Decode String

DESCRIPTION (credit Leetcode.com)

Given an encoded string, write a function to return its decoded string that follows a specific encoding rule: k[encoded_string], where the encoded_string within the brackets is repeated exactly k times. Note that k is always a positive integer. The input string is well-formed without any extra spaces, and square brackets are properly matched. Also, assume that the original data doesn't contain digits other than the ones that specify the number of times to repeat the following encoded_string.

EXAMPLES

Inputs:


s = "3[a2[c]]"

Output:


"accaccacc"

Run your code to see results here

Explanation

We start by initializing our stack, and the variables curr_string and current_number. The stack allows us to account for nested sequences correctly. curr_string represents the current string we currently decoding, and current_number represents the number of times we need to repeat it when the current decode sequence is completed (i.e. when we encounter a closing "]" bracket).


def decodeString(s):
  stack = []
  curr_string = ""
  current_number = 0

  for char in s:
    if char == "[":
      stack.append(curr_string)
      stack.append(current_number)
      curr_string = ""
      current_number = 0
    elif char == "]":
      num = stack.pop()
      prev_string = stack.pop()
      curr_string = prev_string + num * curr_string
    elif char.isdigit():
      current_number = current_number * 10 + int(char)
    else:
      curr_string += char

  return curr_string

decode string

0 / 1

We then iterate over each character in the encoded string, handling each character as follows:

"[": Start of a new sequence

When we encounter an opening bracket, we push the current string curr_string and the current number current_number to the stack and reset curr_string and current_number to empty strings. These values that we pushed onto the stack represent the "context" of the current sequence we are decoding. We use current_number to keep track of the number of times we need to repeat the current string we are about to decode, while curr_string represents the value of the string that will be prepended to the result of the current sequence.


def decodeString(s):
  stack = []
  curr_string = ""
  current_number = 0

  for char in s:
    if char == "[":
      stack.append(curr_string)
      stack.append(current_number)
      curr_string = ""
      current_number = 0
    elif char == "]":
      num = stack.pop()
      prev_string = stack.pop()
      curr_string = prev_string + num * curr_string
    elif char.isdigit():
      current_number = current_number * 10 + int(char)
    else:
      curr_string += char

  return curr_string

[

0 / 2

"]": End of a sequence

When we encounter a closing bracket, we pop the last element from the stack and repeat the current string curr_string by the number of times current_number and append it to the string at the top of the stack.


def decodeString(s):
  stack = []
  curr_string = ""
  current_number = 0

  for char in s:
    if char == "[":
      stack.append(curr_string)
      stack.append(current_number)
      curr_string = ""
      current_number = 0
    elif char == "]":
      num = stack.pop()
      prev_string = stack.pop()
      curr_string = prev_string + num * curr_string
    elif char.isdigit():
      current_number = current_number * 10 + int(char)
    else:
      curr_string += char

  return curr_string

]

0 / 1

We are finished decoding 2[c], so we repeat "c"2 times and prepend "a" and pop the top two values from the stack.

Digit

When char is a digit, we update current_number by multiplying it by 10 and adding the value of the digit. current_number is used to keep track of the number of times we need to repeat the current string we are just about to decode.


def decodeString(s):
  stack = []
  curr_string = ""
  current_number = 0

  for char in s:
    if char == "[":
      stack.append(curr_string)
      stack.append(current_number)
      curr_string = ""
      current_number = 0
    elif char == "]":
      num = stack.pop()
      prev_string = stack.pop()
      curr_string = prev_string + num * curr_string
    elif char.isdigit():
      current_number = current_number * 10 + int(char)
    else:
      curr_string += char

  return curr_string

initialize variables

0 / 2

We need to repeat the result of decoding a2[c] 3 times.

Letter

When we encounter a letter, we append it to the current string curr_string.

