Leetcode 1678: Goal Parser Interpretation

Input: The input consists of a string `command` that can contain the characters 'G', '()', and '(al)'.

Example: command = 'G()(al)'

Constraints:

• 1 <= command.length <= 100

• command consists of 'G', '()', and/or '(al)' in some order.

Output: Return the interpreted string after replacing '()', '(al)', and 'G' according to the specified rules.

Example: Output: 'Goal'

Constraints:

Goal: The goal is to convert the input string into a new string based on the specified patterns.

Steps:

• Iterate through the input string `command` and check each character or sequence of characters.

• If 'G' is encountered, append 'G' to the result.

• If '()' is encountered, append 'o' to the result.

• If '(al)' is encountered, append 'al' to the result.

• Return the concatenated result.

Goal: The input string `command` satisfies the following constraints:

Steps:

• The length of `command` is between 1 and 100.

• The string `command` consists of 'G', '()', and/or '(al)' in some order.

Assumptions:

• The input string `command` will always be valid and will follow the given structure.

• Input: Input: command = 'G()(al)'

• Explanation: The Goal Parser interprets 'G' as 'G', '()' as 'o', and '(al)' as 'al', resulting in the concatenated string 'Goal'.

Approach: To solve this problem, we need to scan through the input string `command`, identify the patterns, and replace them with their corresponding interpretations.

Observations:

• The string can be interpreted by scanning for specific patterns ('G', '()', and '(al)').

• We can use a loop to iterate through the string and handle the patterns accordingly.

Steps:

• Initialize an empty result string.

• Iterate through the input string character by character.

• For each character or sequence, append the corresponding interpretation to the result.

• Return the final concatenated result.

Empty Inputs:

• The input string will always have at least one character.

Large Inputs:

• The input size will not exceed 100 characters, so performance concerns are minimal.

Special Values:

• There are no special values other than the given patterns 'G', '()', and '(al)'.

Constraints:

• The input string will always consist of valid characters.

string interpret(string command) {
    string result= "";
    for(int i=0; i<command.size(); i++)
    {
        if(command[i] == '(')
        {
            if(command[i+1] == ')')
                result += "o";
            if(command[i+1] == 'a' && command[i+2] == 'l')
            {
                result += "al";
                i += 2;
            }
        }
        if(command[i] == 'G')
            result += "G";
    }
    return result;
}

1 : Function Definition

string interpret(string command) {

Define the function 'interpret' that takes a string 'command' as input and returns a modified string.

2 : Variable Initialization

    string result= "";

Initialize an empty string 'result' to store the final modified output.

3 : Loop Setup

    for(int i=0; i<command.size(); i++)

Start a loop to iterate over each character in the input string 'command'.

4 : Condition Check

        if(command[i] == '(')

Check if the current character is an opening parenthesis '(' to begin interpreting a command.

5 : Condition Check

            if(command[i+1] == ')')

Check if the next character is a closing parenthesis ')'. If so, it represents '()' and we append 'o' to the result.

6 : String Manipulation

                result += "o";

Append 'o' to the 'result' string when encountering the '()' pattern.

7 : Condition Check

            if(command[i+1] == 'a' && command[i+2] == 'l')

Check if the characters after '(' are 'al'. If so, append 'al' to the result and skip the next two characters.

8 : String Manipulation

                result += "al";

Append 'al' to the 'result' string when encountering the '(al)' pattern.

9 : Variable Update

                i += 2;

Skip the next two characters ('a' and 'l') after processing the '(al)' pattern.

10 : Condition Check

        if(command[i] == 'G')

Check if the current character is 'G'. If so, append 'G' to the result.

11 : String Manipulation

            result += "G";

Append 'G' to the 'result' string when encountering the 'G' character.

12 : Return Statement

    return result;

Return the final modified string 'result'.

Best Case: O(n)

Average Case: O(n)

Worst Case: O(n)

Description: The algorithm iterates through the string once, processing each character or sequence in constant time.

Best Case: O(n)

Worst Case: O(n)

Description: The space complexity is O(n) due to the space used to store the resulting string.

Leetcode 1678: Goal Parser Interpretation

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