Strings

#include <stdio.h>
 
int main() {
    /* Method 1: initialise directly —  added automatically */
    char name[6] = "HELLO";  /* size 6 = 5 chars + 1 for \0 */
 
    /* Method 2: character by character — must add \0 yourself */
    char city[7];
    city[0] = 'A';
    city[1] = 'U';
    city[2] = 'S';
    city[3] = 'T';
    city[4] = 'I';
    city[5] = 'N';
    city[6] = '\0';  /* MUST add this — or the string has no end */
 
    /* Method 3: let C count the size */
    char lang[] = "Python";  /* compiler sets size to 7 automatically */
 
    printf("Name: %s\n", name);    /* %s prints a string */
    printf("City: %s\n", city);
    printf("Lang: %s\n", lang);
 
    return 0;
}

#include <stdio.h>
 
int main() {
    char name[50];  /* enough space for the input */
 
    printf("Enter your name: ");
    scanf("%s", name);  /* no & needed — array name IS already an address */
 
    printf("Hello, %s!\n", name);
 
    return 0;
}

#include <stdio.h>
#include <string.h>  /* required for all string functions */
 
int main() {
    char s1[20] = "Hello";
    char s2[] = " World";
 
    printf("Length: %lu\n", strlen(s1));    /* Length: 5 */
 
    strcat(s1, s2);                          /* s1 becomes "Hello World" */
    printf("After strcat: %s\n", s1);
 
    printf("Compare: %d\n", strcmp("abc", "abc"));  /* 0 — equal */
    printf("Compare: %d\n", strcmp("abc", "abd"));  /* negative — abc comes before abd */
 
    return 0;
}

#include <stdio.h>
 
int main() {
    char str[] = "Chaduvuko";
    int i = 0;
 
    /* Method 1: use strlen to get the count */
    for (i = 0; str[i] != '\0'; i++) {  /* stop when you hit the null terminator */
        printf("%c ", str[i]);
    }
    printf("\n");
    /* Output: C h a d u v u k o */
 
    /* Method 2: count uppercase and lowercase letters */
    int upper = 0, lower = 0;
    for (i = 0; str[i] != '\0'; i++) {
        if (str[i] >= 'A' && str[i] <= 'Z') {
            upper++;  /* ASCII range for uppercase: 65 to 90 */
        } else if (str[i] >= 'a' && str[i] <= 'z') {
            lower++;  /* ASCII range for lowercase: 97 to 122 */
        }
    }
    printf("Uppercase: %d, Lowercase: %d\n", upper, lower);
    /* Output: Uppercase: 1, Lowercase: 8 */
 
    return 0;
}

Same two-pointer technique as reversing an array from Lesson 3. One pointer at the start, one at the end — swap and move inward until they meet.

#include <stdio.h>
#include <string.h>
 
void reverseString(char str[]) {
    int left = 0;
    int right = strlen(str) - 1;  /* index of last character (before \0) */
    char temp;
 
    while (left < right) {
        temp = str[left];          /* swap characters */
        str[left] = str[right];
        str[right] = temp;
        left++;
        right--;
    }
}
 
int main() {
    char str[] = "Chaduvuko";
    printf("Before: %s\n", str);  /* Chaduvuko */
 
    reverseString(str);
 
    printf("After:  %s\n", str);  /* okuvudahC */
    return 0;
}

A palindrome reads the same forwards and backwards — "madam", "racecar", "level". Compare characters from both ends — if any pair doesn't match, it's not a palindrome.

#include <stdio.h>
#include <string.h>
 
int isPalindrome(char str[]) {
    int left = 0;
    int right = strlen(str) - 1;
 
    while (left < right) {
        if (str[left] != str[right]) {
            return 0;  /* mismatch found — not a palindrome */
        }
        left++;
        right--;
    }
    return 1;  /* all pairs matched — it is a palindrome */
}
 
int main() {
    char s1[] = "madam";
    char s2[] = "hello";
 
    if (isPalindrome(s1)) {
        printf("%s is a palindrome\n", s1);      /* madam is a palindrome */
    }
    if (!isPalindrome(s2)) {
        printf("%s is NOT a palindrome\n", s2);  /* hello is NOT a palindrome */
    }
    return 0;
}

Walk through each character. If it is a, e, i, o, u (uppercase or lowercase) it is a vowel. If it is any other letter it is a consonant. Spaces and digits are neither.

