Simple XML Validator - The Coding Shala

Home >> Programming Questions >> XML Validator

 In this post, we will are going to implement a Simple XML Validator in the Java language.

Simple XML Validator in Java

The input is an ASCII string, we have to write a Java program to check if the given String is valid XML or not. For simplicity purposes, the XML string only has content and tags.

 The output should be a String for whether the text is valid XML or not. If the XML is invalid, output one of the three error strings:
1. "missing closing tag for <start_tag>"
2. "encountered closing tag without matching open tag for </end tag>"
3. "parse error"

Example 1:
Input: <a>some text</a>
Output: valid

Example 2:
Input: <a>
Output: missing closing tag for <a>

Approach
We can use the stack data structure.

Java Program:

class Solution {
	
	public String validate_xml(String xml) {
		
		if(xml == null || xml.isEmpty()) return "valid";
		
		Stack<String> stack = new Stack<>();
		int i = 0;
		while(i < xml.length()) {
			if(xml.charAt(i) == '<') {
				if(i < xml.length() && xml.charAt(i+1) == '/') {
					//closing tag
					i = i+2;
					StringBuilder end = new StringBuilder("<");
					while(xml.charAt(i) != '>') {
						if(xml.charAt(i) == '<') return "parse error";
						end.append(xml.charAt(i));
						i++;
					}
					if(end.length() == 1) return "parse error";
					end.append(">");
					if(stack.peek().equals(end.toString())) {
						stack.pop();
					} else {
						StringBuilder tag = new StringBuilder("</");
						for(int j=1; j<end.length(); j++) {
							tag.append(end.charAt(j));
						}
						return "encountered closing tag without matching open tag for " + tag.toString();
					}
				} else {
					//start tag
					i++;
					StringBuilder sb = new StringBuilder("<");
					while(xml.charAt(i) != '>') {
						if(xml.charAt(i) == '<') return "parse error";
						sb.append(xml.charAt(i));
						i++;
					}
					if(sb.length() == 1) return "parse error";
					sb.append(">");
					stack.push(sb.toString());
				}
			}
			i++;
		}
		
		if(!stack.isEmpty()) {
			return "missing closing tag for " + stack.pop();
		}
		
		return "valid";
	}
	
}


Other Posts You May Like
Please leave a comment below if you like this post or found some errors, it will help me to improve my content.
By

Comments

Post a Comment

Popular Posts from this Blog

Shell Script to Create a Simple Calculator - The Coding Shala

Home >> Scripting >> simple calculator Shell Script to Create a Simple Calculator Write a shell script to create a simple calculator using a switch-case statement? Code:  #! / bin / bash echo "simple calculator" sum = 0 i = "y" echo "enter first number" read n1 echo "enter second number" read n2 while [ $ i = "y" ] do echo "1.Addition" echo "2.Subtraction" echo "3.Multiplication" echo "4.Division" echo "Enter choice" read ch case $ch in 1 ) sum =$(echo " $n1 + $n2" | bc - l) echo "Addition is =" $ sum ;; 2 ) sum =$(echo "$n1 - $n2" | bc - l) echo "Sub is =" $ sum ;; 3 ) sum =$(echo "$n1 * $n2" | bc - l) echo "Mul is =" $ sum ;; 4 ) sum =$(echo "$n1 / $n2" | bc - l) echo "div is =" $ sum ;; * )echo "invalid choice" esac echo "Do y...
Read more

New Year Chaos Solution - The Coding Shala

Home >> Programming Questions >> New Year Chaos New Year Chaos Problem Solution In this post, you will learn how to solve New Year Chaos Problem and implement its solution in Java. Problem:   It's New Year's Day and everyone's in line for the Wonderland rollercoaster ride! There are a number of people queued up, and each person wears a sticker indicating their initial position in the queue. Initial positions increment from at the front of the line to at the back.  Any person in the queue can bribe the person directly in front of them to swap positions. If two people swap positions, they still wear the same sticker denoting their original places in line. One person can bribe at most two others.  For Example, if n=8 and Person 5 bribes Person 4, the queue will look like this: 1,2,3,5,4,6,7,8. Fascinated by this chaotic queue, you decide you must know the minimum number of bribes that took place to get the queue into its current state. Inpu...
Read more

Java Program to Convert Binary to Decimal - The Coding Shala

Home >> Java Programs >> Convert Binary to Decimal  Hey there, welcome back to another post. In this post, we will learn how to convert Binary to Decimal in Java. Java Program to Convert Binary to Decimal You have given a binary number, convert it to a decimal number. Example:  Input: 101 Output: 5 Input: 111 Output: 7 Binary to Decimal Conversion in Java Approach We will the below formula to convert binary to decimal: decimal number = (2^0) * (rightmost digit) + (2^1) * (second rightmost digit) + ...  for example:  binary number = 110 decimal number = 2^0 * 0 + 2^1 * 1 + 2^2 * 1 = 1 * 0 + 2 * 1 + 4 * 1 = 0 + 2 + 4 = 6 Java Program:  import java.util.Scanner; /** * https://www.thecodingshala.com/ */ public class Main { public static void printDecimal( int num) { int decimal = 0; int twos = 1; // 2^0 = 1 initial value while (num > 0) { ...
Read more

N-th Tribonacci Number Solution - The Coding Shala

Home >> Programming >> N-th Tribonacci Number  Hey there, welcome back to another post. In this post, we will learn how to solve the N-th Tribonacci Number problem and will implement its solution in Java. N-th Tribonacci Number Problem Statement The Tribonacci sequence Tn is defined as follows: T0 = 0, T1 = 1, T2 = 1, and Tn+3 = Tn + Tn+1 + Tn+2 for n >= 0. Given n, return the value of Tn.  Example 1:  Input: n = 4 Output: 4 Explanation: T_3 = 0 + 1 + 1 = 2 T_4 = 1 + 1 + 2 = 4 N-th Tribonacci Number Java Solution using Bottom-Up DP This problem is similar to the Fibonacci series. We can solve it using the Bottom-Up approach of dynamic programming. Time Complexity: O(n) Space Complexity: O(n) The space complexity can be reduced to O(1) by using variables to store the previous three values instead of using an array. Java Program:  class Solution { public int tribonacci( int n) { if (n == 0) return 0; ...
Read more

Single Number 2 LeetCode Solution - The Coding Shala

Home >> LeetCode >> Single Number 2  In this post, we will learn how to solve LeetCode's Single Number 2 problem and will implement its solution in Java. Single Number 2 Problem Given an integer array nums where every element appears three times except for one, which appears exactly once. Find the single element and return it. Example 1: Input: nums = [2,2,3,2] Output: 3 Example 2: Input: nums = [0,1,0,1,0,1,99] Output: 99 Single Number 2 Java Solution Approach 1 We can use HashMap to store the count of numbers. Java Program:  class Solution { public int singleNumber ( int [] nums ) { Map < Integer , Integer > map = new HashMap <>(); for ( int num : nums ) { if ( map . containsKey ( num )) map . put ( num , map . get ( num )+ 1 ); else map . put ( num , 1 ); } for ( int num : nums ) { if ( map . get ( num ) == 1 ) return num ; } ...
Read more