Question

Question: Properly convert the solution to this problem from Java to C++

Problem Statement

If you want to write a message anonymously, one way to do it is to cut out letters from headlines in a newspaper and paste them onto a blank piece of paper to form the message you want to write. Given several headlines that you have cut out, determine how many messages from a list you can write using the letters from the headlines. You should only consider each message by itself and not in conjunction with the others, see example 2.

Write the function how_many which takes as parameters a vector<string> headlines containing the headlines which you have cut out as well as a vector<string> messages with the messages you may want to write, and returns an int which is the total number of messages you can write.

Constraints

• All letters that you cut out can be used both as upper or lower case in a message.
• Spaces should be ignored in elements in both headlines and messages.
• headlines will contain between 1 and 50 elements, inclusive.
• messages will contain between 1 and 50 elements, inclusive.
• The length of each element in headlines will be between 1 and 50 characters, inclusive.
• The length of each element in messages will be between 1 and 50 characters, inclusive.
• Each element in headlines will only contain the letters 'A'-'Z', 'a'-'z' and space.
• Each element in messages will only contain the letters 'A'-'Z', 'a'-'z' and space.

Examples

1. headlines =
   
   {"Earthquake in San Francisco",
    "Burglary at musuem in Sweden",
    "Poverty"}
   
   messages =
   
   {"Give me my money back",
    "I am the best coder",
    "TOPCODER"}
   
   Returns: 2
   

   In the first message we have three 'm's, but there are only two 'm's among the headlines (both in the word "museum"), so this message can't be written.

   The second message can be written. Note that the first letter, 'I', only appears as lower case in the headlines, but that's allowed. The last message can also be written, so the method should return 2.

2. headlines = 
   
           {"Programming is fun"}
   
   messages = 
   
           {"program","programmer","gaming","sing","NO FUN"}
   
   Returns: 4
   

   The messages "program", "gaming", "sing" and "NO FUN" can all be written but not "programmer" (no 'e's exist). The method should return 4.

3. headlines = 
   
      {"abcdef","abcdef"}
   
   messages = 
   
      {"AaBbCc","aabbbcc","   ","FADE"}
   
   Returns: 3
   

   All messages except the second one can be written, because it contains three 'b's but there are only two 'b's available. Also note the message only containing spaces - such a message can of course always be written.

Given Function

#include <vector>
#include <string>

int how_many(vector<string> headlines, vector<string> messages) {
// fill in code here
}

Solution (Convert the Java code below to C++).

import java.util.*;

public class Anonymous {
      public int howMany(String[] headlines, String[] messages) {
          
    
          HashMap<Character, Integer> lettersAvailable = countLetters(headlines);

          int wordsWeCanMake = 0; 
          for(String message : messages){
                  String[] oneMessage = {message}; 
                  HashMap<Character, Integer> lettersNeeded = countLetters(oneMessage);
                  boolean canMakeMessage = true;
                  Iterator it = lettersNeeded.entrySet().iterator(); 
                  while (it.hasNext()){
                          Map.Entry pairs = (Map.Entry)it.next();
                          char key = (Character) pairs.getKey(); 
                          if(!(key==' ')){
                                  int numNeeded = (Integer) pairs.getValue();
                                  if(lettersAvailable.containsKey(key)){
                                          int numAvailable = lettersAvailable.get(key); 
                                          canMakeMessage = (numNeeded <= numAvailable) && canMakeMessage;
                                  }
                                  else{
                                          canMakeMessage = false ; 
                                          break;
                                  }
                          }
                          
                      it.remove();
                  }
                  if(canMakeMessage){
                          wordsWeCanMake += 1; 
                  }
          }
          return wordsWeCanMake; 
      }
      
      public HashMap<Character, Integer> countLetters(String[] words){
          HashMap<Character, Integer> counts = new HashMap<Character, Integer>();
          int currentCountOfLetter = 0;
          
          for(String word : words){  
                  for(int i=0; i<word.length(); i++){
                          char letter = Character.toLowerCase(word.charAt(i));
                          if (counts.containsKey(letter)){
                                  currentCountOfLetter = counts.get(letter);
                                  currentCountOfLetter +=1; 
                          }
                          else {
                                  currentCountOfLetter = 1;
                          }
                          counts.put(letter, currentCountOfLetter); 
                  }
          }
          
          return counts; 
      }
      }
}

Answer 1

Name of Proram file Anonymous.cpp

#include<iostream>
#include<vector>
#include<map>
#include<algorithm>
using namespace std;
void countLetters(vector<string> words, map<char,int> &counts) // I have taken the map as input parameter by references just for the optimization purpose
{
map<char,int>:: iterator mit;
// int currentCountOfLetter = 0; no need of this variable in our program
string s;
for(int i=0;i<words.size();i++)
{
s=words[i]; // For each string in the words vector
transform(s.begin(),s.end(),s.begin(),::tolower); // I am using transform method decalred in algorithm header file to conver the whole string in lower case before hand
string::iterator it;
for(it=s.begin();it!=s.end();it++) //I am iterating over strings as it is optimized implementation and it saves time to iterate over indexing
{
char c=*it;
mit=counts.find(c);
if(mit!=counts.end())
{
mit->second++; // increment the count of character already present in the map by one
}
else
{
counts.insert(pair<char,int>(c,1)); // It not prersent in the map then insert in the map with its count 1
}
}
}
}

int howMany(vector<string> headlines, vector<string> messages)
{
char c;
vector<string> v;
int wordsWeCanMake = 0;
bool canMakeMessage;
map<char,int> lettersAvailable;
map<char,int> lettersNeeded;
string m;
map<char,int>::iterator ait;
map<char,int>::iterator nit;
countLetters(headlines,lettersAvailable);
for(int i=0;i<messages.size();i++)
{
canMakeMessage=true;
m=messages[i];
v.push_back(m);
countLetters(v,lettersNeeded);
for(nit=lettersNeeded.begin();nit!=lettersNeeded.end();nit++)
{
c = nit->first;
if(c!=' ')
{
ait = lettersAvailable.find(c);
if(ait!=lettersAvailable.end())
{
if(!(nit->second <= ait->second))
{
canMakeMessage = false;
break;
}
}
else
{
canMakeMessage = false;
break;
}
}
}
v.clear();
lettersNeeded.clear();
if(canMakeMessage)
{
wordsWeCanMake++;
}
}
return wordsWeCanMake;
}

int main() // cross verified different examples by commenting the lines of code and it is working properly, if you want give it a try
{
vector<string> headlines;
headlines.push_back("Earthquake in San Francisco");
headlines.push_back("Burglary at musuem in Sweden");
headlines.push_back("Poverty");
headlines.push_back("Programming is fun");
headlines.push_back("abcdef");
headlines.push_back("abcdef");
vector<string> messages;
messages.push_back("Give me my money back");
messages.push_back("I am the best coder");
messages.push_back("TOPCODER");
messages.push_back("program");
messages.push_back("programmer");
messages.push_back("gaming");
messages.push_back("sing");
messages.push_back("NO FUN");
messages.push_back("AaBbCc");
messages.push_back("aabbbcc");
messages.push_back("FADE");
cout<<howMany(headlines,messages);
return 0;
}

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// I have kept the variables names sames as in java code to understand it better
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