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say hello world with C++ - Solution in Hacker Rank - hackerranksolutions8

  Objective This is a simple challenge to help you practice printing to  stdout . You may also want to complete  Solve Me First  in C++ before attempting this challenge. We’re starting out by printing the most famous computing phrase of all time! In the editor below, use either  printf  or  cout  to print the string  Hello ,World!  to  stdout . The more popular command form is  cout . It has the following basic form: cout<<value_to_print<<value_to_print; Any number of values can be printed using one command as shown. The  printf  command comes from C language. It accepts an optional format specification and a list of variables. Two examples for printing a string are: printf("%s", string);   printf(string); Note that neither method adds a newline. It only prints what you tell it to. Output Format Print   Hello ,World!   to stdout. Sample Output Hello, World! Solution:- //Say Hello, ...

Attribute Parser in C++ – Solution in Hacker Rank - hackerranksolutions8

 

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 :

//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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