Last updated on October 9th, 2024 at 05:52 pm

Here, We see Tag Validator LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

List of all LeetCode Solution

Tag Validator LeetCode Solution

Problem Statement

Given a string representing a code snippet, implement a tag validator to parse the code and return whether it is valid.

A code snippet is valid if all the following rules hold:

1. The code must be wrapped in a valid closed tag. Otherwise, the code is invalid.
2. A closed tag (not necessarily valid) has exactly the following format : <TAG_NAME>TAG_CONTENT</TAG_NAME>. Among them, <TAG_NAME> is the start tag, and </TAG_NAME> is the end tag. The TAG_NAME in start and end tags should be the same. A closed tag is valid if and only if the TAG_NAME and TAG_CONTENT are valid.
3. A valid TAG_NAME only contain upper-case letters, and has length in range [1,9]. Otherwise, the TAG_NAME is invalid.
4. A valid TAG_CONTENT may contain other valid closed tags, cdata and any characters (see note1) EXCEPT unmatched <, unmatched start and end tag, and unmatched or closed tags with invalid TAG_NAME. Otherwise, the TAG_CONTENT is invalid.
5. A start tag is unmatched if no end tag exists with the same TAG_NAME, and vice versa. However, you also need to consider the issue of unbalanced when tags are nested.
6. A < is unmatched if you cannot find a subsequent >. And when you find a < or </, all the subsequent characters until the next > should be parsed as TAG_NAME (not necessarily valid).
7. The cdata has the following format : <![CDATA[CDATA_CONTENT]]>. The range of CDATA_CONTENT is defined as the characters between <![CDATA[ and the first subsequent ]]>.
8. CDATA_CONTENT may contain any characters. The function of cdata is to forbid the validator to parse CDATA_CONTENT, so even it has some characters that can be parsed as tag (no matter valid or invalid), you should treat it as regular characters.

Example 1:
Input: code = “<DIV>This is the first line <![CDATA[<div>]]></DIV>”
Output: true
Explanation: The code is wrapped in a closed tag : <DIV> and </DIV>. The TAG_NAME is valid, the TAG_CONTENT consists of some characters and cdata. Although CDATA_CONTENT has an unmatched start tag with invalid TAG_NAME, it should be considered as plain text, not parsed as a tag. So TAG_CONTENT is valid, and then the code is valid. Thus return true.

Example 2:
Input: code = “<DIV>>> ![cdata[]] <![CDATA[<div>]>]]>]]>>]</DIV>”
Output: true
Explanation: We first separate the code into : start_tag|tag_content|end_tag. start_tag -> “<DIV>” end_tag -> “</DIV>” tag_content could also be separated into : text1|cdata|text2. text1 -> “>> ![cdata[]] “ cdata -> “<![CDATA[<div>]>]]>”, where the CDATA_CONTENT is “<div>]>” text2 -> “]]>>]” The reason why start_tag is NOT “<DIV>>>” is because of the rule 6. The reason why cdata is NOT “<![CDATA[<div>]>]]>]]>” is because of the rule 7.

Example 3:
Input: code = “<A> <B> </A> </B>”
Output: false
Explanation: Unbalanced. If “<A>” is closed, then “<B>” must be unmatched, and vice versa.

1. Tag Validator LeetCode Solution C++

class Solution {
public:
    bool isValid(string code) {
        int i = 0;
        return validTag(code, i) && i == code.size();
    }
private:
    bool validTag(string s, int& i) {
        int j = i;
        string tag = parseTagName(s, j);
        if (tag.empty()) return false;
        if (!validContent(s, j)) return false;
        int k = j + tag.size() + 2; // expecting j = pos of "</" , k = pos of '>'
        if (k >= s.size() || s.substr(j, k + 1 - j) != "</" + tag + ">") return false;
        i = k + 1;
        return true;
    }
    string parseTagName(string s, int& i) {
        if (s[i] != '<') return "";
        int j = s.find('>', i);
        if (j == string::npos || j - 1 - i < 1 || 9 < j - 1 - i) return "";
        string tag = s.substr(i + 1, j - 1 - i);
        for (char ch : tag) {
            if (ch < 'A' || 'Z' < ch) return "";
        }
        i = j + 1;
        return tag;
    }
    bool validContent(string s, int& i) {
        int j = i;
        while (j < s.size()) {
            if (!validText(s, j) && !validCData(s, j) && !validTag(s, j)) break;
        }
        i = j;
        return true;
    }
    bool validText(string s, int& i) {
        int j = i;
        while (i < s.size() && s[i] != '<') { i++; }
        return i != j;
    }
    bool validCData(string s, int& i) {
        if (s.find("<![CDATA[", i) != i) return false;
        int j = s.find("]]>", i);
        if (j == string::npos) return false;
        i = j + 3;
        return true;
    }
};

