C++ Identifying the Frequency of words occurring in a sentence - c++

What is the best STL to use for this task? I've been using Map,
and I couldn't get it to work. I'm not sure how I am supposed to check the number of same words that occur in the sentence for example:
I love him, I love her, he love her.
So I want the program to prompt the user to enter an integer, lets say i enter 3, the output will be love as the same word occurs 3 times in the sentence. But what method to use if I want to do a program like this?
Currently my program prompts for the user to enter the word, and then it shall return how many time that word occurs, which for word love, is 3. but now i want it the other way round. Can it be done? Using which STL will be better?

I assume you use a map to store the number of occurrences.
Well,you first have to understand this,since you are using a map,the key is unique while the stored data may not be unique.
Consider a map, x
with contents
x["I"]=3
x["Love"]=3
x["C"]=5
There is unique a mapping from the key to the value,and not the other way round,if you want this one to one mapping ,i would suggest a different data structure.If you want to use map,and still search for an element,using STL search function or your own.Or you can write your search function.
search().
map<string,int>::iterator ser;
cin>>check;
for(ser=x.begin();ser!=x.end();++ser)
{
if(ser->second==check)
{
cout<<"Word"<<ser->first<<endl;
break;
}
}

First build the mapping from word to count and then build the reverse multi-mapping from that. Finally, you can determine which words occur with a given frequency:
#include <algorithm>
#include <iostream>
#include <iterator>
#include <map>
#include <set>
#include <sstream>
#include <string>
#include <utility>
int main()
{
std::string str("I love him, I love her, he love her");
std::istringstream ss(str);
std::istream_iterator<std::string> begin(ss);
std::istream_iterator<std::string> end;
std::map<std::string, int> word_count;
std::for_each(begin, end, [&](const std::string& s)
{
++word_count[s];
});
std::multimap<int, std::string> count_words;
std::for_each(word_count.begin(), word_count.end(),
[&](const std::pair<std::string, int>& p)
{
count_words.insert(std::make_pair(p.second, p.first));
});
auto its = count_words.equal_range(3);
std::for_each(its.first, its.second,
[](const std::pair<int, std::string>& p)
{
std::cout << p.second << std::endl;
});
}

