Finding the total number of repeat strings in a string C++ -- without maps - c++

Question

My program has to find the total number of repeated strings in a string. I cannot use maps or built in string functions other than length()
Example : string input = "Hello hello Hello"
Hello : 2
I am hitting a roadblock with separating the strings by the spaces and reading them. I can't figure out what to write to make that happen.
What I want to do is create a temp string to compare to the next string and if they are equal to store it in a vector and then read from the vector at the end.
What function could I use to do that?
Here is my code below :
#include<iostream>
#include<string>
#include<vector>
using namespace std;
vector <string> mystring;
int numberString(string const&in)
{
int total = 0;
char temp;
for (int i = 0; i < in.length(); i++)
{
temp = in[i];
if (temp == ' ')
total++;
}
total++;
return total;
}
void findRepeats(string const &in)
{
int numberOfStrings = numberString(in);
int asciiArray[256];
for (int i = 0; i < 256; i++)
asciiArray[i] = 0;
int counter = 0;
string temp = "blank";
while (numberOfStrings != counter)
{
temp = in;
}
}
int main()
{
string input;
cout << "Enter a string : ";
getline(cin, input);
findRepeats(input);
return 0;
}

Answer 1

The straightforward way to count substrings in a whitespace-separated string is to insert them into a map and and track the occurrence count:
std::string input = "Hello hello Hello";
std::istringstream iss(input);
std::map<std::string, size_t> m;
std::string temp;
while(iss >> temp)
{
auto it = m.find(temp);
if(it != std::end(m))
{
++(it->second);
}
else
{
m.insert(std::make_pair(temp, 0));
}
}
//display counts as:
for(auto it = std::begin(m); it != std::end(m); ++it)
{
std::cout<<"string \""<<it->first<<"\" was found "<<it->second<<" times"<<std::endl;
}
The code is untested.

Answer 2

The following code finds duplicate words as long as all words are seperated by a single space:
#include <string>
#include <map>
#include <sstream>
#include <iostream>
using namespace std;
int main()
{
string input = "Hello hello hello";
map<string, int> wordCount;
for (string::size_type p = 0; p < input.size(); )
{
const auto p2 = input.find_first_of(' ', p);
const auto word = input.substr(p, (p == string::npos) ? string::npos : (p2 - p));
++wordCount[word];
if (p2 == string::npos)
break;
p = p2 + 1;
}
for (const auto& it : wordCount)
if (it.second > 1)
std::cout << it.first << " " << it.second << std::endl;
return 0;
}
Be aware that this code does not only find consecutive duplicates. So "a b a" outputs 'a 2'.
The line ++wordCount[word] increments the counter for the word or initializes it to 1 if 'word' is not already found in the map (This is working because the template initializes the value with int() which is guaranteed to initialize to zero)
At the end you get a map with an entry for every unique word (first=Word, second=count)
If you would like to count only consecutive duplicates this piece of code may help you:
#include <string>
#include <map>
#include <sstream>
#include <iostream>
#include <vector>
using namespace std;
int main()
{
string input = "Hello hello hello Hello";
vector<pair<string, int>> wordCount;
for (string::size_type p = 0; p < input.size(); )
{
const auto p2 = input.find_first_of(' ', p);
const auto word = input.substr(p, (p == string::npos) ? string::npos : (p2 - p));
if (wordCount.empty() || wordCount.back().first != word)
wordCount.push_back(make_pair(word, 1));
else
++wordCount.back().second;
if (p2 == string::npos)
break;
p = p2 + 1;
}
for (const auto& it : wordCount)
if (it.second > 1)
std::cout << it.first << " " << it.second << std::endl;
return 0;
}
This code does not use a map because a single word can have different counts depending on its location ("a a b a a a" would output "a 2" and "a 3")
Both examples scan the string word by word with ' ' as the delimiter. You could specify multiple delimiters in find_first_of if you would like to split your string by tab or dot. (input.find_first_of(" \t.", p))