0394. Decode String

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; 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 won't be input like 3a or 2[4].

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"

Example 4:

Input: s = "abc3[cd]xyz"
Output: "abccdcdcdxyz"

Analysis

There are two parts we need to take care of:

1. repeated number: cnt
2. string enclosed by '[]': t

So we need to use two stacks to store the above two pieces of information: stack<int> times which stores the current repeated number and stack<string> stk which stores the current concatenated string.

1. If we see a number, we append the digit to our current partially formed number.
2. If we see '[', we can stop tracking our number string (convert it to an integer and push it to the top of the times stack), and also we need to start pushing the string to stk.

3. If we see ']', we should start "duplicating" times.top() $\times$ t and push that to our stk.

4. Time: $O(n)$

5. Space: $O(n)$

Code

class Solution {
public:
  string decodeString(string s) {
    stack<int> times;
    stack<string> stk;
    int cnt = 0;
    string t = "";
    for (char c : s) {
      if (c <= '9' && c >= '0') {
        cnt = cnt * 10 + c - '0';
      } else if (c == '[') {
        times.push(cnt);
        stk.push(t);
        cnt = 0;
        t = "";
      } else if (c == ']') {
        int time = times.top();
        times.pop();
        for (int i = 0; i < time; ++i) {
          stk.top() += t;
        }
        t = stk.top();
        stk.pop();
      } else {
        t += c;
      }
    }
    return stk.empty() ? t : stk.top();
  }
};