/******************************************************************
Name : Paul Rodgers
Source : HW1.CPP
Compiler : Visual C++ .NET
Action : Program will read in from standard input and determine the
frequency of word lengths found in input. An appropriate
table is also displayed. Maximum word length is 15 characters
words greater then 15 are counted as length 15.
Average word length also displayed.
Note : Words include hyphenated and ones with apostrophes. Words with
apostrophes, i.e. Jim's, will count the apostrophe as part of the
word length. Hyphen is counted if word on same line, else not.
Also an int array is used to hold the number of words with
length associated with matching subscript, with subscript 0
not being used. So subscript 1 corresponds to word length of 1,
subscript 2 to word length of 2 and so on.
------------------------------------------------------------------------*/
#include <iostream>
#include <ctype.h>
#include <iomanip>
using namespace std;
int NextWordLength(void); // function prototypes
void DisplayFrequencyTable(const int Words[]);
const int WORD_LENGTH = 16; // global constant for array
void main()
{
int WordLength; // actual length of word 0 to X
int NumOfWords[WORD_LENGTH] = {0}; // array holds # of lengths of words
WordLength = NextWordLength();
while (WordLength) // continue to loop until no word, i.e. 0
{ // increment length counter
(WordLength <= 14) ? (++NumOfWords[WordLength]) : (++NumOfWords[15]);
WordLength = NextWordLength();
}
DisplayFrequencyTable(NumOfWords);
}
/********************** NextWordLength ********************************
Action : Will determine the length of the next word. Hyphenated words and
words with apostrophes are counted as one word accordingly
Parameters : none
Returns : the length of word, 0 if none, i.e. end of file
-----------------------------------------------------------------------*/
int NextWordLength(void)
{
char Ch;
int EndOfWord = 0, //tells when we have read in one word
LengthOfWord = 0;
Ch = cin.get(); // get first character
while (!cin.eof() && !EndOfWord)
{
while (isspace(Ch) || ispunct(Ch)) // Skips leading white spaces
Ch = cin.get(); // and leading punctation marks
if (isalnum(Ch)) // if character is a letter or number
++LengthOfWord; // then increment word length
Ch = cin.get(); // get next character
if ((Ch == '-') && (cin.peek() == '\n')) //check for hyphenated word over two lines
{
Ch = cin.get(); // don't count hyphen and remove the newline char
Ch = cin.get(); // get next character then on next line
}
if ((Ch == '-') && (isalpha(cin.peek()))) //check for hyphenated word in one line
{
++LengthOfWord; // count the hyphen as part of word
Ch = cin.get(); // get next character
}
if ((Ch == '\'') && (isalpha(cin.peek()))) // check for apostrophe in word
{
++LengthOfWord; // count apostrophe in word length
Ch = cin.get(); // and get next letter
}
if (isspace(Ch) || ispunct(Ch) || cin.eof()) // is it end of word
EndOfWord++;
}
return LengthOfWord;
}
/*********************** DisplayFrequencyTable **************************
Action : Will display the frequency of length of words along with the
average word length
Parameters
IN : Pointer to array holding the frequency of the lengths
Returns : Nothing
Precondition: for loop does not go beyond WORD_LENGTH
------------------------------------------------------------------------*/
void DisplayFrequencyTable(const int Words[])
{
int TotalWords = 0, TotalLength = 0;
cout << "\nWord Length Frequency\n";
cout << "------------ ----------\n";
for (int i = 1; i <= WORD_LENGTH-1; i++)
{
cout << setw(4) << i << setw(18) << Words[i] << endl;
TotalLength += (i*Words[i]);
TotalWords += Words[i];
}
cout << "\nAverage word length is ";
if (TotalLength)
cout << float(TotalLength)/TotalWords << endl;
else
cout << 0 << endl;
}

#include<iostream>
#include<string>
#include<vector>
#include<cstddef>
#include<map>
using std::cout;
using std::cin;
using std::string;
using std::endl;
using std::vector;
using std::map;
int main() {
cout << "Please enter a string: " << endl;
string str;
getline(cin, str, '\n');
size_t str_len = str.size();
cout << endl << endl;
size_t i = 0, j = 0;
bool pop = false;
map<string, int> myMap;
for (size_t k = 0; k < str_len-1; k++) {
if (((k == 0) && isalpha(str[0])) || (!(isalpha(str[k-1])) && isalpha(str[k])))
i = k;
if ( isalpha(str[k]) && !(isalpha(str[k+1])) ) {
j = k;
pop = true;
}
if ( (k == str_len-2) && isalpha(str[k+1]) ) {
j = k+1;
pop = true;
}
if ( (i <= j) && pop ) {
string tmp = str.substr(i, j-i+1);
cout << tmp << '\t';
myMap[tmp]++;
pop = false;
}
}
cout << endl << endl;
map<string, int>::iterator itr, end = myMap.end();
for (itr = myMap.begin(); itr != end; itr++)
cout << itr->first << "\t - - - - - \t" << itr->second << endl;
cout << endl;
return 0;
}

