Attribute Parser in C++ | Hacker Rank Solution

Hello there, today we are going to solve Attribute Parser Hacker Rank Solution in C++.

Table Of Contents 👊

• Problem
• Input Format
• Constraints
• Output Format
• Solution

Problem

This challenge works with a custom-designed markup language HRML. In HRML, each element consists of a starting and ending tag, and there are attributes associated with each tag. Only starting tags can have attributes. We can call an attribute by referencing the tag, followed by a tilde, '~' and the name of the attribute. The tags may also be nested.

The opening tags follow the format:

<tag-name attribute1-name = "value1" attribute2-name = "value2" ...>

The closing tags follow the format:

</tag-name>

The attributes are referenced as:

tag1~value  
tag1.tag2~name

Given the source code in HRML format consisting of N lines, answer Q queries. For each query, print the value of the attribute specified. Print "Not Found!" if the attribute does not exist.

Example

HRML listing
<tag1 value = "value">
<tag2 name = "name">
<tag3 another="another" final="final">
</tag3>
</tag2>
</tag1>

Queries
tag1~value
tag1.tag2.tag3~name
tag1.tag2~value

Here, tag2 is nested within tag1, so attributes of tag2 are accessed as tag1.tag2~<attribute>. Results of the queries are:

Query                 Value
tag1~value            "value"
tag1.tag2.tag3~name   "Not Found!"
tag1.tag2.tag3~final  "final"

Input Format

The first line consists of two space separated integers, N and Q. N specifies the number of lines in the HRML source program. Q specifies the number of queries.

The following N lines consist of either an opening tag with zero or more attributes or a closing tag. There is a space after the tag-name, attribute-name, '=' and value.There is no space after the last value. If there are no attributes there is no space after tag name.

Q queries follow. Each query consists of string that references an attribute in the source program.More formally, each query is of the form tagi1. tagi2. tagi3 . . . tagim ~ attr - name where m >= 1 and tagi1. tagi2. tagi3 . . . tagim are valid tags in the input.

Constraints

• 1 <= N <= 20
• 1 <= Q <= 20
• Each line in the source program contains, at most, 200 characters.
• Every reference to the attributes in the Q queries contains at most 200 characters.
• All tag names are unique and the HRML source program is logically correct, i.e. valid nesting.
• A tag can may have no attributes.

Output Format

Print the value of the attribute for each query. Print "Not Found!" without quotes if the attribute does not exist.

Sample Input 

4 3
<tag1 value = "HelloWorld">
<tag2 name = "Name1">
</tag2>
</tag1>
tag1.tag2~name
tag1~name
tag1~value

Sample Output 

Name1
Not Found!
HelloWorld

Solution - Attribute Parser in C++

#include <bits/stdc++.h>
using namespace std;

int main()
{
int n, q,i;
cin>>n>>q;
string temp;
vector<string> hrml;
vector<string> quer;
cin.ignore();

for(i=0;i<n;i++)
{
    getline(cin,temp);
    hrml.push_back(temp);
}
for(i=0;i<q;i++)
{
    getline(cin,temp);
    quer.push_back(temp);
}

map<string, string> m;
vector<string> tag;

for(i=0;i<n;i++)
{
    temp=hrml[i];
    temp.erase(remove(temp.begin(), temp.end(), '\"' ),temp.end());
    temp.erase(remove(temp.begin(), temp.end(), '>' ),temp.end());

    if(temp.substr(0,2)=="</")
    {
        tag.pop_back();
    }
    else
    {
        stringstream ss;
        ss.str("");
        ss<<temp;
        string t1,p1,v1;
        char ch;
        ss>>ch>>t1>>p1>>ch>>v1;
        string temp1="";
        if(tag.size()>0)
        {
            temp1=*tag.rbegin();
            temp1=temp1+"."+t1;
        }
        else
            temp1=t1;
        tag.push_back(temp1);
        m[*tag.rbegin()+"~"+p1]=v1;
        while(ss)
        {
            ss>>p1>>ch>>v1;
            m[*tag.rbegin()+"~"+p1]=v1;
        }
    }

}

for(i=0;i<q;i++){
    if (m.find(quer[i]) == m.end())
        cout << "Not Found!\n";
    else
        cout<<m[quer[i]]<<endl;
}
return 0;

}

Disclaimer: The above Problem (Attribute Parser) is generated by Hacker Rank but the Solution is Provided by Sloth Coders. This tutorial is only for Educational and Learning Purpose.