Attribute Parser in C++ – Hacker Rank Solution

Attribute Parser in C++ - Hacker Rank Solution
Attribute Parser in C++ – Hacker Rank Solution

Problem

We have defined our own 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>

For example:

<tag1 value = "HelloWorld">
<tag2 name = "Name1">
</tag2>
</tag1>

The attributes are referenced as:

tag1~value
tag1.tag2~name

You are given the source code in HRML format consisting of N lines. You have to answer Q  queries. Each query asks you to print the value of the attribute specified. Print “Not Found!” if there isn’t any such attribute.


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.tagi4……tagim ~attr-name where m>=1 and tagi1.tagi2….tagim are valid tags in the input.

Constraints :

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

Output Format :

Print the value of the attribute for each query. Print “Not Found!” without quotes if there is no such attribute in the source program.


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


Attribute Parser in C++ – Hacker Rank Solution

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#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 in C++ ) is generated by Hackerrank but the Solution is Provided by Chase2Learn. This tutorial is only for Educational and Learning purposes. Authority if any of the queries regarding this post or website fill the following contact form thank you.

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