LeetCode 394. Decode String

394. 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 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].

 

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Solution

The key requirements of this problem are as follows:

1. Value of s is made of the k[string] format.
2. Decode the string by multiplying string k times.
3. Return final decoded string.

To satisfy this requirements, I used a stack approach with multiple lists.

The elements of s are digits, characters and brackets.

 

First, while iterating though the string, append each element to the result list as long as the element is not a closing bracket.

 

ex)
s = "21[abc]3[cd]ef"
result = ["2", "1", "[", "a" , "b", "c"]

 

Next, when a closing bracket is encountered, move characters from result to the chars list while result[-1] != "[", and pop each character from result.

 

 

ex)
s = "21[abc]3[cd]ef"
result = ["2", "1", "["]
chars = ["c", "b", "a"]

 

 

Since the stack stores elements in reverse order, reverse chars to restore the correct order.

 

Afther that, pop the opening bracket.

Then, while result is not empty and result[-1].isdigit() is true, build the digit string using str_num = result.pop() + str_num.

 

The reason why adding digits this way is that stack pops elements in reverse order.

For example, 21 is stored as "2" then "1", so "1" is popped first.

 

ex)
s = "21[abc]3[cd]ef"
result = []
chars = ["a", "b", "c"]
str_num = '21'

 

 

Convert str_num into an integer and multiply the decoded string.

Using ''.join(chars) * num , the string "abc" is repeated 21 times and appended to result.

 

 

ex)
result = ["abc", "abc"...]

 

 

After finishing the iteration, return the final decoded string.

 

 

class Solution(object):
    def decodeString(self, s):
        result = []
        for i in s:
            if i != "]":
                result.append(i)
            else:
                char = []
                while result[-1] != "[":
                    char.append(result.pop())
                char.reverse()
                result.pop()

                str_num = ''
                while result and result[-1].isdigit():
                    str_num = result.pop() + str_num

                num = int(str_num)if str_num else 1
                temp = ''.join(char) * num
                for i in temp:
                    result.append(i)

        return ''.join(result)

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