Count frequency of digits in string - c

Question

I need to implement a function that can count the number of digits in a string. So for numbers but also for somehting like: aD23b. If I could make it work...it should look like:
Input: 0912302
Output:
0: 2
1: 1
2: 2
3: 1
4: 0
5: 0
6: 0
7: 0
8: 0
9: 1
At this point I can't code anything that works unfortunately...My basic idea is: Use a loop to check every character from Input, if it's a digit, store it in a second array (let's say frequency). The problems I have are that I need to somehow convert every character into a integer or somehow be able to count how
often each digits appears... I was hoping this might work but it doesn't at all:
I forgot to mention I'm a beginner in programming so I would really appreciate if you could give me tips and explanations.
void calc_occurrences(int s[], int occurrences[])
{
int i = 0;
int j;
int count = 0;
while (s[i] != '\0') {
if (isdigit(s[i])) {
for (j = 0; occurrences[j] != '\0'; j++) {
occurrences[j] = s[i];
}
}
i++;
for (j = i + 1; s[j] != '\0'; j++) {
if (isdigit(s[i]) == isdigit(s[j])) {
count++;
occurrences[j] = 0;
}
}
if(occurrences[i] != 0) {
occurrences[i] = count;
}
}
}

Answer 1

Make an array to count the frequency of each relevant character.
Something like this:
#include <stdio.h>
void count_freq(char* str, int freq[10])
{
int i = 0;
while(str[i]) // Loop to end of string
{
if (str[i] >= '0' && str[i] <= '9') // Check that the character is in range
{
++freq[str[i]-'0']; // notice the -'0' to get in range 0..9
}
++i;
}
}
int main(void) {
int freq[10] = {0}; // Array to count occurence
char str[] = "0034364hh324h34"; // Input string
count_freq(str, freq); // Calculate frequency
for (int i=0; i < 10; ++i) // Print result
{
printf("%d: %d\n", i, freq[i]);
}
return 0;
}
Output:
0: 2
1: 0
2: 1
3: 4
4: 4
5: 0
6: 1
7: 0
8: 0
9: 0

Answer 2

You can have an integer array with size 10 with 0 stored in all indices. Then, when you spot a digit, you can increment the number in the respective index.
In example, when you see a "0", you can do arr[0]++;.
Also, you may check if a character is a digit with the isdigit() function.

Answer 3

PS : I know, I'm answering an old post, but I was doing some challenges on HackerRank, and I managed to solve this nearly exact problem, in case it might help someone since I've used dynamic allocation on my code.
/* Problem: hackkerrank.com/challenges/frequency-of-digits-1/problem */
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main(void)
{
char *s;
int *arr;
int i;
i = 0;
s = (char*)malloc(sizeof(char));
scanf("%s", s);
arr = (int*)malloc(10 * sizeof(int));
while(i < 10)
{
*(arr + i) = 0;
i++;
}
i = 0;
while (i < strlen(s))
{
if (*(s + i) >= '0' && *(s + i) <= '9')
{
(*(arr + (*(s + i) - '0'))) += 1;
}
i++;
}
i = 0;
while (i < 10)
{
printf("%d ", *(arr + i)); // As HackerRank problem wanted the output format.
// printf("%d: %d\n", i, *(arr + i)); As you wanted it
i++;
}
return (0);
}