Count the number of occurrences of each letter in string

Question

How can I count the number of occurrences in c of each letter (ignoring case) in the string? So that it would print out letter: # number of occurences, I have c

Answer 1

You can use the following code.

main()
{
    int i = 0,j=0,count[26]={0};
    char ch = 97;
    char string[100]="Hello how are you buddy ?";
    for (i = 0; i < 100; i++)
    {
        for(j=0;j<26;j++)
            {
            if (tolower(string[i]) == (ch+j))
                {
                    count[j]++;
                }
        }
    }
    for(j=0;j<26;j++)
        {

            printf("\n%c -> %d",97+j,count[j]);

    }

}

Hope this helps.

Answer 2

#include <stdio.h>
#include <string.h>
void main()
{
    printf("PLEASE ENTER A STRING\n");
    printf("GIVE ONLY ONE SPACE BETWEEN WORDS\n");
    printf("PRESS ENETR WHEN FINISHED\n");

    char str[100];
    int arr[26]={0};
    char ch;
    int i;

    gets(str);
    int n=strlen(str);

    for(i=0;i<n;i++)
    {
        ch=tolower(str[i]);
        if(ch>=97 && ch<=122)   
        {
            arr[ch-97]++;
        }
    }
    for(i=97;i<=122;i++)
        printf("%c OCCURS %d NUMBER OF TIMES\n",i,arr[i-97]);   
    return 0;
}

Answer 3

#include<stdio.h>

void frequency_counter(char* str)
{
    int count[256] = {0};  //partial initialization
    int i;

    for(i=0;str[i];i++)
        count[str[i]]++;

    for(i=0;str[i];i++) {
        if(count[str[i]]) {
            printf("%c %d \n",str[i],count[str[i]]);
            count[str[i]]=0;
        }
    }
}

void main()
{
    char str[] = "The quick brown fox jumped over the lazy dog.";
    frequency_counter(str);
}

Answer 4

int charset[256] = {0};
int charcount[256] = {0};

for (i = 0; i < 20; i++)
{
    for(int c = 0; c < 256; c++)
    {
        if(string[i] == charset[c])
        {
           charcount[c]++;
        }
    }
}

charcount will store the occurence of any character in the string.

Answer 5

//This is JavaScript Code.

function countWordOccurences()
{
    // You can use array of words or a sentence split with space.
    var sentence = "The quick brown fox jumped over the lazy dog.";
    //var sentenceArray = ['asdf', 'asdf', 'sfd', 'qwr', 'qwr'];
    var sentenceArray = sentence.split(' ', 1000); 
    var output;
    var temp;
    for(var i = 0; i < sentenceArray.length; i++) {
        var k = 1;
        for(var j = i + 1; j < sentenceArray.length; j++) {
            if(sentenceArray[i] == sentenceArray[j])
                    k = k + 1;
        }
        if(k > 1) {
            i = i + 1;
            output = output + ',' + k + ',' + k;
        }
        else
            output = output + ',' + k;
    }
    alert(sentenceArray + '\n' + output.slice(10).split(',', 500));
}

You can see it live --> http://jsfiddle.net/rammipr/ahq8nxpf/

Answer 6

Here is the C code with User Defined Function:

/* C Program to count the frequency of characters in a given String */

#include <stdio.h>
#include <string.h>

const char letters[] = "abcdefghijklmnopqrstuvwxzy";

void find_frequency(const char *string, int *count);

int main() {
    char string[100];
    int count[26] = { 0 };
    int i;

    printf("Input a string: ");
    if (!fgets(string, sizeof string, stdin))
        return 1;

    find_frequency(string, count);

    printf("Character Counts\n");

    for (i = 0; i < 26; i++) {
        printf("%c\t%d\n", letters[i], count[i]);
    }
    return 0;
}

void find_frequency(const char *string, int *count) {
    int i;
    for (i = 0; string[i] != '\0'; i++) {
        p = strchr(letters, string[i]);
        if (p != NULL) {
            count[p - letters]++;
        }
    }
}