C++, count repeated words in the string and display - c++

Question

I have string say "walk talk, can't won't Won't woN'T talk." I want to count the reapeated words and display.
Note: it is not case sensitive.
I have used delimeter
strtok(string, ",.;:\"!? -_\n\t*()##=+");
and saved it in
char *temp[100];
Now how can I check for repeatation of words? And display as below
3 won't
2 talk
1 can't
1 walk
it should display from highest repeat to lowest. And if the repeatation is same then display alphabetic order.
Sorry for my bad english.

Answer 1

Use a std::string to hold the result of the strtok(). Then create a std::map<string, int> to hold the count of the times the string (the key) has occurred.
You can populate the map with:
std::map<string, int> myMap;
myMap[tokenizedWord]++; //Increase count of word.
You can then cycle through the map content and print out wherever the integer value is greater than 2.
for (std::map<string, int>::iterator iter = myMap.begin(); iter != myMap.end(); ++iter)
{
if (iter->second > 1)
std::cout << "Duplicated word: " << iter->first << " count = " << iter->second;
}
I'll let you figure out how to traverse it in order. You can put the values in a vector or something and use std::sort before printing or whatever else you like. Maps, unfortunately, are associative containers and you can't sort them as it breaks their internal ordering.
Background Info on std::map
A map is an associative array meaning that every key maps to a specific value, and keys are unique. You can actually create a multimap where keys are not unique, so that's why this is important.
Basically, since keys are unique, you can access or create an element just by using the key as the array index.
For example:
//Create a map and insert a couple things into it - prices of meat?
std::map<string, float> myMap;
myMap["Chicken"] = 4.99;
myMap["Turkey"] = 6.99;
//Retrieve the price of something using the key.
std::cout << "Chicken costs " << myMap["Chicken"] << std::end;
You can do standard insertion and location operations on a map too, but the associative array syntax is just simpler, so why bother? :)
PS: To fully answer your comment, just in case, the ++ at the end of myMap[tokenizedWord]++ is just saying to increment the value of the integer value stored for that key by 1. You could just as well do myMap[tokenizedWord] = myMap[tokenizedWord] + 1 OR you could also do myMap[tokenizedWord] += 1.

Answer 2

a complete implementation of your problem (Let me know if you want a sample code for sorting):
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <errno.h>
#define ARRAY_ELEMS_COUNT(A) sizeof(A)/sizeof(*A)
typedef struct _word_t
{
char *word;
int occurr_count;
struct _word_t *next;
} word_t;
typedef struct _word_list_t
{
struct _word_t *head;
struct _word_t *tail;
int elems_count;
} word_list_t;
/* Creation of the words list */
word_list_t *make_list(void)
{
word_list_t *w_list = (word_list_t *)malloc(sizeof (struct _word_list_t));
if (w_list == NULL)
{
fprintf(stderr, "malloc faild --> %s\n", strerror(errno));
return NULL;
}
w_list->head = w_list->tail = NULL;
w_list->elems_count = 0;
return w_list;
}
int list_word_lookup(word_list_t *w_list, char *word)
{
word_t *temp_word = w_list->head;
while(temp_word)
{
if (strcmp(temp_word->word, word) == 0)
{
/* We got it before, increment the count */
temp_word->occurr_count++;
return 1;
}
else
{
temp_word = temp_word->next;
}
}
return 0;
}
/* Adding new words to the list of words if they are not present, otherwise increment their occurrence count */
/* TODO : Sort the list using Merge sort for performance */
int adding_to_list(word_list_t *w_list, char *word)
{
int return_status = 0;
char *tmp_word = (char *)malloc(sizeof(char)*(strlen(word) + 1));
word_t *new_word = (word_t *)malloc(sizeof(struct _word_t));
/* Empty list */
if (w_list->head == NULL)
{
strcpy(tmp_word, word);
new_word->word = tmp_word;
new_word->occurr_count = 1;
w_list->head = w_list->tail = new_word;
w_list->head->next = NULL;
w_list->elems_count++;
}
else
{
/* The list is not empty */
/* Checking if the word exist in the list */
return_status = list_word_lookup(w_list, word);
if (return_status == 1)
{
fprintf(stdout, "WE got this word before --> increment count\n");
}
else
{
strcpy(tmp_word, word);
new_word->word = tmp_word;
new_word->occurr_count = 1;
w_list->tail->next = new_word;
w_list->tail = new_word;
w_list->tail->next = NULL;
}
}
return 0;
}
void words_list_dump(word_list_t *w_list)
{
word_t *temp;
for (temp = w_list->head; temp; temp = temp->next) {
fprintf(stdout, "Word : %s -- Count = %d\n", temp->word, temp->occurr_count);
}
}
/* Destroying all words */
void free_words(word_list_t *w_list)
{
word_t *temp;
for (temp = w_list->head; temp; temp = temp->next) {
/* Freeing the word string */
free(temp->word);
/* Freeing the word */
free(temp);
}
w_list->head = NULL;
w_list->tail = NULL;
}
/* Destroying the words list */
void free_words_list(word_list_t *w_list)
{
if (!w_list)
{
return;
}
free_words(w_list);
free(w_list);
}
/* TODO : create a function that converts your input text to a char ** array, so you can pass it to adding_to_list */
/* For testing */
int main(int argc, char **argv)
{
const char *string[] = {"Hello", "World", "Stackoverflow", "C", "Hello", "C", "WORDS", "words", "List", "list", "Hello", "World", "Count"};
word_list_t *my_list = make_list();
int i;
for (i = 0; i < ARRAY_ELEMS_COUNT(string); i++)
adding_to_list(my_list, string[i]);
words_list_dump(my_list);
free_words_list(my_list);
return 0;
}