Related

Code gets stuck in loop (but loop itself doesn't keep going). C++ string

I am currently self-studying C++ with Schaum's outline book (which covers mostly C contents, or so I've been told, but whatever) and I have encountered some trouble with problem 9.8.
You are supposed to count the number of appearances of every different word in a given c++ string, for which I assumed each word was separated from the next one by a white space, a newline or a dot or coma (followed in these two last cases by another white space).
My code is the following:
#include <iostream>
#include <string>
using namespace std;
int main()
{ string s;
cout << "Enter text (enter \"$\" to stop input):\n";
getline(cin,s,'$');
string s2 = s, word;
int ini = 0, last, count_word = 0;
int count_1 = 0, count_2 = 0, count_3 = 0;
cout << "\nThe words found in the text, with its frequencies, are the following:\n";
for (ini; ini < s.length(); )
{ // we look for the next word in the string (at the end of each iteration
// ini is incremented in a quantity previous_word.length()
last = ini;
cout << "1: " << ++count_1 << endl;
while(true)
{ if (s[last] == ' ') break;
if (s[last] == '\n') break;
if (s[last] == ',') break;
if (s[last] == '.') break;
if (last > s.length()-1 ) break;
++last;
cout << "2: " << ++count_2 << endl;
}
--last; // last gives the position within s of the last letter of the current word
// now me create the word itself
word = s.substr(ini,last-ini+1); //because last-ini is word.length()-1
int found = s2.find(word);
while( found != s2.length() ) // the loop goes at least once
++count_word;
s2.erase(0,found+word.length()); // we erase the part of s2 where we have already looked
found = s2.find(word);
cout << "3: " << ++count_3 << endl;
cout << "\t["<<word<<"]: " << count_word;
++last;
s2 = s;
s2.erase(0,ini + word.length()); // we do this so that in the next iteration we don't look for
// the new word where we know it won't be.
if (s[last] == ' ' || s[last] == '\n') ini = last + 1;
if (s[last] == ',' || s[last] == '.') ini = last + 2;
count_word = 0;
}
}
When I ran the program nothing was sshown on screen, so I figured out that one of the loops must had been stuck (that is why I defined the variables count_1,2 and 3, to know if this was so).
However, after correctly counting the number of iterations for the fist word to be found, nothing else is printed and all I see is the command prompt (I mean the tiny white bar) and I cannot even stop the program by using ctrl z.
This is a very complicated method for a very simple problem. You can just use a stringstream to extract each word seperated by a white space. You then just take the extracted word and increment the word counter using a std::map<std::string, int>.
My take on this:
#include <iostream>
#include <map>
#include <string>
#include <sstream>
int main() {
std::map<std::string, int> word_to_count;
std::string in;
std::getline(std::cin, in);
std::stringstream s(in);
std::string temp_word;
while (s >> temp_word) {
word_to_count[temp_word]++;
}
for (const auto& x : word_to_count) {
std::cout << x.first << ": " << x.second << std::endl;
}
return 0;
}
input
hello world hello world test
Output
hello: 2
test: 1
world: 2
Keep in mind this is just one of many possible solutions, so just take this as inspiration :).

