Given an encoded string, return it's 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].


Examples:

s = "3[a]2[bc]", return "aaabcbc".
s = "3[a2[c]]", return "accaccacc".
s = "2[abc]3[cd]ef", return "abcabccdcdcdef".

Source: leetcode | Difficulty: MEDIUM_HARD
Tags: String algorithms

Data structures

Complete search

Starter code: class Solution:
    def decodeString(self, s: str) -> str:
        

Solutions:
class Solution:

	def decodeString(self, s):
		stack = []
		stack.append(['', 1])
		num = ''
		for ch in s:
			if ch.isdigit():
				num += ch
			elif ch == '[':
				stack.append(['', int(num)])
				num = ''
			elif ch == ']':
				(st, k) = stack.pop()
				stack[-1][0] += st * k
			else:
				stack[-1][0] += ch
		return stack[0][0]


class Solution:

	def decodeString(self, s):
		m = len(s)
		if m == 0:
			return ''
		result = []
		for i in s:
			if i != ']':
				result.append(i)
			else:
				char_temp = []
				r1m = len(result)
				for j in range(r1m - 1, -1, -1):
					if result[j] != '[':
						char_temp.insert(0, result.pop())
					else:
						result.pop()
						break
				digit_char = []
				r2m = len(result)
				for j in range(r2m - 1, -1, -1):
					if result[j].isdigit():
						digit_char.insert(0, result.pop())
					else:
						break
				result += char_temp * int(''.join(digit_char))
		return ''.join(result)


class Solution:

	def decodeString(self, s):
		stack_num = []
		stack_str = []
		num = ''
		string = ''
		for c in s:
			if c.isdigit():
				if num == '':
					stack_str.append(string)
					string = ''
				num += c
			elif c == '[':
				stack_num.append(int(num))
				num = ''
			elif c == ']':
				string = stack_str.pop() + string * stack_num.pop()
			else:
				string += c
		return string


class Solution:

	def decodeString(self, s):
		count = ''
		res = ''
		count_stack = []
		res_stack = []
		for i in range(len(s)):
			if ord('0') <= ord(s[i]) <= ord('9'):
				count += s[i]
			elif s[i] == '[':
				count_stack.append(int(count) if count else 0)
				res_stack.append(res)
				count = ''
				res = ''
			elif s[i] == ']':
				res = res_stack.pop() + res * count_stack.pop()
			else:
				res += s[i]
		return res


class Solution:

	def decodeString(self, s):
		if len(s) == 0:
			return ''
		idx = 0
		while s[idx].isdigit():
			idx += 1
		if idx > 0:
			k = int(s[:idx])
			idx += 1
			idx_from = idx
			bracket_count = 1
			while bracket_count > 0:
				if s[idx] == '[':
					bracket_count += 1
				elif s[idx] == ']':
					bracket_count -= 1
				idx += 1
			return self.decodeString(s[idx_from:idx - 1]) * k + self.decodeString(s[idx:])
		return s[0] + self.decodeString(s[1:])