HTML Entity Parser

HTML entity parser is the parser that takes HTML code as input and replace all the entities of the special characters by the characters itself.

The special characters and their entities for HTML are:

• Quotation Mark: the entity is " and symbol character is ".
• Single Quote Mark: the entity is ' and symbol character is '.
• Ampersand: the entity is & and symbol character is &.
• Greater Than Sign: the entity is > and symbol character is >.
• Less Than Sign: the entity is < and symbol character is <.
• Slash: the entity is ⁄ and symbol character is /.

Given the input text string to the HTML parser, you have to implement the entity parser.

Return the text after replacing the entities by the special characters.

Example 1:

Input: text = "& is an HTML entity but &ambassador; is not." Output: "& is an HTML entity but &ambassador; is not." Explanation: The parser will replace the & entity by &

Example 2:

Input: text = "and I quote: "..."" Output: "and I quote: "...""

HTML Entity Parser

This problem requires parsing a string and replacing HTML entities with their corresponding characters. Here's a breakdown of a solution in Python.

Naive Solution

A straightforward approach is to iterate through the string and check for each known HTML entity. If an entity is found, replace it with its corresponding character. This can be done using a series of if statements or a dictionary lookup.

def entity_parser_naive(text):
    text = text.replace(""", '"')
    text = text.replace("'", "'")
    text = text.replace("&", '&')
    text = text.replace(">", '>')
    text = text.replace("&amp;lt;", '<')
    text = text.replace("&amp;frasl;", '/')
    return text

Optimal Solution

An optimal solution improves on the naive approach by using a dictionary to store the entities and their corresponding characters. This reduces the number of comparisons needed.

def entity_parser_optimal(text):
    entity_map = {
        """: '"',
        "'": "'",
        "&": '&',
        ">": '>',
        "&amp;lt;": '<',
        "&amp;frasl;": '/'
    }
    result = ""
    i = 0
    while i < len(text):
        found = False
        for entity, char in entity_map.items():
            if text[i:].startswith(entity):
                result += char
                i += len(entity)
                found = True
                break
        if not found:
            result += text[i]
            i += 1
    return result

Big(O) Run-time Analysis

• Naive Solution: The replace method has a time complexity of O(n) where n is the length of the string. Since we are calling it a fixed number of times (6 times in this case), the overall time complexity is still O(n).
• Optimal Solution: In the optimal solution, for each character in the input string, we iterate through the entity_map. In the worst-case scenario, no entities are found, and we iterate through the entire string once. In the best-case scenario, all characters are part of an entity. The startswith method is O(k), where k is the length of the entity. Since the number of entities is fixed (6), the overall time complexity is O(n * k), which simplifies to O(n) because k is constant.

Big(O) Space Usage Analysis

• Naive Solution: The replace method creates a new string each time it is called. However, since these are intermediate strings and the number of calls is fixed, the overall space complexity is O(n).
• Optimal Solution: The entity_map stores a fixed number of entities. The result string grows proportionally to the length of the input string in the worst case. Thus, the overall space complexity is O(n).

Edge Cases

1. Empty String:
   • If the input string is empty, the function should return an empty string without any errors.
2. No Entities Present:
   • If the input string contains no HTML entities, the function should return the original string.
3. Overlapping Entities:
   • The algorithm should handle cases where entities might overlap or be adjacent to each other correctly.
4. Partial Entities:
   • If the string contains a partial entity (e.g., &am), it should not be replaced.
5. Multiple Consecutive Entities:
   • The algorithm should handle multiple consecutive entities correctly (e.g., && should become &&).

Here is how we can handle these cases:

def entity_parser_robust(text):
    entity_map = {
        """: '"',
        "'": "'",
        "&": '&',
        ">": '>',
        "&amp;lt;": '<',
        "&amp;frasl;": '/'
    }
    result = ""
    i = 0
    while i < len(text):
        found = False
        for entity, char in entity_map.items():
            if text[i:].startswith(entity):
                result += char
                i += len(entity)
                found = True
                break
        if not found:
            result += text[i]
            i += 1
    return result