c++ count words in array

I need to write a function that gets a string and count how many words there are in the string and how many letters. And then calculate the average of it.
A word in a string is a sequence of letters and numbers separated by one or more spaces.
First of all I have to check if the string is correct. The string must contain only lowercase letters, uppercase letters, and numbers only.
i didnt menage to count all sort of words correctly and also my function doesnt count the last letter.
#include <iostream>
using namespace std;
#include <string.h>
#define SIZE 50
float checkString(char string[]) {
float wordCounter = 0;
float letterCounter = 0;
bool isLegit = true;
int i = 0;
while (isLegit) {
if (((string[i] >= 48 && string[i] <= 57) ||
(string[i] >= 65 && string[i] <= 90) ||
(string[i] >= 97 && string[i] <= 122 ))) {
for (int j = 0; j <= strlen(string); j++) {
if ((string[j - 1] != ' ' && string[j] == ' ' &&
string[i + 1] != ' ')
|| j == (strlen(string) - 1)) {
wordCounter++;
}
else if (string[j] != ' ') {
letterCounter++;
cout << string[j];
}
}
cout << " The avareage is : " << (letterCounter /
wordCounter) << endl;
isLegit = false;
}
else {
return -1;
isLegit = false;
}
}
cout << "Number of words " << wordCounter << endl;
cout << "Number of letters " <<letterCounter << endl;
}
int main() {
char string[SIZE];
cout << "please enter a sentence " << endl;
cin.getline(string, SIZE);
checkString(string);
}
Instead of using char[] for strings, I suggest that you use std::string which can grow and shrink dynamically. It's one of the most common types to use in the standard C++ library. You can also make use of stringstreams which lets you put a string inside it and then you can extract the contents of the stringstream using >>, just like when reading from std::cin.
Example with comments in the code:
#include <iostream>
#include <sstream> // std::stringstream
#include <string> // std::string
// use std::string instead of a char[]
float checkString(const std::string& string) {
// put the string in a stringstream to extract word-by-word
std::istringstream is(string);
unsigned words = 0;
unsigned letters = 0;
std::string word;
// extract one word at a time from the stringstream:
while(is >> word) {
// erase invalid characters:
for(auto it = word.begin(); it != word.end();) {
// Don't use magic numbers. Put the character literals in the code so
// everyone can see what you mean
if((*it>='0' && *it<='9')||(*it>='A' && *it<='Z')||(*it>='a' && *it<='z')) {
// it was a valid char
++it;
} else {
// it was an invalid char, erase it
it = word.erase(it);
}
}
// if the word still has some characters in it, make it count:
if(word.size()) {
++words;
letters += word.size();
std::cout << '\'' << word << "'\n"; // for debugging
}
}
std::cout << "Number of words " << words << "\n";
std::cout << "Number of letters " << letters << "\n";
std::cout << "The average number of letters per word is "
<< static_cast<float>(letters) / words << '\n';
return 0.f; // not sure what you are supposed to return, but since the function
// signature says that you should return a float, you must return a float.
}
int main() {
checkString(" Hello !!! World, now let's see if it works. ");
}
I would like to add an additional answer. This answer is based on "more-modern" C++ and the usage of algorithms. You want to solve 3 tasks:
Check, if string is OK and matched to your expectations
Count the number of words in the given string
