Grokking the Coding Interview: Patterns for Coding Questions
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Solution: Decode String
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Problem Statement

You have a string that represents encodings of substrings, where each encoding is of the form k[encoded_string], where k is a positive integer, and encoded_string is a string that contains letters only.

Your task is to decode this string by repeating the encoded_string k times and return it. It is given that k is always a positive integer.

Examples

    • Input: "3[a3[c]]"
    • Expected Output: "acccacccaccc"
    • Justification: The inner 3[c] is decoded as ccc, and then a is appended to the front, forming acc. This is then repeated 3 times to form acccacccaccc.
    • Input: "2[b3[d]]"
    • Expected Output: "bdddbddd"
    • Justification: The inner 3[d] is decoded as ddd, and then b is appended to the front, forming bddd. This is then repeated 2 times to form bddd bddd.
    • Input: "4[z]"
    • Expected Output: "zzzz"
    • Justification: The 4[z] is decoded as z repeated 4 times, forming zzzz.

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].

Solution

  • Understanding the Problem:

    • The problem involves decoding a string that contains patterns where a number is followed by a string in brackets.
    • The number indicates how many times the string in brackets should be repeated.

  • Approach:

    • Use a stack to keep track of the characters in the string.
    • Iterate through the string character by character.
    • When a number is encountered, calculate the complete number.
    • When an opening bracket [ is encountered, push the calculated number to the stack.
    • When a closing bracket ] is encountered, pop elements from the stack until a number is encountered and form the substring to be repeated.
    • Multiply the substring with the number and push the result back to the stack.

  • Handling Nested Brackets:

    • The stack will naturally handle nested brackets as it will continue popping elements until a number is encountered, forming the substring for the innermost bracket first.

Algorithm Walkthrough

  • Given Input: "3[a3[c]]"
  • Steps:
    • Initialize an empty stack.
    • Iterate through the string:
      • Encounter 3, push 3 to the stack.
      • Encounter [, do nothing.
      • Encounter a, push a to the stack.
      • Encounter 3, push 3 to the stack.
      • Encounter [, do nothing.
      • Encounter c, push c to the stack.
      • Encounter ], pop c and 3, form ccc and push it back to the stack.
      • Encounter ], pop ccc, a, and 3, form acccacccaccc and push it back to the stack.
    • The final stack contains acccacccaccc as the only element, which is the decoded string.

Code

Python3
Python3
. . . .

Complexity Analysis

  • Time Complexity: O(n), where n is the length of the input string. This is because we are iterating through the string once and processing each character.
  • Space Complexity: O(n), where n is the length of the input string. In the worst case, the stack will store all the characters of the input string.

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