## Algorithm

Problem Name: Attribute Parser

In this HackerRank Functions in C++ programming problem solution,

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:

The closing tags follow the format:

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

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~`. Results of the queries are:

```Query                 Value
tag1~value            "value"
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 -atr  - 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 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.tag2~name
tag1~name
tag1~value
``````

Sample Output

``````Name1
HelloWorld
``````

## Code Examples

### #1 Code Example with C++ Programming

```Code - C++ Programming```

``````
#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())
else
cout << m[quer[i]] << endl;
}
return 0;
}
``````
Copy The Code &

Input

cmd
4 3 tag1.tag2~name tag1~name tag1~value

Output

cmd