LeetCode
LeetCode 394 Decode String - Medium
394. Decode String -- Medium
394. Decode String — Medium
Problem
- 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; 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"
Solution
### using stack:
class Solution:
def decodeString(self, s: str) -> str:
stack = []
multiplier = 0
res = ''
for char in s:
if char.isdigit():
multiplier = multiplier*10 + int(char)
elif char == '[':
stack.append((multiplier, res))
multiplier = 0
res = ''
elif char.isalpha():
res += char
elif char == ']':
curMultiplier, prevString = stack.pop()
res = prevString + curMultiplier * res
return res
# using recursion:
class Solution:
def decodeString(self, s: str) -> str:
if not s or len(s) == 0:
return s
result, position = self.dfs(0,s,0,'')
return result
def dfs(self, position, s, prev_num, prev_str):
while position < len(s):
while s[position].isdigit():
prev_num = prev_num*10 + int(s[position])
position += 1
if s[position] == "[":
#reset the prev_str
returned_str, ending_pos = self.dfs(position+1, s, 0, "")
#backtrack
prev_str = prev_str + returned_str*prev_num
position = ending_pos
prev_num = 0
#return the result
elif s[position] == ']':
return prev_str, position
else:
prev_str += s[position]
position += 1
return prev_str, position