Solution


def decodeString(s):
  stack = []
  curr_string = ""
  current_number = 0

  for char in s:
    if char == "[":
      stack.append(curr_string)
      stack.append(current_number)
      curr_string = ""
      current_number = 0
    elif char == "]":
      num = stack.pop()
      prev_string = stack.pop()
      curr_string = prev_string + num * curr_string
    elif char.isdigit():
      current_number = current_number * 10 + int(char)
    else:
      curr_string += char

  return curr_string

decode string

0 / 20

Complexity Analysis

Time Complexity: O(n). Each character in the string is processed once. Space Complexity:O(n) where n is the length of the input array for the size of the stack.

Next: Longest Valid Parentheses

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Decode String

DESCRIPTION (credit Leetcode.com)

EXAMPLES

Inputs:


s = "3[a2[c]]"

Output:


"accaccacc"

Run your code to see results here

Explanation


def decodeString(s):
  stack = []
  curr_string = ""
  current_number = 0

  for char in s:
    if char == "[":
      stack.append(curr_string)
      stack.append(current_number)
      curr_string = ""
      current_number = 0
    elif char == "]":
      num = stack.pop()
      prev_string = stack.pop()
      curr_string = prev_string + num * curr_string
    elif char.isdigit():
      current_number = current_number * 10 + int(char)
    else:
      curr_string += char

  return curr_string

decode string

0 / 1

We then iterate over each character in the encoded string, handling each character as follows:

"[": Start of a new sequence


def decodeString(s):
  stack = []
  curr_string = ""
  current_number = 0

  for char in s:
    if char == "[":
      stack.append(curr_string)
      stack.append(current_number)
      curr_string = ""
      current_number = 0
    elif char == "]":
      num = stack.pop()
      prev_string = stack.pop()
      curr_string = prev_string + num * curr_string
    elif char.isdigit():
      current_number = current_number * 10 + int(char)
    else:
      curr_string += char

  return curr_string

[

0 / 2

"]": End of a sequence


def decodeString(s):
  stack = []
  curr_string = ""
  current_number = 0

  for char in s:
    if char == "[":
      stack.append(curr_string)
      stack.append(current_number)
      curr_string = ""
      current_number = 0
    elif char == "]":
      num = stack.pop()
      prev_string = stack.pop()
      curr_string = prev_string + num * curr_string
    elif char.isdigit():
      current_number = current_number * 10 + int(char)
    else:
      curr_string += char

  return curr_string

]

0 / 1

We are finished decoding 2[c], so we repeat "c"2 times and prepend "a" and pop the top two values from the stack.

Digit


def decodeString(s):
  stack = []
  curr_string = ""
  current_number = 0

  for char in s:
    if char == "[":
      stack.append(curr_string)
      stack.append(current_number)
      curr_string = ""
      current_number = 0
    elif char == "]":
      num = stack.pop()
      prev_string = stack.pop()
      curr_string = prev_string + num * curr_string
    elif char.isdigit():
      current_number = current_number * 10 + int(char)
    else:
      curr_string += char

  return curr_string

initialize variables

0 / 2

We need to repeat the result of decoding a2[c] 3 times.

Letter

When we encounter a letter, we append it to the current string curr_string.

Solution


def decodeString(s):
  stack = []
  curr_string = ""
  current_number = 0

  for char in s:
    if char == "[":
      stack.append(curr_string)
      stack.append(current_number)
      curr_string = ""
      current_number = 0
    elif char == "]":
      num = stack.pop()
      prev_string = stack.pop()
      curr_string = prev_string + num * curr_string
    elif char.isdigit():
      current_number = current_number * 10 + int(char)
    else:
      curr_string += char

  return curr_string

decode string

0 / 20

Complexity Analysis

Time Complexity: O(n). Each character in the string is processed once. Space Complexity:O(n) where n is the length of the input array for the size of the stack.

Next: Longest Valid Parentheses

Your account is free and you can post anonymously if you choose.

Sort By

Loading comments...

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