C Basics

C String Problems and Solutions: From Basics to Advanced Logic

In the previous section we discussed the basics of C Strings. Now let us take a look at some real-life application Problems.

1. Write a C program to count vowels and consonants in a string.

				
					#include <stdio.h>

int main() {
    char str[100];
    int i, vowel = 0, constraint = 0;
    printf("Enter a string:");
    fgets(str, sizeof(str), stdin);

    for (i = 0; str[i] != '\0'; i++) {
        if (str[i]=='a'||str[i]=='e'||str[i]=='i'||str[i]=='o'||str[i]=='u'||
            str[i]=='A'||str[i]=='E'||str[i]=='I'||str[i]=='O'||str[i]=='U')
            vowel++;
        else if ((str[i]>='a'&&str[i]<='z') || (str[i]>='A'&&str[i]<='Z'))
            constraint++;
    }
    printf("Vowels = %d\nConsonants = %d\n", vowel, constraint);
    return 0;
}

// Enter a string: qubrica
// Vowels = 3
// Consonants = 4

Index (i)	Character str[i]	Category	Logic	Vowels	Consonants
0	'q'	Consonant	else if	0	1
1	'u'	Vowel	if	1	1
2	'b'	Consonant	else if	1	2
3	'r'	Consonant	else if	1	3
4	'i'	Vowel	if	2	3
5	'c'	Consonant	else if	2	4
6	'a'	Vowel	if	3	4
7	'\n'	Newline	Neither	3	4
8	'\0'	Null	Terminate	3	4

2. Write a C program to find the frequency of a character in a string.

				
					#include <stdio.h>
int main() {
    char input_text[120], target;
    int index, occurrences = 0;

    printf("Please provide a sentence: ");
    scanf("%[^\n]", input_text);

    printf("Which character are you looking for? ");
    scanf(" %c", &target);

    // Iterating through the text 
    for (index = 0; input_text[index] != '\0'; index++) {
        if (input_text[index] == target) {
            occurrences++;
        }
    }
    printf("Total count of '%c': %d\n", target, occurrences);
    return 0;
}

// Please provide a sentence: qubrica
// Which character are you looking for? i
// Total count of 'i': 1

index	Character	Comparison (== 'i')	Action	occurrences
0	'q'	False	Skip	0
1	'u'	False	Skip	0
2	'b'	False	Skip	0
3	'r'	False	Skip	0
4	'i'	True	Increment	1
5	'c'	False	Skip	1
6	'a'	False	Skip	1
7	'\0'	N/A	Terminate	1

3. Write a C program to remove all spaces from a string.

				
					#include <stdio.h>

int main() {
    char str[100];
    int i, j = 0;

    printf("Enter a string: ");
    scanf("%[^\n]", str);

    for (i = 0; str[i] != '\0'; i++) {
        if (str[i] != ' ')
            str[j++] = str[i];
    }
    // Null-terminate the modified string
    str[j] = '\0';

    printf("String without spaces: %s\n", str);
    return 0;
}

// Enter a string: C Program
// String without spaces: CProgram

Index	str[i]	Condition (!= ' ')	Action	Resulting str (First j chars)	j value
0	'C'	True	str[0] = 'C'	C	1
1	" "	False	Skip	C	1
2	'P'	True	str[1] = 'P'	CP	2
3	'r'	True	str[2] = 'r'	CPr	3
4	'o'	True	str[3] = 'o'	CPro	4
5	'g'	True	str[4] = 'g'	CProg	5
6	'r'	True	str[5] = 'r'	CProgr	6
7	'a'	True	str[6] = 'a'	CProgra	7
8	'm'	True	str[7] = 'm'	CProgram	8
9	'm'	True	str[8] = 'm'	CProgramm	9
10	'i'	True	str[9] = 'i'	CProgrammi	10
11	'n'	True	str[10] = 'n'	CProgrammin	11
12	'g'	True	str[11] = 'g'	CProgramming	12
13	'\0'	N/A	Terminate	CProgramming\0	12

