LeetCode #1410 — MEDIUM

HTML Entity Parser

Move from brute-force thinking to an efficient approach using hash map strategy.

The Problem

Problem Statement

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

Example 2:

Input: text = "and I quote: &quot;...&quot;"
Output: "and I quote: \"...\""

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: 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 &frasl; 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map

Example 1

"&amp; is an HTML entity but &ambassador; is not."

Example 2

"and I quote: &quot;...&quot;"

Step 02

Core Insight

What unlocks the optimal approach

Search the string for all the occurrences of the character '&'.
For every '&' check if it matches an HTML entity by checking the ';' character and if entity found replace it in the answer.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1410: HTML Entity Parser
class Solution {
    public String entityParser(String text) {
        Map<String, String> d = new HashMap<>();
        d.put(""", "\"");
        d.put("&apos;", "'");
        d.put("&amp;", "&");
        d.put("&gt;", ">");
        d.put("&lt;", "<");
        d.put("&frasl;", "/");
        StringBuilder ans = new StringBuilder();
        int i = 0;
        int n = text.length();
        while (i < n) {
            boolean found = false;
            for (int l = 1; l < 8; ++l) {
                int j = i + l;
                if (j <= n) {
                    String t = text.substring(i, j);
                    if (d.containsKey(t)) {
                        ans.append(d.get(t));
                        i = j;
                        found = true;
                        break;
                    }
                }
            }
            if (!found) {
                ans.append(text.charAt(i++));
            }
        }
        return ans.toString();
    }
}

// Accepted solution for LeetCode #1410: HTML Entity Parser
func entityParser(text string) string {
	d := map[string]string{
		""":  "\"",
		"&apos;":  "'",
		"&amp;":   "&",
		"&gt;":    ">",
		"&lt;":    "<",
		"&frasl;": "/",
	}
	var ans strings.Builder
	i, n := 0, len(text)

	for i < n {
		found := false
		for l := 1; l < 8; l++ {
			j := i + l
			if j <= n {
				t := text[i:j]
				if val, ok := d[t]; ok {
					ans.WriteString(val)
					i = j
					found = true
					break
				}
			}
		}
		if !found {
			ans.WriteByte(text[i])
			i++
		}
	}

	return ans.String()
}

# Accepted solution for LeetCode #1410: HTML Entity Parser
class Solution:
    def entityParser(self, text: str) -> str:
        d = {
            '"': '"',
            '&apos;': "'",
            '&amp;': "&",
            "&gt;": '>',
            "&lt;": '<',
            "&frasl;": '/',
        }
        i, n = 0, len(text)
        ans = []
        while i < n:
            for l in range(1, 8):
                j = i + l
                if text[i:j] in d:
                    ans.append(d[text[i:j]])
                    i = j
                    break
            else:
                ans.append(text[i])
                i += 1
        return ''.join(ans)

// Accepted solution for LeetCode #1410: HTML Entity Parser
struct Solution;

impl Solution {
    fn entity_parser(text: String) -> String {
        text.replace(""", "\"")
            .replace("&apos;", "'")
            .replace("&frasl;", "/")
            .replace("&lt;", "<")
            .replace("&gt;", ">")
            .replace("&amp;", "&")
    }
}

#[test]
fn test() {
    let text = "&amp; is an HTML entity but &ambassador; is not.".to_string();
    let res = "& is an HTML entity but &ambassador; is not.".to_string();
    assert_eq!(Solution::entity_parser(text), res);
    let text = "and I quote: "..."".to_string();
    let res = "and I quote: \"...\"".to_string();
    assert_eq!(Solution::entity_parser(text), res);
    let text = "Stay home! Practice on Leetcode :)".to_string();
    let res = "Stay home! Practice on Leetcode :)".to_string();
    assert_eq!(Solution::entity_parser(text), res);
    let text = "x &gt; y &amp;&amp; x &lt; y is always false".to_string();
    let res = "x > y && x < y is always false".to_string();
    assert_eq!(Solution::entity_parser(text), res);
    let text = "leetcode.com&frasl;problemset&frasl;all".to_string();
    let res = "leetcode.com/problemset/all".to_string();
    assert_eq!(Solution::entity_parser(text), res);
}

// Accepted solution for LeetCode #1410: HTML Entity Parser
function entityParser(text: string): string {
    const d: Record<string, string> = {
        '"': '"',
        '&apos;': "'",
        '&amp;': '&',
        '&gt;': '>',
        '&lt;': '<',
        '&frasl;': '/',
    };

    let ans: string = '';
    let i: number = 0;
    const n: number = text.length;

    while (i < n) {
        let found: boolean = false;
        for (let l: number = 1; l < 8; ++l) {
            const j: number = i + l;
            if (j <= n) {
                const t: string = text.substring(i, j);
                if (d.hasOwnProperty(t)) {
                    ans += d[t];
                    i = j;
                    found = true;
                    break;
                }
            }
        }

        if (!found) {
            ans += text[i++];
        }
    }

    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × l)

Space

O(l)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan the rest of the array looking for a match. Two nested loops give n × (n−1)/2 comparisons = O(n²). No extra space since we only use loop indices.

HASH MAP

O(n) time

O(n) space

One pass through the input, performing O(1) hash map lookups and insertions at each step. The hash map may store up to n entries in the worst case. This is the classic space-for-time tradeoff: O(n) extra memory eliminates an inner loop.

Shortcut: Need to check “have I seen X before?” → hash map → O(n) time, O(n) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.