Count the number of letters
Calculate the ratio of words/letters
For all this you may use existings algorithms from the C++ standard library. In the attached example code, you will see a one-liner for each task.
The statements are somehow very simple, so that I will not explain much more. If there should be a question, I am happy to answer.
Please see here one possible example code:
#include <iostream>
#include <string>
#include <iterator>
#include <regex>
#include <algorithm>
#include <tuple>
#include <cctype>
std::regex re("\\w+");
std::tuple<bool, int, int, double> checkString(const std::string& str) {
// Check if string consists only of allowed values, spaces or alpha numerical
bool stringOK{ std::all_of(str.begin(), str.end(), [](const char c) { return std::isalnum(c) || std::isspace(c); }) };
// Count the number of words
int numberOfWords{ std::distance(std::sregex_token_iterator(str.begin(),str.end(), re, 1), {}) };
// Count the number of letters
int numberOfLetters{ std::count_if(str.begin(), str.end(), isalnum) };
// Return all calculated values
return std::make_tuple(stringOK, numberOfWords, numberOfLetters, static_cast<double>(numberOfWords)/ numberOfLetters);
}
int main() {
// Ask user to input string
std::cout << "Please enter a sentence:\n";
// Get string from user
if (std::string str{}; std::getline(std::cin, str)) {
// Analyze string
auto [stringOk, numberOfWords, numberOfLetters, ratio] = checkString(str);
// SHow result
std::cout << "\nString content check: " << (stringOk ? "OK" : "NOK") << "\nNumber of words: "
<< numberOfWords << "\nNumber of letters: " << numberOfLetters << "\nRatio: " << ratio << "\n";
}
return 0;
}
Of course there are many more other possible solutions. But, because of the simplicity of this solution, I showed this variant.

Counting Letters in Word

I am having troubles how to start creating program which counts letters in words in .txt file, and print it if letter number is bigger than 20.. Does someone know how to start or have some similar program to help me..Thank you
In general,I start programs by:
#include <iostream>
void Pause();
int main()
{
std::cout << "My program\n";
Pause();
return 0;
}
void Pause()
{
std::cout << "\n\nPaused. Press ENTER to continue.\n";
std::cin.ignore(100000, '\n');
}
After I get the above working, I add content.
Assuming the ASCII encoding, you could make some assumptions:
int counts[256] = {0};
char c;
while (my_file.read(&c, 1))
{
counts[c]++;
}
The above loop will count all occurrences of the letters in the text file. The next task would be to print the data:
for (int i = ' ' + 1; i < 127; ++i)
{
std::cout << "'" << ((char) i) << "': " << counts[i] << \n;
}
This will print out the counts for all printable characters, starting with the character after a space.
This is a popular beginning assignment, so you should be able to find more examples on the internet.
Edit 1: For counts of words
The std::string and std::map are your friends.
std::string word;
std::map<std::string, int> database;
while (my_file >> word)
{
std::map<std::string, int>::iterator iter;
iter = database.find(word);
if (iter != database.end())
{
map[word]++;
}
else
{
map[word] = 1;
}
}
The above loop builds a database of [word, count] pairs. If the word exists, the count is incremented.
The next part is to iterate through the database and only print the words that have a length > 20. This is left as an exercise for the reader.

