[LeetCode] Decode String

Problem: 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 10^5.

Example:

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

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

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

Approach: If you look at the example, the thought of using stack immediately comes into the mind as it's just like calculating postfix statements. You can understand the algorithm by just looking at the implementation. Nothing complex here.

Implementation in C#:

        public string DecodeString(string s)

    {

        int length = s?.Length ?? 0;

        if (length <= 1)

        {

            return s;

        }

        Stack<string> stack = new Stack<string>();

        string currStr = "";

        foreach(char ch in s)

        {

            if (ch == '[')

            {

                stack.Push(currStr);

                stack.Push("[");

                currStr = "";

            }

            else if (ch == ']')

            {

                while (stack.Peek() != "[")

                {

                    currStr = stack.Pop() + currStr;

                }

                stack.Pop();

                int freq = int.Parse(stack.Pop());

                string result = "";

                for (int i = 0; i < num; ++i)

                {

                    result += currStr;

                }

                stack.Push(result);

                currStr = "";

            }

            else if (char.IsDigit(ch))

            {

                if (currStr == "" || char.IsDigit(currStr[0]))

                {

                    currStr += ch;

                }

                else

                {

                    stack.Push(currStr);

                    currStr = "";

                    currStr += ch;

                }

            }

            else

            {

                currStr += ch;

            }

        }

        string ans = currStr;

        while (stack.Count != 0)

        {

            ans = stack.Pop() + ans;

        }

        return ans;

    }

Complexity: O(n)