2. Tag Validator LeetCode Solution Java

class Solution {
      public boolean isValid(String code) {
        int length = code.length();
        Stack S=new Stack();
        for (int i = 0; i < length; i++) {
            if(i>0 && S.empty())
                return false;
            if(code.startsWith("<![CDATA[",i)){
                int pos=code.indexOf("]]>",i+9);
                if (pos==-1)
                    return false;
                else
                    i=pos+2;
            }
            else if(code.startsWith("</",i)){
                int pos=code.indexOf('>',i+2);
                if (pos==-1) {
                    return false;
                }
                String to_be_matched=code.substring(i+2,pos);
                if(S.empty())
                   return false;
                if(!to_be_matched.equals((String)S.pop()))
                    return false;
                i=pos;
            }
            else if (code.charAt(i) == '<') {
                int pos=code.indexOf('>',i+1);
                if (pos==-1)
                    return false;
                String to_be_pushed=code.substring(i+1, pos);
                if(!(to_be_pushed.length() >=1 && to_be_pushed.length() <=9))
                    return false;
                for(int k=0;k<to_be_pushed.length();k++) {
                    char temp = to_be_pushed.charAt(k);
                    if (!(temp >= 65 && temp <= 90))
                        return false;
                }
                S.push(to_be_pushed);
                i=pos;
            }
            else
            {
                int pos=code.indexOf('<',i);
                if (pos!=-1) {
                    i=pos-1;
                }
            }
        }
        if(S.empty())
            return true;
        else
            return false;
    }
}

3. Tag Validator LeetCode Solution JavaScript

var isValid = function(code) {
    if (code === 't') return false;
    while (/<!\[CDATA\[.*?\]\]>/.test(code)) {
        code = code.replace(/<!\[CDATA\[.*?\]\]>/g, 'c');
    }
    while (/<([A-Z]{1,9})>([^<]*)<\/(\1)>/.test(code)) {
        code = code.replace(/<([A-Z]{1,9})>([^<]*)<\/(\1)>/g, 't');
    }
    return code === 't';
};

4. Tag Validator LeetCode Solution Python

class Solution(object):
    def isValid(self, code):
        code = code.replace("<![CDATA[", "?")
        if code[0] == "?": return False
        n = len(code)
        stack = []
        i = 0
        while i < n:
            if code[i] == "<":
                isClosingTag = False
                tagName = []
                i += 1
                if i < n and code[i] == "/":
                    isClosingTag = True
                    i += 1
                while i < n and code[i] != ">":
                    tagName.append(code[i])
                    i += 1
                if len(tagName) < 1 or len(tagName) > 9: return False
                tagName = "".join(tagName)
                if tagName.isalpha() and tagName.isupper():
                    if isClosingTag: 
                        if stack and stack[-1] == tagName: 
                            stack.pop()
                            if i != n - 1 and not stack: return False
                        else: return False
                    else: stack.append(tagName)
                else: return False
            elif code[i] == "?":
                i += 1
                while (i + 2 < n) and not (code[i] == "]" and code[i + 1] == "]" and code[i + 2] == ">"): i += 1
                if i + 2 < n and code[i] == "]" and code[i + 1] == "]" and code[i + 2] == ">": i += 3
            else: 
                if i != n - 1 and not stack: return False
                i += 1
        return not stack