LeetCode #591 — HARD

Tag Validator

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

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:

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 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.
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 = "<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.

Constraints:

1 <= code.length <= 500
code consists of English letters, digits, '<', '>', '/', '!', '[', ']', '.', and ' '.

Step 01

Brute Force Baseline

Problem summary: 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 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. 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

Baseline thinking

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

Pattern signal: Stack

Example 1

"<DIV>This is the first line <![CDATA[<div>]]></DIV>"

Example 2

"<DIV>>>  ![cdata[]] <![CDATA[<div>]>]]>]]>>]</DIV>"

Example 3

"<A>  <B> </A>   </B>"

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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

Largest constraint values

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 #591: Tag Validator
class Solution {
    public boolean isValid(String code) {
        Deque<String> stk = new ArrayDeque<>();
        for (int i = 0; i < code.length(); ++i) {
            if (i > 0 && stk.isEmpty()) {
                return false;
            }
            if (code.startsWith("<![CDATA[", i)) {
                i = code.indexOf("]]>", i + 9);
                if (i < 0) {
                    return false;
                }
                i += 2;
            } else if (code.startsWith("</", i)) {
                int j = i + 2;
                i = code.indexOf(">", j);
                if (i < 0) {
                    return false;
                }
                String t = code.substring(j, i);
                if (!check(t) || stk.isEmpty() || !stk.pop().equals(t)) {
                    return false;
                }
            } else if (code.startsWith("<", i)) {
                int j = i + 1;
                i = code.indexOf(">", j);
                if (i < 0) {
                    return false;
                }
                String t = code.substring(j, i);
                if (!check(t)) {
                    return false;
                }
                stk.push(t);
            }
        }
        return stk.isEmpty();
    }

    private boolean check(String tag) {
        int n = tag.length();
        if (n < 1 || n > 9) {
            return false;
        }
        for (char c : tag.toCharArray()) {
            if (!Character.isUpperCase(c)) {
                return false;
            }
        }
        return true;
    }
}

// Accepted solution for LeetCode #591: Tag Validator
func isValid(code string) bool {
	var stk []string
	for i := 0; i < len(code); i++ {
		if i > 0 && len(stk) == 0 {
			return false
		}
		if strings.HasPrefix(code[i:], "<![CDATA[") {
			n := strings.Index(code[i+9:], "]]>")
			if n == -1 {
				return false
			}
			i += n + 11
		} else if strings.HasPrefix(code[i:], "</") {
			if len(stk) == 0 {
				return false
			}
			j := i + 2
			n := strings.IndexByte(code[j:], '>')
			if n == -1 {
				return false
			}
			t := code[j : j+n]
			last := stk[len(stk)-1]
			stk = stk[:len(stk)-1]
			if !check(t) || last != t {
				return false
			}
			i += n + 2
		} else if strings.HasPrefix(code[i:], "<") {
			j := i + 1
			n := strings.IndexByte(code[j:], '>')
			if n == -1 {
				return false
			}
			t := code[j : j+n]
			if !check(t) {
				return false
			}
			stk = append(stk, t)
			i += n + 1
		}
	}
	return len(stk) == 0
}

func check(tag string) bool {
	n := len(tag)
	if n < 1 || n > 9 {
		return false
	}
	for _, c := range tag {
		if c < 'A' || c > 'Z' {
			return false
		}
	}
	return true
}

# Accepted solution for LeetCode #591: Tag Validator
class Solution:
    def isValid(self, code: str) -> bool:
        def check(tag):
            return 1 <= len(tag) <= 9 and all(c.isupper() for c in tag)

        stk = []
        i, n = 0, len(code)
        while i < n:
            if i and not stk:
                return False
            if code[i : i + 9] == '<![CDATA[':
                i = code.find(']]>', i + 9)
                if i < 0:
                    return False
                i += 2
            elif code[i : i + 2] == '</':
                j = i + 2
                i = code.find('>', j)
                if i < 0:
                    return False
                t = code[j:i]
                if not check(t) or not stk or stk.pop() != t:
                    return False
            elif code[i] == '<':
                j = i + 1
                i = code.find('>', j)
                if i < 0:
                    return False
                t = code[j:i]
                if not check(t):
                    return False
                stk.append(t)
            i += 1
        return not stk

// Accepted solution for LeetCode #591: Tag Validator
impl Solution {
    pub fn is_valid(code: String) -> bool {
        fn check(tag: &str) -> bool {
            let n = tag.len();
            n >= 1 && n <= 9 && tag.as_bytes().iter().all(|b| b.is_ascii_uppercase())
        }

        let mut stk = Vec::new();
        let mut i = 0;
        while i < code.len() {
            if i > 0 && stk.is_empty() {
                return false;
            }
            if code[i..].starts_with("<![CDATA[") {
                match code[i + 9..].find("]]>") {
                    Some(n) => {
                        i += n + 11;
                    }
                    None => {
                        return false;
                    }
                };
            } else if code[i..].starts_with("</") {
                let j = i + 2;
                match code[j..].find('>') {
                    Some(n) => {
                        let t = &code[j..j + n];
                        if !check(t) || stk.is_empty() || stk.pop().unwrap() != t {
                            return false;
                        }
                        i += n + 2;
                    }
                    None => {
                        return false;
                    }
                };
            } else if code[i..].starts_with("<") {
                let j = i + 1;
                match code[j..].find('>') {
                    Some(n) => {
                        let t = &code[j..j + n];
                        if !check(t) {
                            return false;
                        }
                        stk.push(t);
                    }
                    None => {
                        return false;
                    }
                };
            }
            i += 1;
        }
        stk.is_empty()
    }
}

// Accepted solution for LeetCode #591: Tag Validator
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #591: Tag Validator
// class Solution {
//     public boolean isValid(String code) {
//         Deque<String> stk = new ArrayDeque<>();
//         for (int i = 0; i < code.length(); ++i) {
//             if (i > 0 && stk.isEmpty()) {
//                 return false;
//             }
//             if (code.startsWith("<![CDATA[", i)) {
//                 i = code.indexOf("]]>", i + 9);
//                 if (i < 0) {
//                     return false;
//                 }
//                 i += 2;
//             } else if (code.startsWith("</", i)) {
//                 int j = i + 2;
//                 i = code.indexOf(">", j);
//                 if (i < 0) {
//                     return false;
//                 }
//                 String t = code.substring(j, i);
//                 if (!check(t) || stk.isEmpty() || !stk.pop().equals(t)) {
//                     return false;
//                 }
//             } else if (code.startsWith("<", i)) {
//                 int j = i + 1;
//                 i = code.indexOf(">", j);
//                 if (i < 0) {
//                     return false;
//                 }
//                 String t = code.substring(j, i);
//                 if (!check(t)) {
//                     return false;
//                 }
//                 stk.push(t);
//             }
//         }
//         return stk.isEmpty();
//     }
// 
//     private boolean check(String tag) {
//         int n = tag.length();
//         if (n < 1 || n > 9) {
//             return false;
//         }
//         for (char c : tag.toCharArray()) {
//             if (!Character.isUpperCase(c)) {
//                 return false;
//             }
//         }
//         return true;
//     }
// }

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)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.

MONOTONIC STACK

O(n) time

O(n) space

Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.

Shortcut: Each element pushed once + popped once → O(n) amortized. The inner while-loop does not make it O(n²).

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.