Answer 3

Here is an answer using strtok but without std::map. In one pass of string, every word in is checked against previous words and repeats are counted.
#include <iostream>
using std::cin;
using std::cout;
using std::endl;
#include <string>
using std::string;
#include <vector>
using std::vector;
#include <cstring>
using std::tolower;
int main()
{
char *strin;
string inputstr;
vector<string> svec;
vector<int> cvec;
char *pch;
int unique_word_count=0;
while(getline(cin,inputstr))
{
//token-ize the string
//First string
strin = &inputstr[0];
pch = std::strtok(strin," ,-");
bool unique_word_found = true;
//subsequent words
while (pch != NULL)
{
string word(pch);
for(string::size_type i=0; i < word.size(); i++)
word[i]=tolower(word[i]);
//first word
//just add to svec and no comparisons
if(unique_word_count==0)
{
svec.push_back(word);
cvec.push_back(1);
cvec[unique_word_count++]=1; //init count of first word
//next word
pch = std::strtok(NULL, " ,-");
unique_word_found = true; //reset flag
continue;
}
//start comparing with other words currently in string vector
//do not do this if only 1 word present
vector<string>::iterator iter=svec.begin();
while(iter < svec.end())
{
if(word == *iter)
{
//match found
cvec[iter-svec.begin()]++; //increment count of that word
unique_word_found = false;
}
iter++;
}
if(unique_word_found)
{
//add to unique word list and increment count
svec.push_back(word);
cvec.push_back(1);
cvec[unique_word_count++]=1;
}
//next word
pch = std::strtok(NULL, " ,-");
unique_word_found = true; //reset flag
}
}
cout << "Word" << " ---> " << "Occurences" << endl;
for(vector<string>::size_type i=0; i < svec.size(); i++)
{
cout << svec[i] << " ---> " << cvec[i] << endl;
}
return 0;
}

Answer 4

The general strategy can be as follows:
Sanitize the input (convert all characters to lower case, remove unwanted punctuation, etc.)
Walk through the input
Add each character to a string, finalizing when a space is encountered
Add the string to a key-value structure. The string is the key. If this is a new entry not already contained in the structure, set the value 1. Otherwise set it to the current value + 1 (so as to count the number of times encountered so far).
Repeat for each word
Walk through the key-value structure and print each entry.