Valid Nested Tags — Problem Statement & Solution Guide
Quick Answer & Algorithm Key Takeaway
Master coding challenges related to Stack and solve the Valid Nested Tags problem optimally.
O(n)O(k)Problem Description
Valid Nested Tags
Given an array of strings tags, each element is either an opening tag formatted as "<name>" or a closing tag formatted as "</name>", where name consists solely of lowercase English letters. The task is to decide whether the entire sequence is properly nested. A sequence is proper if every opening tag is matched by a later closing tag with the identical name and tags close in the exact reverse order of their opening. Return true when the sequence satisfies these rules; otherwise return false.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Valid Nested Tags"
WHY DOES IT MATTER?
The stack-based validation pattern is essential because it efficiently handles nested structures, which are ubiquitous in programming, data formats, and system design. It provides a clear and intuitive way to model hierarchical relationships, ensuring that elements are properly matched and closed. This pattern is foundational for solving a wide range of problems, including balanced parentheses, expression evaluation, and syntax parsing, making it a critical skill for any software engineer.
OPTIMIZATION CHALLENGE
The key insight that reduces time and space complexity is the use of a stack to maintain the current nesting context, allowing each tag to be processed in constant time. By leveraging the LIFO property, the algorithm avoids the need for recursive searches or complex matching logic, resulting in linear time complexity O(n) and space complexity O(k), where k is the maximum depth of nesting. This optimization is critical for handling large-scale data efficiently.
REAL-WORLD CONNECTION
A practical real-world analogy is the way a web browser parses HTML. When the browser encounters an opening tag, it pushes it onto a stack to track the current context. When it encounters a closing tag, it checks if it matches the top of the stack and pops it if so. This process ensures that the HTML document is well-formed and that elements are properly nested. Similarly, in distributed systems, message queues often use stack-like structures to manage task dependencies, ensuring that tasks are completed in the correct order.
In an interview, it is important to clearly articulate the intuition behind the stack-based approach, emphasizing the LIFO property and how it models the nesting structure. You should also discuss edge cases and potential optimizations, such as handling self-closing tags or reducing memory usage. Demonstrating a deep understanding of the problem and its practical applications will set you apart from other candidates.
COMPLEXITY AT A GLANCE
O(n)O(k)Core Theory — Why This Approach?
The problem of validating nested tags is a classic application of the Stack data structure, which operates on the Last-In-First-Out (LIFO) principle. The core theoretical underpinning is that valid nesting requires a strict reverse order of closure: the most recently opened tag must be the first one closed. This mirrors the mathematical property of well-formed parentheses or brackets. When an opening tag is encountered, it is pushed onto the stack, representing an 'active' context. When a closing tag appears, it must match the top of the stack; if it does, the tag is popped, indicating the context has been resolved. If the stack is empty when a closing tag is encountered, or if the stack is not empty after processing all tags, the sequence is invalid. This approach ensures that every opening tag has a corresponding closing tag in the correct hierarchical order.
Naive approaches, such as using two pointers or recursive matching without a stack, often fail on large inputs due to their quadratic time complexity O(n^2) or higher. For instance, a recursive solution that searches for the matching closing tag for each opening tag can degrade significantly when tags are deeply nested or when there are many mismatched tags, leading to stack overflow errors or excessive computation. In contrast, the stack-based approach achieves linear time complexity O(n), as each tag is processed exactly once, and each push and pop operation is O(1). This efficiency is critical for handling large-scale data, such as validating HTML documents or XML feeds, where performance is paramount.
The optimal paradigm for this problem is the stack-based validation algorithm, which leverages the LIFO property to maintain the current nesting context. This approach is not only efficient but also intuitive, as it directly models the real-world behavior of nested structures. By using a stack, the algorithm ensures that tags are closed in the correct order, preventing mismatches and ensuring the integrity of the nested sequence. This paradigm is widely applicable to other problems involving balanced parentheses, bracket matching, and expression evaluation, making it a fundamental tool in a programmer's toolkit.
Interview Questions on This Problem
Q1How would you modify the stack-based approach to handle self-closing tags like '<br/>' or '<img/>' in HTML validation?
To handle self-closing tags, you would first check if the tag ends with '/>' before processing it as an opening or closing tag. If it is a self-closing tag, you can ignore it entirely since it does not require a matching closing tag. This modification ensures that the stack only tracks tags that need to be closed, maintaining the correctness of the validation logic. For example, in a loop, you would add a condition: if (tag.endsWith('/>')) continue; This approach is commonly used in HTML parsers and is a practical extension of the basic stack-based validation algorithm.
Q2What are the edge cases you should consider when validating nested tags, and how would you handle them in your code?
Key edge cases include: 1) An empty array, which should return true since there are no tags to validate. 2) A single opening tag without a closing tag, which should return false. 3) A single closing tag without an opening tag, which should return false. 4) Mismatched tags, such as '<a></b>', which should return false. To handle these, you would check if the stack is empty when processing a closing tag (return false if so) and check if the stack is empty after processing all tags (return true if so). Additionally, you would ensure that the top of the stack matches the closing tag before popping. These checks ensure robustness and correctness in the validation logic.
Q3How would you optimize the space complexity of the stack-based approach if the input size is extremely large and memory is constrained?
To optimize space complexity, you can use a more memory-efficient data structure for the stack, such as a linked list or a dynamic array with pre-allocated capacity. Additionally, you can avoid storing the entire tag string in the stack by storing only the tag name or a hash of the tag name, reducing memory usage. For example, instead of pushing the full string '<name>', you can push only 'name' or a hash value. This approach reduces the space complexity from O(n) to O(k), where k is the maximum depth of nesting, which is often much smaller than n. This optimization is particularly useful in memory-constrained environments, such as embedded systems or mobile applications.
