C program to count number of vowels and consonants in a String

Given a string and write a C program to count the number of vowels and consonants in this string. 

Examples:

Input: str = "geeks for geeks"
Output:
Vowels: 5
Consonants: 8

Input: str = "abcdefghijklmnopqrstuvwxyz"
Output:
Vowels: 5
Consonants: 21

Using For Loops

• Take the string as input
• Take each character from this string to check
• If this character is a vowel, increment the count of vowels
• Else increment the count of consonants.
• Print the total count of vowels and consonants in the end.

Below is the implementation of the above approach: 

C

// C program to count the number of
// vowels and consonants in a string

#include <stdio.h>

// Function to count number
// of vowels and consonant
void count_vowels_and_consonant(char* str)
{
    // Declare the variable vowels and consonant
    int vowels = 0, consonants = 0;

    int i;
    char ch;

    // Take each character from this string to check
    for (i = 0; str[i] != '\0'; i++) {

        ch = str[i];

        // If this character is a vowel,
        // increment the count of vowels
        if (ch == 'a' || ch == 'e'
            || ch == 'i' || ch == 'o'
            || ch == 'u' || ch == 'A'
            || ch == 'E' || ch == 'I'
            || ch == 'O' || ch == 'U')
            vowels++;

        // If this character is a space
        // skip it
        else if (ch == ' ')
            continue;

        else
            // Else increment the count of consonants
            consonants++;
    }

    // Print the total count of vowels and consonants
    printf("\nVowels: %d", vowels);
    printf("\nConsonants: %d", consonants);
}

// Driver function.
int main()
{
    char* str = "geeks for geeks";
    printf("String: %s", str);

    count_vowels_and_consonant(str);

    return 0;
}

Output

String: geeks for geeks
Vowels: 5
Consonants: 8

The complexity of the above method

Time Complexity: O(1)

Auxiliary Space: O(1)

Using Recursion

Below is the C program count number of vowels and consonants in a String using recursion:

C

// C program to count the number of
// vowels and consonants in a string
#include <stdio.h>
#include <string.h>

// Function to count vowels and consonants
void stringcount(char* s)
{
    static int i, vowels = 0, consonants = 0;

    if (!s[i]) {
        printf("\nVowels: %d\n", vowels);
        printf("Consonants: %d\n", consonants);
        return;
    }
    else {
        if ((s[i] >= 65 && s[i] <= 90)
            || (s[i] >= 97 && s[i] <= 122)) {
            if (s[i] == 'a' || s[i] == 'e' || s[i] == 'i'
                || s[i] == 'o' || s[i] == 'u' || s[i] == 'A'
                || s[i] == 'E' || s[i] == 'I' || s[i] == 'O'
                || s[i] == 'U')
                vowels++;
            else
                consonants++;
        }
        i++;
        stringcount(s);
    }
}

// Driver code
int main()
{
    char s[1000] = "Geeks for Geeks";
    printf("String: %s", s);

    stringcount(s);
}

Output

String: Geeks for Geeks
Vowels: 5
Consonants: 8

Using Pointers and While Loop

Below is the C program to count the number of vowels and consonants in a string using pointers and while loop:

C

// C program to count the number of
// vowels and consonants in a string
#include <stdio.h>
#include <string.h>

// Driver code
int main()
{
    char s[1000] = "geeks for geeks", *p;
    int vowels = 0, consonants = 0;

    p = s;

    while (*p) {
        if ((*p >= 65 && *p <= 90)
            || (*p >= 97 && *p <= 122)) {
            if (*p == 'a' || *p == 'e' || *p == 'i'
                || *p == 'o' || *p == 'u' || *p == 'A'
                || *p == 'E' || *p == 'I' || *p == 'O'
                || *p == 'U')
                vowels++;
            else
                consonants++;
        }
        p++;
    }

    printf("String: %s", s);
    printf("\nVowels: %d\n", vowels);
    printf("Consonants: %d\n", consonants);

    return 0;
}

Output

String: geeks for geeks
Vowels: 5
Consonants: 8