Program that detects which letter is the most common - c

Question

I am a beginner in C, I am trying to make a program which detects which letter is the most common of max 10 letters.
Here is what I've got so far:
char one = 'a'; //0110 0001
char check[10];
scanf("%s", &check);
char *ptr;
int a = 0;int b = 0;int c = 0;int d = 0;int e = 0;int f = 0;int g = 0;int h = 0;int i = 0;int j = 0;int k = 0;int l = 0;int m = 0;int n = 0;int o = 0;int p = 0;int q = 0;int r = 0;int s = 0;int t = 0;int u = 0;int v = 0;int w = 0;int x = 0;int y = 0;int z = 0;
if (check[0]=='a'){
a += 1;
if (a> b && a> c && a> d && a> e && a> f && a> g && a> h && a> i && a> j && a> k && a> l && a> m && a> n && a> o && a> p && a> q && a> r && a> s && a> t && a> u && a> v && a> x && a> y ){
printf("A is the most common letter);
}
}
The 'if' statement is only for the first letter that is entered and it is only checking the letter a.
Here is where I need help, how do I optimize that? How can I make a loop instead of having all that a>b && a>c ... etc. Also if it is possible to declare a lot of similar variables in a shorter way? Just generally how do I keep short and am I doing something wrong?
Thank you.

Answer 1

Yes you are doing a lot of things wrong.
this part of code
int a = 0;int b = 0;int c = 0;int d = 0;int e = 0;int f = 0;int g = 0;int h = 0;int i = 0;int j = 0;int k = 0;int l = 0;int m = 0;int n = 0;int o = 0;int p = 0;int q = 0;int r = 0;int s = 0;int t = 0;int u = 0;int v = 0;int w = 0;int x = 0;int y = 0;int z = 0;
is something horrific.
Use an array, like that:
int letter[25];
letter[0] will be your a, letter[1] will be your b ... letter[49] will be your z.
this part of code
if (check[0]=='a'){
a += 1;
is a bad way to do what you want.
you should do it like this:
int i;
for(i = 0; i < 10; ++i)
{
if(check[i] >= 'A' && check[i] <= 'Z') // Check if letter is uppercase.
++letter[check[i] - 'A']; // 'A' == 65, But our array is from 0 to 49.
if(check[i] >= 'a' && check[i] <= 'z') // Check if letter is lowercase.
++letter[check[i] - 'a']; // 'a' == 97. Note that 'A' is not 'a'.
}
This will check which character is most common, and stores it in the letter array.

Answer 2

You can certainly do it your way and compare every letter with every other. But usually you do this in two phases
count how often the letters occur
int letters[26];
int i, n = strlen(check), max;
memset(letters, 0, sizeof(letters));
for (i = 0; i < n; ++i) {
char c = tolower(check[i]);
letters[c - 'a']++;
}
pick the highest one
max = 0;
for (i = 1; i < 26; ++i)
if (letters[i] > letters[max])
max = i;
printf("%c is the most common letter\n", max + 'a');

Answer 3

Impressive work :)
int counts[26]; // We are expecting 26 letters
char check[10];
char *ptr;
scanf("%s", check);
memset(counts, 0, sizeof(counts)); // zero all array values
for (ptr = check; *ptr; ptr++)
{
char ch = *ptr;
if (isalpha(ch)) // ignore non-alphas
{
ch = tolower(ch);
counts[ch - 'a']++;
}
}
You can add best index lookup code from dreamzor answer

Answer 4

You need to use loops and arrays. Just store the amount of letters you've got in an array and then compare current the letter amount to best amount:
char check[10];
scanf("%s", &check);
const int checkSize = strlen(check);
int numberOfLetters[26];
for(int i = 0; i < 26; ++i)
numberOfLetters[i] = 0;
int bestLetterIndex = -1;
for(int i = 0; i < checkSize; ++i) {
int letterIndex = check[i] - 'a'; // getting index from ASCII code
numberOfLetters[letterIndex]++;
if(i == 0 || // first letter, index not found yet
numberOfLetters[letterIndex] > numberOfLetters[bestLetterIndex]) {
bestLetterIndex = letterIndex;
}
}
printf("Most common letter is %c", (char)(bestLetterIndex + 'a'));