Examples
Input
["<a>","<b>","</b>","</a>"]
Output
true
Explanation: 1. "<a>" pushes 'a' onto the stack. 2. "<b>" pushes 'b'. 3. "</b>" matches top of stack ('b') and pops it. 4. "</a>" matches remaining top ('a') and pops it. Stack is empty at the end, so the nesting is valid.
Input
["<x>","<y>","</x>","</y>"]
Output
false
Explanation: 1. "<x>" pushes 'x'. 2. "<y>" pushes 'y'. 3. "</x>" expects top of stack to be 'x' but top is 'y', mismatch → invalid.
Input
["<a>","</a>","</b>"]
Output
false
Explanation: 1. "<a>" pushes 'a'. 2. "</a>" matches and pops 'a'. 3. "</b>" attempts to close 'b' but the stack is empty, so there is no matching opening tag → invalid.
Constraints
- 1 <= tags.length <= 100000
- 3 <= tags[i].length <= 20
- tag name contains only lowercase letters and length between 1 and 10
Optimal Approach & Strategy
The optimized approach uses a stack to maintain the current nesting context, pushing opening tags and popping them when a matching closing tag is encountered. This results in a time complexity of O(n) and space complexity O(k), where k is the maximum depth of nesting.
Brute Force Approach
A brute-force approach would involve using recursion to search for the matching closing tag for each opening tag, resulting in a time complexity of O(n^2) or higher. This approach is inefficient and can lead to stack overflow errors for deeply nested structures.
Code Solutions
const fs = require('fs');
const input = fs.readFileSync(0,'utf8').trim().split(/\s+/);
if(input.length===0){process.exit(0);}
let idx=0;
const n = parseInt(input[idx++]);
const tags = input.slice(idx, idx+n);
function isProperlyNested(tags){
const stack = [];
for(const t of tags){
if(t.length>=2 && t[1] !== '/'){
// opening tag
const name = t.slice(1, -1);
stack.push(name);
}else if(t.length>=3 && t[1] === '/'){
// closing tag
if(stack.length===0) return false;
const name = t.slice(2, -1);
if(stack[stack.length-1] !== name) return false;
stack.pop();
}else{
return false; // malformed
}
}
return stack.length===0;
}
console.log(isProperlyNested(tags) ? 'true' : 'false');#include <bits/stdc++.h>
using namespace std;
bool isProperlyNested(const vector<string>& tags) {
vector<string> st;
for(const string& t: tags){
if(t.size()>=2 && t[1]!='/'){
// opening tag like <name>
string name = t.substr(1, t.size()-2);
st.push_back(name);
}else if(t.size()>=3 && t[1]=='/'){
// closing tag like </name>
if(st.empty()) return false;
string name = t.substr(2, t.size()-3);
if(st.back()!=name) return false;
st.pop_back();
}else{
return false; // malformed
}
}
return st.empty();
}
int main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n; if(!(cin>>n)) return 0;
vector<string> tags(n);
for(int i=0;i<n;++i) cin>>tags[i];
cout << (isProperlyNested(tags) ? "true" : "false");
return 0;
}
import java.io.*;
import java.util.*;
public class Main {
public static boolean isProperlyNested(String[] tags) {
Deque<String> stack = new ArrayDeque<>();
for (String t : tags) {
if (t.length() >= 2 && t.charAt(1) != '/') {
// opening tag <name>
String name = t.substring(1, t.length() - 1);
stack.push(name);
} else if (t.length() >= 3 && t.charAt(1) == '/') {
// closing tag </name>
if (stack.isEmpty()) return false;
String name = t.substring(2, t.length() - 1);
if (!stack.peek().equals(name)) return false;
stack.pop();
} else {
return false; // malformed
}
}
return stack.isEmpty();
}
public static void main(String[] args) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String line = br.readLine();
if (line == null) return;
int n = Integer.parseInt(line.trim());
String[] tags = new String[n];
for (int i = 0; i < n; i++) tags[i] = br.readLine().trim();
System.out.println(isProperlyNested(tags) ? "true" : "false");
}
}
import sys
def is_properly_nested(tags):
stack = []
for t in tags:
if len(t) >= 2 and t[1] != '/':
# opening tag like <name>
name = t[1:-1]
stack.append(name)
elif len(t) >= 3 and t[1] == '/':
# closing tag like </name>
if not stack:
return False
name = t[2:-1]
if stack[-1] != name:
return False
stack.pop()
else:
return False # malformed
return not stack
def main():
data = sys.stdin.read().strip().split()
if not data:
return
n = int(data[0])
tags = data[1:1+n]
print(str(is_properly_nested(tags)).lower())
if __name__ == "__main__":
main()
const fs = require('fs');
const input = fs.readFileSync(0,'utf8').trim().split(/\s+/);
if(input.length===0){process.exit(0);}
let idx=0;
const n = parseInt(input[idx++]);
const tags = input.slice(idx, idx+n);
function isProperlyNested(tags){
const stack = [];
for(const t of tags){
if(t.length>=2 && t[1] !== '/'){
// opening tag
const name = t.slice(1, -1);
stack.push(name);
}else if(t.length>=3 && t[1] === '/'){
// closing tag
if(stack.length===0) return false;
const name = t.slice(2, -1);
if(stack[stack.length-1] !== name) return false;
stack.pop();
}else{
return false; // malformed
}
}
return stack.length===0;
}
console.log(isProperlyNested(tags) ? 'true' : 'false');
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