591. Tag Validator

Problem

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:

• The code must be wrapped in a valid closed tag. Otherwise, the code is invalid.

• 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 TAGNAME in start and end tags should be the same. A closed tag is valid if and only if the TAGNAME and TAG_CONTENT are valid.

• A valid TAG_NAME only contain upper-case letters, and has length in range [1,9]. Otherwise, the TAG_NAME is invalid.

• 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.

• 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.

• 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).

• 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 ]]>.

• 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 = "This is the first line ]]>"
Output: true
Explanation: 
The code is wrapped in a closed tag :  and . 
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 = ">>  ![cdata[]] ]>]]>]]>>]"
Output: true
Explanation:
We first separate the code into : start_tag|tag_content|end_tag.
start_tag -> ""
end_tag -> ""
tag_content could also be separated into : text1|cdata|text2.
text1 -> ">>  ![cdata[]] "
cdata -> "]>]]>", where the CDATA_CONTENT is "]>"
text2 -> "]]>>]"
The reason why start_tag is NOT ">>" is because of the rule 6.
The reason why cdata is NOT "]>]]>]]>" is because of the rule 7.

Example 3:

Input: code = "  **    **"
Output: false
Explanation: Unbalanced. If "" is closed, then "" must be unmatched, and vice versa.

  Constraints:

• 1 <= code.length <= 500

• code consists of English letters, digits, '<', '>', '/', '!', '[', ']', '.', and ' '.

Solution

class Solution {
    public boolean isValid(String code) {
        Stack<String> stack = new Stack<>();
        for(int i = 0; i < code.length();){
            if(i>0 && stack.isEmpty()) return false;
            if(code.startsWith("<![CDATA[", i)){
                int j = i+9;
                i = code.indexOf("]]>", j);
                if(i < 0) return false;
                i += 3;
            }else if(code.startsWith("</", i)){
                int j = i + 2;
                i = code.indexOf('>', j);
                if(i < 0 || i == j || i - j > 9) return false;
                for(int k = j; k < i; k++){
                    if(!Character.isUpperCase(code.charAt(k))) return false;
                }
                String s = code.substring(j, i++);
                if(stack.isEmpty() || !stack.pop().equals(s)) return false;
            }else if(code.startsWith("<", i)){
                int j = i + 1;
                i = code.indexOf('>', j);
                if(i < 0 || i == j || i - j > 9) return false;
                for(int k = j; k < i; k++){
                    if(!Character.isUpperCase(code.charAt(k))) return false;
                }
                String s = code.substring(j, i++);
                stack.push(s);
            }else{
                i++;
            }
        }
        return stack.isEmpty();
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).