Counting occurrence of each character in string C++ - c++

Question

So i wanted to count all the characters in a string and categorized them in vowels, consonants, and specials character. For example;
Enter string: sentence example ..
Vowels: e(5) a(1)
Consonants: s(1) n(1) t(1) c(1) x(1) m(1) p(1) l(1)
Specials: blank space .(2)
Here's coding:
void characterType(string input)
{
int vowel = 0;
int consonant = 0;
int special = 0;
int n = input.size();
int freq[26];
memset(freq, 0, sizeof(freq));
for (int i = 0; i < n; i++)
{
freq[input[i] - 'a']++;
}
cout<<"Vowels: ";
for (int i = 0; i < n; i++)
{
char character = input[i];
if(isalpha(character))
{
character = tolower(character);
if (character == 'a' || character == 'e' || character == 'i' || character == 'o' || character == 'u')
{
cout<<input[i]<<freq[input[i] - 'a']<<" ";
}
}
}
cout<<endl;
cout<<"Consonants: ";
for (int i = 0; i < n; i++)
{
char character = input[i];
if(isalpha(character))
{
character = tolower(character);
if (character != 'a' || character != 'e' || character != 'i' || character != 'o' || character != 'u')
{
cout<<input[i]<<freq[input[i] - 'a']<<" ";
}
}
}
cout<<endl;
cout<<"Specials: ";
for (int i = 0; i < n; i++)
{
char character = input[i];
if(!isalpha(character))
{
if(character == ' ')
{
cout<<"[black space]"<<freq[input[i] - 'a']<<" ";
}
else
cout<<input[i]<<freq[input[i] - 'a']<<" ";
}
}
}
And heres what ive got so far:
How do i make it not repeat the same character and why does special characters is not counting?

Answer 1

Since you want to do this and it's not some type of assignment, here's how I would approach the problem, using modern C++ features:
#include <iostream>
#include <map>
#include <algorithm>
using namespace std;
int main() {
std::map<char, std::size_t> occurrance;
std::string input{"This is A long string with lots of characters in it. AAA$#!#$!^^! "};
auto is_vowel = [] (char c) -> bool
{
auto lc = std::tolower(c);
return
c == 'a' ||
c == 'e' ||
c == 'i' ||
c == 'o' ||
c == 'u';
};
auto is_special = [] (char c) -> bool
{
// use ascii table to find only "non-special characters"
if(c < '0') return true;
if(c > '9' && c < 'A') return true;
if(c > 'Z' && c < 'a') return true;
if(c > 'z') return true;
return false;
};
auto vowels = std::count_if(
input.begin(),
input.end(),
is_vowel);
auto special = std::count_if(
input.begin(),
input.end(),
is_special);
for(auto c : input)
{
occurrance[c] += 1;
}
std::cout << "Vowels: " << vowels << '\n';
std::cout << "Special: " << special << '\n';
for(auto [c, count] : occurrance)
{
std::cout << c << " -> " << count << '\n';
}
return 0;
}
https://ideone.com/YVaDiI

Answer 2

This does the job, although you can make the output prettier... and take the uppercase letters into account.
void printCharWithFreq(string str)
{
int v=0, c=0, s=0;
// size of the string 'str'
int n = str.size();
// 'freq[]' implemented as hash table
int freq[SIZE];
// initialize all elements of freq[] to 0
memset(freq, 0, sizeof(freq));
// accumulate freqeuncy of each character in 'str'
for (int i = 0; i < n; i++)
freq[str[i] - 'a']++;
// traverse 'str' from left to right
for (int i = 0; i < n; i++) {
// if frequency of character str[i] is not
// equal to 0
if (freq[str[i] - 'a'] != 0) {
// print the character along with its
// frequency
if(str[i] == 'a' || str[i] == 'e' || str[i] == 'i' || str[i] == 'o' || str[i] == 'u' ||
str[i] == 'A' || str[i] == 'E' || str[i] == 'I' || str[i] == 'O' || str[i] == 'U')
{ cout<<"Vowel: " << str[i] << freq[str[i] - 'a'] << " "<<endl; v+=freq[str[i] - 'a']; }
else if(str[i] == ' ' || str[i] == '.')
{ cout<<"Specials"<<endl; s++; }
else
{ cout<<"Consonant: " << str[i] << freq[str[i] - 'a'] << " "<<endl; c+=freq[str[i] - 'a']; }
// update frequency of str[i] to 0 so
// that the same character is not printed
// again
freq[str[i] - 'a'] = 0;
}
}
cout<<"Number of vowels: "<<v<<endl;
cout<<"Number of consonants: "<<c<<endl;
cout<<"Number of specials "<<s<<endl;
}
Then test it in the main function like
int main() {
string str = "some text . .";
printCharWithFreq(str);
return 0;
}
Basically, frequency is updated so it doesn't print the same character again, as it says in the comments. You also made a mistake in your code where you wrote:
character != 'a' || character != 'e' || character != 'i' || character != 'o' || character != 'u'
It is a consonant only if the input is different than ALL of the vowels, meaning you have to put &&, not ||.
As for the special characters, they can be counted with a simple counter.
The rest, I think, is clear from the comments in the code.
I adapted the code from the following source:
https://www.geeksforgeeks.org/print-characters-frequencies-order-occurrence/