#include <bits/stdc++.h>
using namespace std;

/*
  Problem 394. Decode String
  Medium
  
  Given an encoded string, return its decoded string.

  The encoding rule is: k[encoded_string], where the encoded_string
  inside the square brackets is being repeated exactly k times. Note
  that k is guaranteed to be a positive integer.

  You may assume that the input string is always valid; there are no
  extra white spaces, square brackets are well-formed, etc. Furthermore,
  you may assume that the original data does not contain any digits and
  that digits are only for those repeat numbers, k. For example, there
  will not be input like 3a or 2[4].

  The test cases are generated so that the length of the output will
  never exceed 105.

  Example 1:

  Input: s = "3[a]2[bc]"
  Output: "aaabcbc"
  Example 2:

  Input: s = "3[a2[c]]"
  Output: "accaccacc"
  Example 3:

  Input: s = "2[abc]3[cd]ef"
  Output: "abcabccdcdcdef"
 
  Constraints:

  1 <= s.length <= 30
  s consists of lowercase English letters, digits, and square brackets '[]'.
  s is guaranteed to be a valid input.
  All the integers in s are in the range [1, 300].
*/

class Solution {
public:
    deque<string> sStack; // stack is a wrapper around deque
    stack<int> kStack;
    
    ////////////////////////////////////////
    // parseK
    int parseK(const string& s, size_t index)
    {
        deque<int> que;
        while (s.at(index) != '[')
            que.push_back(s.at(index++) - '0');
        
        int k = 0, i = 0;
        while (!que.empty())
        {
            int digit = que.back();
            k += digit * pow(10, i++);
            que.pop_back();
        }
        kStack.push(k);
        return index;
    }

    int parseKNoQue(const string& s, size_t index)
    {
        int k = 0;
        size_t i = s.find_first_of('[', index) - index - 1;
        while(s.at(index) != '[')
        {
            int digit = s.at(index++) - '0';
            k += digit * pow(10, i--);
        }
        kStack.push(k);
        return index;
    }

    string decodeString(string s) {
        size_t index = 0;
        string result;
        while (index < s.size())
        {
            if (isdigit(s.at(index)))
            {
                int k = 0;
                size_t exp = s.find_first_of('[', index) - index - 1;
                while(s.at(index) != '[')
                {
                    int digit = s.at(index++) - '0';
                    k += digit * pow(10, exp--);
                }
                kStack.push(k);
            }
            else if (s.at(index) == '[')
            {
                cout << "push result: " << result << endl;
                sStack.push_front(result);
                result.clear();
                index++;
            }
            else if (s.at(index) == ']')
            {
                cout << "index: " << index << " result: " << result << " stack: "; for (auto i: sStack) {cout << i << " ";} cout << endl;
                int k = kStack.top();
                string next = sStack.front();
                for (int i = 0; i < k; i++)
                    next += result;
                result = next;
                kStack.pop();
                sStack.pop_front();
                index++;
            }
            else if (s.at(index) >= 'a' && s.at(index) <= 'z')
                result += s.at(index++);
        }
        return result;
    }
};

int main()
{
    vector<tuple<string,string>> testCases = {
        {"3[a]2[bc]","aaabcbc"},
        {"3[a2[c]]","accaccacc"}, // acc repeats
        {"xxx3[a2[c]]","accaccacc"}, // xxx then acc repeats
        {"2[abc]3[cd]ef","abcabccdcdcdef"},
        {"12[a]","aaaaaaaaaaaa"},
        {"1[x2[y3[z]]]","xyzzzyzzz"}, // xyzzz repeats
        {"3[a2[z]]","azzazzazz"}, // azz repeats
        {"3[a2[z1[f]]]","azfzfazfzfazfzf"}, // azfzf repeats
    };

    for (auto& tc: testCases)
    {
        Solution s;
        string result = s.decodeString(get<0>(tc));
        cout << "   input: " << get<0>(tc) << endl;
        cout << "  output: " << result << endl;
        cout << "expected: " << get<1>(tc) << endl;
        cout << "-------------------\n";        
    }

    
    string s = "abcde4[";
    int index = 5;
    Solution ss;
    cout << "parseK: " << ss.parseK(s, index) << " kStack.top: " << ss.kStack.top() << endl;
    Solution sss;
    cout << "parseKNoQue: " << sss.parseKNoQue(s, index) << " kStack.top: " << sss.kStack.top() << endl;
}