4. Write a C program to check whether one string is a substring of another.

				
					#include <stdio.h>

int main() {
    char str[100], sub[100];
    int i, j, found = 0;

    printf("Enter main string: ");
    scanf("%[^\n]", str);
    printf("Enter substring: ");
    scanf(" %[^\n]", sub);

    for (i = 0; str[i] != '\0'; i++) {
        for (j = 0; sub[j] != '\0'; j++) {
            if (str[i + j] != sub[j])
                break;
        }
        if (sub[j] == '\0') {
            found = 1;
            break;
        }
    }

    if (found)
        printf("Substring found\n");
    else
        printf("Substring not found\n");

    return 0;
}

// Enter main string: qubrica
// Enter substring: rica
// Substring found

Outer Index (i)	str[i]	Inner Index (j)	sub[j]	Comparison (str[i+j] == sub[j])	Action	found
0	'q'	0	'r'	'q' == 'r' (False)	break inner loop	0
1	'u'	0	'r'	'u' == 'r' (False)	break inner loop	0
2	'b'	0	'r'	'b' == 'r' (False)	break inner loop	0
3	'r'	0	'r'	'r' == 'r' (True)	Continue inner	0
3		1	'i'	'i' == 'i' (True)	Continue inner	0
3		2	'c'	'c' == 'c' (True)	Continue inner	0
3		3	'a'	'a' == 'a' (True)	Continue inner	0
3		4	'\0'	N/A	Inner loop ends	1

5. Write a C program to convert all lowercase letters in a string to uppercase.

				
					#include <stdio.h>

int main() {
    char str[100];
    int i;

    printf("Enter a string: ");
    scanf("%[^\n]", str);

    for (i = 0; str[i] != '\0'; i++) {
        // Check if the character is a lowercase letter
        if (str[i] >= 'a' && str[i] <= 'z')
            str[i] = str[i] - 32; 
    }

    printf("Uppercase string: %s\n", str);
    return 0;
}

// Enter a string: qubrica
// Uppercase string: QUBRICA

Index (i)	Character str[i]	ASCII Value	Condition ('a' to 'z')	Calculation (-32)	New Character
0	'q'	113	True	113 - 32 = 81	'Q'
1	'u'	117	True	117 - 32 = 85	'U'
2	'b'	98	True	98 - 32 = 66	'B'
3	'r'	114	True	114 - 32 = 82	'R'
4	'i'	105	True	105 - 32 = 73	'I'
5	'c'	99	True	99 - 32 = 67	'C'
6	'a'	97	True	97 - 32 = 65	'A'
7	'\0'	0	False	N/A	Terminate

6. Write a C program to count the number of words in a string.

				
					console.log( 'Code is Poetry' );#include <stdio.h>

int main() { 
    char str[100];
    int i, count = 0;

    printf("Enter a string: ");
    scanf("%[^\n]", str);

    for (i = 0; str[i] != '\0'; i++) {
        // Increment count when a space is followed by a character
        if (str[i] == ' ' && str[i+1] != ' ' && str[i+1] != '\0')
            count++;
    }

    // Account for the first word if the string doesn't start with a space
    if (str[0] != ' ' && str[0] != '\0')
        count++;

    printf("Number of words = %d\n", count);
    return 0;
}

//Enter a string: I love Qubrica
// Number of words = 3

Index	Character	Logic Check	Result	count
0	'i'	str[0] != ' '	First word detected	1
1	' '	str[1]==' ' && str[2]=='l'	Space-to-Character transition	2
2 - 5	love	str[i] != ' '	Characters skipped (no space)	2
6	' '	str[6]==' ' && str[7]=='q'	Space-to-Character transition	3
7 - 13	qubrica	str[i] != ' '	Characters skipped (no space)	3
14	'\0'	str[14] == '\0'	Loop termination	3