Q: C++ - Reversing a string (sentence and word) using a class

For my code, I am trying to create a class with two functions that:
Display a cstring where each word is reversed
Display an entire cstring reversed
My two test sentences are "Hi There" and "To Be", so the output is:
erehT iH
eB oT
iH erehT
oT eB
Here is my code:
#include <iostream>
#include <cstring>
using namespace std;
class cStringType {
public:
char sentenceInput[80]; //Member variable
void reverse_sentence(); //Member function
void reverse_words(); //Member function
}; //Bottom of cStringType
int main()
{
cStringType sentence1, sentence2;
//Objects declared of cStringType
cout << "Please enter a sentence!\n" << endl;
cin.get(sentence1.sentenceInput, 79, '\n');
cin.ignore(80, '\n');
cout << "\nPlease enter another sentence!\n" << endl;
cin.get(sentence2.sentenceInput, 79, '\n');
cout << "\nThe first sentence reversed: ";
sentence1.reverse_sentence();
cout << endl;
cout << "The second sentence where each word is reversed: ";
sentence2.reverse_words();
cout << endl;
cout << endl;
cout << "The first sentence where each word is reversed: ";
sentence1.reverse_words();
cout << endl;
cout << "The second sentence reversed: ";
sentence2.reverse_sentence();
cout << endl;
return 0;
}
void cStringType::reverse_sentence()
{
char reverse_sentence;
//Reverse entire sentence using loop
for (int i = 0; i < strlen(sentenceInput) / 2; i++)
{
//Reverse the sentence using the length of the
//variable in the class
reverse_sentence = sentenceInput[i];
//First get the user input
//Set your variable equal to the variable in the class
sentenceInput[i] = sentenceInput[strlen(sentenceInput) - i - 1];
//Then reverse the characters and word order
//Starts from the last character in the array
//and goes backwards to 0
sentenceInput[strlen(sentenceInput) - i - 1] = reverse_sentence;
//Set the variable equal to the result
//sentenceInput is now the reverse of the user input in main
}
cout << sentenceInput << endl;
//Output of the new sentence
}
void cStringType::reverse_words()
{
int beginning, end, j = 0;
char reverse_words;
//Reverse each word separately using loop
for (int i = 0; i <= strlen(sentenceInput); i++)
//Get the length of the sentence in the class
{
if (sentenceInput[i] == ' ' || sentenceInput[i] == '\0')
//Check for spaces or null characters
//This allows only the letters of each word to be
//reversed, not the entire sentence
{
for (beginning = j, end = i - 1;
beginning < (i + j) / 2; beginning++, end--)
//j is the beginning of the array; increases
//i is the end of the array; decreases
{
reverse_words = sentenceInput[beginning];
//Set a variable equal to the first
//word in the original user input
sentenceInput[beginning] = sentenceInput[end];
//Set the first letter of a word equal to
//the last letter of a word
sentenceInput[end] = reverse_words;
//Set the result equal to the variable
//sentenceInput is now the user input where each
//word is reversed
}
}
j = i + 1;
}
cout << sentenceInput << endl;
//Output of the new sentence
}
When I try to run the code, the output becomes something like this:
Please enter a sentence!
Hi There
Please enter another sentence!
To Be
The first sentence reversed: erehT iH
The second sentence where each word is reversed: oT eB
The first sentence where each word is reversed: There Hi
The second sentence reversed: Be To
I tried fixing it, but to no avail. The output is never correct.
Is there some way to fix this issue? Or better yet, to simplify the code? I believe the issue is with the code in the function.
The main problem with your code is that it's using the same buffer for both transformations. In other words: you are reversing the words in the same string which you've already reversed entirely. So you need to have another copy of the original string to do these independently.
Regarding simplifying your code you need to define a function that would reverse a string given a pointer and size or begin and end pointers. Then you can use this function on your entire string or on every word you find while searching for a space character:
char *begin = sentenceInput; //points to the beginning of the word
char *end = sentenceInput + strlen(sentenceInput);
for (char *it = begin; it != end; ++it)
if (*it == ' ') {
reverse(begin, it);
begin = it + 1;
}
reverse(begin, end); //reverse the last word
The reverse function can be either std::reverse, which can be used in the above code and on the entire string as follows:
std::reverse(sentenceInput, sentenceInput + strlen(sentenceInput))
or you can create a similar function like this:
void reverse(char *begin, char *end)
{
--end; //point to the last character instead of one-past-last
while (begin < end)
std::swap(*begin++, *end--);
}
I would suggest using stack for it, it is a natural way of looking at it.
so
#include <stack>
and then the function would be like that
void cStringType::reverse_words()
{
int beginning, end, j = 0;
char reverse_words;
stack<char> lastWord;
//Reverse each word separately using loop
for (int i = 0; i <= strlen(sentenceInput); i++)
//Get the length of the sentence in the class
{
if (sentenceInput[i] == ' ' || sentenceInput[i] == '\0')
//Check for spaces or null characters
//This allows only the letters of each word to be
//reversed, not the entire sentence
{
//we want to print the last word that was parsed
while(!lastWord.empty())
{
//we print in the reverse order the word by taking off the stack char by char
cout<< lastWord.top();
lastWord.pop();
}
cout<<" ";
}
//if the letter is not space or end of string then push it on the stack
else
lastWord.push(sentenceInput[i]);
j = i + 1;
}
cout << sentenceInput << endl;
//Output of the new sentence
}

