LeetCode 394. Decode String

Description

https://leetcode.com/problems/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"

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].

Explanation

Using recursion to solve the problem. Whenever we see a left bracket ‘[‘, we use a helper function to build a substring repeating with the precedent number of times.

Python Solution

class Solution:
    def decodeString(self, s: str) -> str:
        if not s:
            return ""
        
        result, position = self.helper(s, 0)
        
        return result
    
    
    def helper(self, s, position):
        result = ""
        
        number = 0
        
        while position < len(s):
            char = s[position]
            
            if char.isdigit():
                number = number * 10 + int(char)
            elif char == '[':
                substring, position = self.helper(s, position + 1)
                result += substring * number
                number = 0
            elif char == ']':
                return result, position
            else:
                result += char
            
            position += 1
                
        return result, position
            

• Time Complexity: ~N
• Space Complexity: ~N