HTML Entity Parser

HTML Entity Parser - Problem

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

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 &frasl; and symbol character is /

Given an 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.

Input & Output

Example 1 — Basic Entities

$ Input: text = "& is an HTML entity but &ambassador; is not."

› Output: "& is an HTML entity but &ambassador; is not."

💡 Note: & is a valid entity that gets replaced with &, but &ambassador; is not a valid HTML entity so it stays unchanged

Example 2 — Multiple Entities

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

› Output: "and I quote: \"...\""

💡 Note: " entities are replaced with quotation marks to form: and I quote: "..."

Example 3 — No Valid Entities

$ Input: text = "Stay home! Practice on Leetcode :)"

› Output: "Stay home! Practice on Leetcode :)"

💡 Note: No HTML entities present, so the string remains unchanged

Constraints

1 ≤ text.length ≤ 10⁵
The string may contain any possible characters out of all the 256 ASCII characters.

Visualization

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Asked in

a Amazon 15 f Facebook 12 G Google 8 M Microsoft 6

The key insight is to use a hash map for O(1) entity lookup while scanning the string once. When encountering '&', check if a valid entity starts there. Best approach is single pass with hash map. Time: O(n), Space: O(1)

Common Approaches

✓ Greedy

⏱️ Time: N/A Space: N/A

Multiple String Replacements

⏱️ Time: O(n × k) Space: O(n)

Go through each of the 6 HTML entities and use string replacement to convert them to their corresponding characters. This requires multiple passes through the string.

Single Pass with Hash Map

⏱️ Time: O(n) Space: O(1)

Scan the string once, and when we encounter '&', check if the following characters form a valid entity using hash map lookup. If yes, replace and skip ahead; if no, keep the '&' as is.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char* solution(char* text) {
    int len = strlen(text);
    char* result = malloc((len + 1) * sizeof(char));
    int j = 0;
    
    for (int i = 0; i < len; ) {
        if (text[i] == '&') {
            // Check for HTML entities
            if (i + 5 <= len && strncmp(&text[i], "&quot;", 6) == 0) {
                result[j++] = '"';
                i += 6;
            }
            else if (i + 5 <= len && strncmp(&text[i], "&apos;", 6) == 0) {
                result[j++] = '\'';
                i += 6;
            }
            else if (i + 4 <= len && strncmp(&text[i], "&amp;", 5) == 0) {
                result[j++] = '&';
                i += 5;
            }
            else if (i + 3 <= len && strncmp(&text[i], "&gt;", 4) == 0) {
                result[j++] = '>';
                i += 4;
            }
            else if (i + 3 <= len && strncmp(&text[i], "&lt;", 4) == 0) {
                result[j++] = '<';
                i += 4;
            }
            else if (i + 6 <= len && strncmp(&text[i], "&frasl;", 7) == 0) {
                result[j++] = '/';
                i += 7;
            }
            else {
                // Not a valid entity, copy the character
                result[j++] = text[i++];
            }
        }
        else {
            // Regular character, copy it
            result[j++] = text[i++];
        }
    }
    
    result[j] = '\0';
    return result;
}

int main() {
    char text[10001];
    if (fgets(text, sizeof(text), stdin) != NULL) {
        // Remove trailing newline if present
        text[strcspn(text, "\n")] = 0;
        
        char* result = solution(text);
        printf("%s\n", result);
        free(result);
    }
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

⚡ Linearithmic Space

28.0K Views

Medium Frequency

~15 min Avg. Time

890 Likes

Ln 1, Col 1

Smart Actions

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