Anagram generator for C++ (not using STL)

I am trying to create an anagram solver just using a very basic, procedural approach. I am finding out that I probably should have done this using classes, but now it is too late and my assignment is about due. Any suggestions on how to figure this out would be great!
Basically, this is what the algorithm should do:
Get all words in the dictionary; store them in a container
Get a word from the user; quit if appropriate
Get all permutations of the word that the user entered
Strip the word the user entered from the permutations
Strip all words in the permutation collection that aren't also in the dictionary I collected in part 1
Now for the last step, I must make sure that I don't display duplicate anagrams (i.e. anagrams which contain the same letter, such as "loop"). I cannot seem to get this check to work, which is noted below with under the TODO comment block.
Any suggestions would be awesome!!
#include <iostream>
#include <fstream>
#include <string>
//
// Change size below to accomodate more anagrams and dictionary words
//
#define MAX_ANGM_SIZE 4096
#define MAX_WORD_SIZE 1048576
using namespace std;
//
// Determines whether anagram is valid or not; will not display word
// which user entered or words not contained in dictionary
//
bool isValidAnagram(string word, string userWord,
string dictionary[], unsigned int listIdx)
{
for(unsigned int idx = 0; idx < listIdx; ++idx)
{
if(word == userWord)
return false;
else if (word == dictionary[idx])
return true;
}
return false;
}
//
// Determines whether user's word is contained in the dictionary
// or not
//
bool isValidWord(string word, string dictionary[],
unsigned int listIdx)
{
for(unsigned int idx = 0; idx < listIdx; ++idx)
{
if(word == dictionary[idx])
return true;
}
return false;
}
//
// TODO:This function should test for duplicate anagrams and return
// true if duplicates are found.
//
bool isRepeated(string anagrams[], unsigned int anaIdx)
{
for(unsigned int idx = anaIdx; idx != 0; --idx)
{
if(anagrams[idx] == anagrams[anaIdx])
return true;
else
return false;
}
return false;
}
//
// Only display elements in array which aren't blank and don't
// display duplicate anagrams; notify user if no anagrams
// were found.
//
void displayAnagrams(string anagrams[], unsigned int next)
{
int flag = 0;
for (unsigned int idx = 0; idx < next; ++idx)
{
if((anagrams[idx] != "") || (!(isRepeated(anagrams, idx))))
{
if(idx == 1)
cout << " Anagrams: ";
if(idx > 0)
flag = 1;
cout << anagrams[idx] << " ";
}
else
continue;
}
if(flag == 0)
cout << " no anagrams found" << endl;
}
static void swap(char &c1, char &c2)
{
char temp = c1;
c1 = c2;
c2 = temp;
}
//
// Pass in word to be altered, the userWord for comparison, the array to store
// anagrams, the dictionary for comparison, the count for the number of anagrams
// and the count for number of dictionary words
//
static void permute(string word, string userWord, int k, string anagrams[],
string dictionary[], unsigned int &next, unsigned int listIdx)
{
if(k == word.length()-1)
{
if(isValidAnagram(word, userWord, dictionary, listIdx))
anagrams[next] = word;
++next;
}
else
{
for(int idx = k; idx < word.length(); ++idx)
{
swap(word[k], word[idx]);
permute(word, userWord, k+1, anagrams, dictionary, next, listIdx);
}
}
}
//
// Create container to store anagrams, validate user's word in dictionary, get all
// of the anagrams, then display all valid anagrams
//
void getAnagrams(string word, string dictionary[], unsigned int listIdx)
{
string anagrams[MAX_ANGM_SIZE];
unsigned int next = 0;
if(isValidWord(word, dictionary, listIdx))
{
permute(word, word, 0, anagrams, dictionary, next, listIdx);
}
else
{
cerr << " \"" << word << "\"" << " is not a valid word" << endl;
return;
}
displayAnagrams(anagrams, next);
}
//
// Read in dictionary file, store contents of file in a list, prompt
// the user to type in words to generate anagrams
//
int main()
{
string file;
string word;
string quit = "quit";
string dictionary[MAX_WORD_SIZE];
unsigned int idx = 0;
cout << "Enter a dictionary file: ";
cin >> file;
cout << "Reading file \"" << file << "\"" << endl;
cout << endl;
ifstream inFile(file.c_str());
if(!(inFile.is_open()))
{
cerr << "Can't open file \"" << file << "\""
<< endl;
exit(EXIT_FAILURE);
}
while(!inFile.eof())
{
inFile >> dictionary[idx];
++idx;
}
inFile.close();
while(true)
{
cout << "Enter a word: ";
cin >> word;
if(word == quit) break;
getAnagrams(word, dictionary, idx);
cout << endl;
}
return 0;
}
You may want to rethink your step (3). If the user enters a 12-letter word you have 479,001,600 permutations of it which will probably be impractical to assemble all at once (and if that's not, then a 16-letter word will be...).
Instead, try thinking about how you could store the words and look up potential anagrams in a way that doesn't require you to do that.
Edit: I get that ability to solve largeish words may not be your biggest concern at this point, but it might actually make your fourth and fifth steps easier if you do them by assembling the set of valid words rather than starting with all possibilities and removing all the ones that don't match. 'Removing' an item from an array is a bit awkward since you have to shuffle all the following items up to fill in the gap (this is exactly the kind of thing that STL manages for you).
Better algorithm : don't store your word, but store a tupple containing (your word, sorted letters). Moreover, you sort that big storage by the second key (hint, you could use a sqlite database to do the work for you and use an index (can't be unique!))
E.g. to store
"Florent", "Abraham","Zoe"
you would store in memory
("aaabhmr", "abraham"),("eflnort","florent"),("eoz","zoe")
When you got your word from your user, you just use same "sorting letter inside word" algorithm.
Then you look for that pattern in your storage, and you find all anagrams very quickly (log(size of dictionary)) as it's sorted. Of course, original words are the second elements of your tuple.
You can do that using classes, standard structures, a database, up to you to choose the easiest implementation (and the one fitting your requirements)