1678. Goal Parser Interpretation

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String

Time: O(n)

Space: O(n)

Problem Overview

The goal parser maps fixed tokens to output characters: G stays G, () becomes o, (al) becomes al, and ()!

Intuition

The goal parser maps fixed tokens to output characters: G stays G, () becomes o, (al) becomes al, and ()! becomes !. Scan left to right and match the longest pattern first so (al) is not mistaken for ().

Algorithm

1Initialize empty result and index i = 0.
2While i < length: if s[i] == G, append G and i++.
3Else if s starts with "(al)" at i, append "al", i += 4.
4Else if s starts with "()" at i, check next char for !.
5If "()!", append ! and i += 4; else append o and i += 2.
6Return result string.

Example Walkthrough

Input: command = "G()(al)"

1.G -> G. () -> o. (al) -> al.

Output: "Goal"

Common Pitfalls

•Match "(al)" before bare "()" — order of pattern checks matters.
•Input is guaranteed valid — no error handling needed.
•StringBuilder avoids repeated string concatenation in loops.

1678.cs

// Approach: Scan the command string; 'G'→'G', '()'→'o', '(al)'→'al'.
// Time: O(n) Space: O(n)

public class Solution
{
    public string Interpret(string command)
    {
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < command.Length;)
        {
            if (command[i] == 'G')
            {
                sb.Append("G");
                ++i;
            }
            else if (command[i + 1] == ')')
            {
                sb.Append("o");
                i += 2;
            }
            else
            {
                sb.Append("al");
                i += 4;
            }
        }
        return sb.ToString();
    }
}

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