C Program to Find Frequency of Character in String

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In this tutorial, you will learn and get code about finding the frequency of a character in a given string. That is, how many times a particular character present in a given string. For example, if user enters the string as codescracker, and wants to check for the frequency of a character say c. Then it will be 3. Because c occurs 3 times in the given string codescracker

The question is, write a program in C to find frequency of any given character in a given string. Both string and character will be entered by user (at run-time). The answer to this question is:

#include<stdio.h>
#include<conio.h>
int main()
{
    int i, count=0;
    char str[100], ch;
    printf("Enter the String: ");
    gets(str);
    printf("Enter any character (present in string) to find its frequency: ");
    scanf("%c", &ch);
    for(i=0; str[i]!='\0'; i++)
    {
        if(ch==str[i])
            count++;
    }
    printf("\nFrequency of %c = %d", ch, count);
    printf("\n\n%c occurs %d times in %s", ch, count, str);
    getch();
    return 0;
}

As the above program is written under Code::Blocks IDE, therefore after successful build and run, here is its sample run:

Now enter any string say codescracker and press ENTER key. Then enter any character that presents inside the string codescracker say c and press ENTER again, to see the frequency of c in codescracker:

Program Explained

Get string input from user using gets() function
Get character input from user using scanf() function
Now We've created a for loop to check each and every character of the string, that is from the 0^th character of the string to the null terminated character '\0' of the string
Whenever the character matches with the character at any place of the string, then increment the count variable's value
As count variable initially initialized to 0, therefore count holds the value as total times the character presents inside the string
For example, let's suppose the string is codescracker (in str variable) and the character is c (in ch variable)
Therefore, str[0] holds c, str[1] holds o, str[2] holds d, str[3] holds e, ....., str[9] holds k, str[10] holds e, and str[11] holds r
The character c presents at the index number 0, 5, and 8. Total of 3 times.
Therefore the count variable gets incremented 3 times. As the condition ch==str[i] gets true three times
Because ch holds c and str[0], str[5], and str[8] also holds c.
Therefore, the frequency of c in the given string say codescracker will be 3

Find Frequency of each Character in a String

Now let's modify the above program and in place of finding the frequency of one particular character, let's find frequency of each and every character present in the given string:

#include<stdio.h>
#include<conio.h>
#include<string.h>
int main()
{
    char str[100], ch;
    int i, j, k, len, count=0;
    printf("Enter the string: ");
    gets(str);
    len = strlen(str);
    printf("\n");
    for(i=0; i<len; i++)
    {
        ch = str[i];
        for(j=0; j<len; j++)
        {
            if(ch==str[j])
            {
                count++;
                for(k=j; k<(len-1); k++)
                {
                    str[k] = str[k+1];
                }
                len--;
                str[len] = '\0';
                j--;
            }
        }
        printf("%c = %d times\n", ch, count);
        count=0;
        i--;
    }
	getch();
	return 0;
}

Here is the sample run:

c program frequency of each character string

Now supply any string say we love codescracker and press ENTER to see the frequency of all the character one by one as shown in the screenshot given below:

Program Explained

Here we have used one outer loop that runs until the null terminated character ('\0') occurs in the string
And in inner loop we have used the same loop with another variable
Before inner loop We've initialized the initial character of the string to the ch variable
And then comes inside the inner loop
In this loop we have performed checking operation that where-where ch's value is present inside the string, if present then we have incremented the count variable and then shifted one index back all the character of the string, that is str[1] is shifted to str[0], str[2] shifted to str[1], and so on.
As the character is counted up, therefore character at that particular place has not any further use, therefore we have shifted and deleted that character from the string using the second inner for loop using another variable
After breaking out of this loop, decrement the string length say len here, and initialized null terminated character ('\0') to the last index of the string
And then decrement the second inner for loop variable and continue to run until the condition of first inner for loop evaluates to be false
After this, print the output and initialized count as 0 and decrement the value of outer loop
In this way frequency of all the character is calculated and printed one by one

Print Frequency with Remaining Character

Here is another program that list out all the character's frequency along with remaining character in string:

#include<stdio.h>
#include<conio.h>
#include<string.h>
int main()
{
    char str[100], ch;
    int i, j, k, len, count=0;
    printf("Enter the string: ");
    gets(str);
    len = strlen(str);
    printf("\n");
    printf("Character\tCount\t\tString Left\n\n");
    for(i=0; i<len; i++)
    {
        ch = str[i];
        for(j=0; j<len; j++)
        {
            if(ch==str[j])
            {
                count++;
                for(k=j; k<(len-1); k++)
                    str[k] = str[k+1];
                len--;
                str[len] = '\0';
                j--;
            }
        }
        printf("%c occurs \t%d times\t\t%s\n", ch, count, str);
        count=0;
        i--;
    }
    getch();
    return 0;
}

Here is its sample run:

Now supply any string say codescracker and press enter to check it out:

find frequency of character in string c program

Same Program in Other Languages

C Online Test

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