Character Counter in C++? - c++

Question

I'm trying to find a way to display all the characters in a string and the number of times they occur.
This is what I have so far:
//Any unused includes are part of the default code
#include <iostream>
#include <ctime>
#include <cstdlib>
#include <cmath>
#include <string>
using namespace std;
int main()
{
string st = "";
cout << "Input a sentence: " << endl;
getline(cin, st);
int index = 0;
int index2 = 0;
int counters[26] = {0};
for(int i = 0; i < st.length(); i++)
{
int counter = 0;
index = st.find(st[i],0);
for(int j = 0; j < st.length(); j++)
{
index2 = st.find(st[j]);
if(index == index2)
{
counter++;
}
}
cout << st[i] << ": " << counter << endl;
}
//cout << st[i] <<": " << counters[st[i] - 'a'] << endl;
return 0;
}
and I return this:
Input a sentence:
hello
h: 1
e: 1
l: 2
l: 2
o: 1
so I kind of have something but I can't figure out how to make the letters not repeat more than once. I know that I need to store them in an array but it's out of my ken.

Answer 1

You were very close, nice try! I liked the approach with the counters array, where every cell would represent the frequency of a letter in the given string.
So, just go and update this array, as this answer implies How to get character's position in alphabet in C language?, without the plus one they mention there, since you want the index of the letter in your array. In other words, for 'a', you need 0, 'b', you need 1 and so on.
Then, in the printing phase, just use the above link's suggestion in the reverse way. When you print the i-th non-zero element of counters, print the i-th element of the element, which will reveal the letter in question.
Putting all together, you get:
#include <iostream>
#include <ctime>
#include <cstdlib>
#include <cmath>
#include <string>
using namespace std;
int main()
{
string st = "";
cout << "Input a sentence: " << endl;
getline(cin, st);
int index = 0;
int index2 = 0;
int counters[26] = {0};
for(size_t i = 0; i < st.length(); i++)
{
int counter = 0;
index = st.find(st[i],0);
for(size_t j = 0; j < st.length(); j++)
{
index2 = st.find(st[j]);
if(index == index2)
{
counter++;
}
}
counters[st[i] - 'a'] = counter; // update 'counters' array
}
for(int i = 0; i < 26; ++i)
if(counters[i] != 0) // print non-zero counters
cout << (char)(i + 'a') << ": " << counters[i] << endl;
return 0;
}
Output:
e: 1
h: 1
l: 2
o: 1

Answer 2

I would do something like this:
#include <iostream>
#include <map>
#include <string>
int main()
{
std::string st;
std::cout << "Input a sentence: " << std::endl;
std::getline(std::cin, st);
std::map<char, int> m;
for (const char c : st)
++m[c];
for (const std::pair<char, int>& me : m)
std::cout << me.first << ": " << me.second << std::endl;
}

Answer 3

lechuga2000 beat me to the post, but this is my suggestion:
#include <iostream>
#include <map>
#include <string>
int main()
{
std::string input_sentence = "Now is the time for all good men to come to the aid of the party.";
/*
std::cout << "Input a sentence: " << std::endl;
std::getline(std::cin, input_sentence);
*/
std::map<char, int> character_counts;
for (const auto character : input_sentence)
++character_counts[character];
for (const auto counted_character : character_counts)
std::cout << counted_character.first << ": " << counted_character.second << '\n';
return 0;
}
And here is the output:
: 15
.: 1
N: 1
a: 3
c: 1
d: 2
e: 6
f: 2
g: 1
h: 3
i: 3
l: 2
m: 3
n: 1
o: 8
p: 1
r: 2
s: 1
t: 7
w: 1
y: 1