Decode String in C++

Suppose we have an encoded string; we have to return its decoded string. The rule for encoding is: k[encoded_string], this indicates where the encoded_string inside the square brackets is being repeated exactly k times. We can assume that the original data does not contain any numeric characters and that digits are only for those repeat numbers, k. So if the input is like “1[ba]2[na]”, then the output will be “banana”.

To solve this, we will follow these steps −

create one empty stack, set i := 0
while i < size of a string
- if s[i] is ‘]’
  - res := delete element from the stack and take only the string that is inside the square brackets.
  - n := 0
  - while stack is not empty, and stack top is one numeric character, then accommodate the numbers and form the actual integer as n
  - for j in range 1 to n
    - for x in range 0 to size of res
      - insert res[x] into the stack
- otherwise insert s[i] into the stack
- increase i by 1
ans := an empty string
while stack is not empty
- ans := stack top element + ans
- pop from stack
return ans

Example(C++):

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

Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
   string decodeString(string s) {
      stack <char> st;
      int i = 0;
      while(i<s.size()){
         if(s[i] == ']'){
            string res = "";
            while(st.top()!='['){
               res = st.top() + res;
               st.pop();
            }
            st.pop();
            int n = 0;
            int x = 1;
            while(!st.empty() && st.top()>='0' && st.top()<='9'){
               n = n + (st.top()-'0')*x;
               x*=10;
               st.pop();
            }
            for(int j = 1; j <= n; j++){
               for(int x = 0; x < res.size();x++){
                  st.push(res[x]);
               }
            }
         }
         else{
            st.push(s[i]);
         }
         i++;
      }
      string ans ="";
      while(!st.empty()){
         ans = st.top() + ans;
         st.pop();
      }
      return ans;
   }
};
main(){
   Solution ob;
   cout << ob.decodeString("1[ba]2[na]");
}