#include <stdio.h>
 
int main() {
    char str[] = "Chaduvuko Platform";
    int vowels = 0, consonants = 0;
    int i;
 
    for (i = 0; str[i] != '\0'; i++) {
        char c = str[i];
 
        /* convert to lowercase for easier checking */
        if (c >= 'A' && c <= 'Z') {
            c = c + 32;  /* ASCII trick: 'A'(65) + 32 = 'a'(97) */
        }
 
        if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u') {
            vowels++;
        } else if (c >= 'a' && c <= 'z') {
            consonants++;  /* it is a letter but not a vowel */
        }
        /* spaces, digits, symbols — ignored */
    }
 
    printf("Vowels: %d\n", vowels);        /* Vowels: 6 */
    printf("Consonants: %d\n", consonants);  /* Consonants: 11 */
    return 0;
}

Two strings are anagrams if they contain the exact same characters in any order. "listen" and "silent" are anagrams. The approach: count how many times each letter appears in string 1, then subtract for string 2. If all counts reach zero — anagram.

#include <stdio.h>
#include <string.h>
 
int isAnagram(char s1[], char s2[]) {
    /* first check: if lengths differ they cannot be anagrams */
    if (strlen(s1) != strlen(s2)) {
        return 0;
    }
 
    int freq[26] = {0};  /* one slot for each letter a-z, all start at 0 */
    int i;
 
    /* count up for every character in s1 */
    for (i = 0; s1[i] != '\0'; i++) {
        freq[s1[i] - 'a']++;  /* 'a'-'a'=0, 'b'-'a'=1, 'c'-'a'=2 ... */
    }
 
    /* count down for every character in s2 */
    for (i = 0; s2[i] != '\0'; i++) {
        freq[s2[i] - 'a']--;
    }
 
    /* if any count is non-zero, the strings are not anagrams */
    for (i = 0; i < 26; i++) {
        if (freq[i] != 0) {
            return 0;
        }
    }
    return 1;  /* all counts are zero — anagram confirmed */
}
 
int main() {
    char s1[] = "listen";
    char s2[] = "silent";
 
    if (isAnagram(s1, s2)) {
        printf("'%s' and '%s' ARE anagrams\n", s1, s2);  /* ARE anagrams */
    } else {
        printf("NOT anagrams\n");
    }
    return 0;
}

A new word starts when you see a non-space character after a space (or at the very beginning). Track whether the previous character was a space — that is all you need.

#include <stdio.h>
 
int countWords(char str[]) {
    int count = 0;
    int inWord = 0;  /* flag: are we currently inside a word? */
    int i;
 
    for (i = 0; str[i] != '\0'; i++) {
        if (str[i] != ' ' && inWord == 0) {
            inWord = 1;  /* just entered a new word */
            count++;
        } else if (str[i] == ' ') {
            inWord = 0;  /* just left a word */
        }
    }
    return count;
}
 
int main() {
    char sentence[] = "Learn DSA on Chaduvuko";
    printf("Word count: %d\n", countWords(sentence));  /* Word count: 4 */
    return 0;
}

Walk through the string once. Every time you find the target character, replace it. Single pass — O(n).

#include <stdio.h>
 
void replaceChar(char str[], char find, char replace) {
    int i;
    for (i = 0; str[i] != '\0'; i++) {
        if (str[i] == find) {
            str[i] = replace;
        }
    }
}
 
int main() {
    char str[] = "data engineering";
    printf("Before: %s\n", str);       /* data engineering */
 
    replaceChar(str, 'a', '@');
 
    printf("After:  %s\n", str);       /* d@t@ engineering */
    return 0;
}