
            st.push(i);
         }
         else{
            ret.push_back(i);
            while(!st.empty()){
               ret.push_back(st.top());
               st.pop();
            }
         }
      }
      st.push(s.size() + 1);
      while(!st.empty()){
         ret.push_back(st.top());
         st.pop();
      }
      return ret;
   }
};
main(){
   Solution ob;
   print_vector(ob.findPermutation("DIID"));
}