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



Solution :


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//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;
}





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1 Comments
  • Anonymous
    Anonymous Monday, July 11, 2022

    Great and probably the best solution of this problem !

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