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Decode String

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DESCRIPTION (inspired by Leetcode.com)

Given an encoded string s, write a function to return its decoded string.

The encoding rule is k[encoded_string], where the encoded_string inside the square brackets is repeated exactly k times. k is always a positive integer, and the brackets can be nested.

You can assume the input is always well-formed: there are no extra spaces, every square bracket is properly matched, and digits only ever appear to specify a repeat count k (so you won't see input like 3a or 2[4]).

Constraints:

1 <= s.length
s consists of lowercase English letters, digits, and the square brackets [ and ].
All repeat counts k are positive integers and may have more than one digit (for example, 10[a]).
The input string is guaranteed to be valid.

Example 1:

Inputs:

s = "3[a]2[bc]"

Output:

"aaabcbc"

(Explanation: 3[a] decodes to "aaa" and 2[bc] decodes to "bcbc".)

Example 2:

Inputs:

s = "3[a2[c]]"

Output:

"accaccacc"

(Explanation: the inner 2[c] becomes "cc", so a2[c] is "acc", which is then repeated 3 times.)

Example 3:

Inputs:

s = "2[abc]3[cd]ef"

Output:

"abcabccdcdcdef"

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're 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).

Visualization


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

An opening bracket means we're about to decode a nested sequence, so we save our progress and start fresh. We push curr_string (everything we've built up to this point) and current_number (how many times the sequence we're about to read will repeat) onto the stack, then reset both to "" and 0.

Think of it as leaving a bookmark. In 3[a2[c]], by the time we hit the second [ we've already read the 3, built "a", and read the 2. We stash "a" and 2 on the stack so we can pick them back up later, then start decoding the inner c with a clean slate.

Visualization


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

A closing bracket means the current sequence is finished, so we go back to the bookmark we left. We pop the two values we saved on the way in, the repeat count first and then the string that came before the bracket. We repeat curr_string that many times, glue it onto the previous string, and that becomes our new curr_string.

Back in 3[a2[c]], the first ] finishes the inner sequence: we pop 2 and "a", repeat "c" twice to get "cc", then prepend "a" for "acc". The next ] pops 3 and the empty string we saved at the start, repeating "acc" three times for "accaccacc".

Visualization


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.

Visualization


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

Visualization

Try these examples:


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

For this problem, let n be the length of the input string and S be the length of the decoded output string.

Test Your Knowledge

Complexity Analysis

Time Complexity: O(n + S) While we iterate through the input string once in O(n) time, constructing the repeated strings takes time proportional to the total number of characters in the final decoded result. The decoded string can be much larger than the input (e.g., '3[a2[c]]' decodes to 'accaccacc').

Space Complexity: O(S) We need to store the decoded string, and in the worst case, the stack can also grow proportional to the output size when dealing with nested sequences.

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Stack

Decode String

DESCRIPTION (inspired by Leetcode.com)

Explanation

"[": Start of a new sequence

"]": End of a sequence

Digit

Letter

Solution

Complexity Analysis

Test Your Knowledge

Complexity Analysis

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