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Goal Parser Interpretation - Leetcode Solution

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LeetCode:  Goal Parser Interpretation Leetcode Solution

Difficulty: Easy

Topics: string  

Problem Statement:

You own a Goal Parser that can interpret a string command. The command consists of an alphabet of "G", "()" and/or "(al)" in some order. The Goal Parser will interpret "G" as the string "G", "()" as the string "o", and "(al)" as the string "al". The interpreted strings are then concatenated in the original order.

Given the string command, return the Goal Parser's interpretation of command.

Example:

Input: command = "G()(al)" Output: "Goal" Explanation: The Goal Parser interprets the command as follows: G -> "G" () -> "o" (al) -> "al" The final concatenated result is "Goal".

Solution:

The problem statement is quite simple. We have to implement a Goal Parser that can interpret a string command consisting of G, (), and/or (al) characters. The Goal Parser will interpret G as the string "G", () as the string "o", and (al) as the string "al". The interpreted strings are then concatenated in the original order to get the final output.

The solution to this problem is quite simple and can be implemented in a single pass. We can start iterating from the beginning of the input string and interpret each character as follows:

  1. If the current character is G, simply append "G" to the output string.

  2. If the current character is (, check the next character. If it is ), append "o" to the output string. If the next two characters are "al", append "al" to the output string.

  3. If the current character is any other character besides G and (, simply ignore it.

Finally, return the output string as the result.

Here is the Python code implementation of the solution:

def interpret(command: str) -> str: output = "" i = 0 while i < len(command): if command[i] == "G": output += "G" i += 1 elif command[i] == "(": if command[i+1] == ")": output += "o" i += 2 elif command[i+1:i+3] == "al": output += "al" i += 4 else: i += 1 return output

Explanation:

  • We initialize an empty string variable called output to store the interpreted characters of the input string.

  • We also initialize the index variable i to 0 which we will use during iteration.

  • We use a while loop to iterate over each character in the input string until we reach the end of the string.

  • If the current character is G, we simply append "G" to the output string and increment i by 1 to move to the next character.

  • If the current character is (, we check the next character. If it is ), we append "o" to the output string and increment i by 2 to move to the next character. If the next two characters are "al", we append "al" to the output string and increment i by 4 to move to the next character.

  • If the current character is any other character besides G and (, we simply ignore it and increment i by 1 to move to the next character.

  • Finally, we return the output string as the result.

Time Complexity:

The time complexity of the solution is O(n) where n is the length of the input string.

Space Complexity:

The space complexity of the solution is O(n) where n is the length of the input string. The space complexity is dominated by the output string variable.

Goal Parser Interpretation Solution Code

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