String Reversal Approaches in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of string reversal using two-pointer swap

What is the output of the following C code that reverses a string using two pointers swapping characters?

DSA C

void reverse(char *str) {
    int left = 0;
    int right = strlen(str) - 1;
    while (left < right) {
        char temp = str[left];
        str[left] = str[right];
        str[right] = temp;
        left++;
        right--;
    }
}

int main() {
    char s[] = "hello";
    reverse(s);
    printf("%s", s);
    return 0;
}

A olleh

B hello

C ohlle

D Syntax error

💡 Hint

Think about swapping characters from start and end moving towards the center.

The code swaps characters from the start and end moving inward, reversing the string 'hello' to 'olleh'.

❓ Predict Output

intermediate

Output of string reversal using recursion

What is the output of this C code that prints a string in reverse using recursion?

DSA C

#include <stdio.h>
void reversePrint(char *str) {
    if (*str == '\0') return;
    reversePrint(str + 1);
    putchar(*str);
}

int main() {
    char s[] = "abc";
    reversePrint(s);
    return 0;
}

A abc

B cba

C Syntax error

D Runtime error

💡 Hint

The function calls itself with the next character before printing the current one.

The recursion goes to the end of the string, then prints characters on the way back, reversing the output.

🧠 Conceptual

advanced

Why is in-place reversal more memory efficient?

Which reason best explains why reversing a string in-place (modifying the original string) is more memory efficient than creating a new reversed string?

A It uses a linked list internally

B It compresses the string during reversal

C It avoids allocating extra memory for a new string copy

D It uses recursion which saves memory

💡 Hint

Think about memory allocation for new data versus modifying existing data.

In-place reversal changes the original string without needing extra space for another string, saving memory.

🔧 Debug

advanced

Identify the error in this string reversal code

What error does this C code produce when trying to reverse a string literal?

DSA C

void reverse(char *str) {
    int left = 0;
    int right = strlen(str) - 1;
    while (left < right) {
        char temp = str[left];
        str[left] = str[right];
        str[right] = temp;
        left++;
        right--;
    }
}

int main() {
    char *s = "hello";
    reverse(s);
    printf("%s", s);
    return 0;
}

A Segmentation fault (runtime error)

B Compilation error

C Output: olleh

D No output

💡 Hint

String literals are stored in read-only memory in C.

The code tries to modify a string literal which is stored in read-only memory, causing a runtime crash.

🚀 Application

expert

Result of reversing a palindrome string in-place

Given the string "madam", what is the state of the string after applying an in-place reversal function?

A Syntax error

B madma

C madam\0

D madam

💡 Hint

A palindrome reads the same forwards and backwards.

Reversing a palindrome string results in the same string because it reads identically both ways.

Practice

1. What does the string reversal operation do to the string "hello"?

easy

A. It changes it to "olleh"

B. It changes it to "hello" (no change)

C. It removes all vowels

D. It converts all letters to uppercase

Solution

1. Step 1: Understand string reversal

   Reversing a string means writing its characters in the opposite order.

2. Step 2: Apply reversal to "hello"

   The characters 'h', 'e', 'l', 'l', 'o' become 'o', 'l', 'l', 'e', 'h'.

3. Final Answer:

   "olleh" -> Option A

4. Quick Check:

   Reverse("hello") = "olleh" [OK]

Hint: Reverse means flip characters order [OK]

Common Mistakes:

• Thinking reversal changes letters, not order
• Confusing reversal with uppercase conversion
• Assuming reversal removes characters

2. Which of the following C code snippets correctly reverses a string str in place?

easy

A. for(int i = 0; i < strlen(str)/2; i++) { char temp = str[i]; str[i] = str[strlen(str)-1-i]; str[strlen(str)-1-i] = temp; }

B. for(int i = 0; i < strlen(str); i++) { str[i] = str[strlen(str)-1-i]; }

C. strrev(str);

D. for(int i = 0; i < strlen(str); i++) { str[i] = toupper(str[i]); }

Solution

1. Step 1: Understand in-place reversal logic

   Swap characters from start and end moving towards center, stopping at half length.

2. Step 2: Analyze for(int i = 0; i < strlen(str)/2; i++) { char temp = str[i]; str[i] = str[strlen(str)-1-i]; str[strlen(str)-1-i] = temp; }

   It swaps str[i] with str[length-1-i] for i from 0 to half length, correctly reversing the string.

3. Final Answer:

   Option A code correctly reverses string in place -> Option A

4. Quick Check:

   Swapping pairs from ends reverses string [OK]

Hint: Swap pairs from ends to middle [OK]

Common Mistakes:

• Overwriting characters without swapping
• Using full length instead of half length in loop
• Confusing string reversal with uppercase conversion

3. What is the output of the following C code?

char str[] = "abcde";
int i = 0, j = 4;
while(i < j) {
  char temp = str[i];
  str[i] = str[j];
  str[j] = temp;
  i++;
  j--;
}
printf("%s", str);

medium

A. abcde

B. Error: infinite loop

C. edcde

D. edcba

Solution

1. Step 1: Trace the while loop swaps

   Swap str[0] and str[4]: 'a' and 'e' -> str = "ebcda"; i=1, j=3 Swap str[1] and str[3]: 'b' and 'd' -> str = "edcba"; i=2, j=2 Loop ends as i == j.

2. Step 2: Print the reversed string

   After swaps, str is "edcba" which is printed.

3. Final Answer:

   edcba -> Option D

4. Quick Check:

   Swapped pairs reverse string [OK]

Hint: Swapping ends moves characters inward [OK]

Common Mistakes:

• Not updating i and j correctly
• Confusing loop condition causing infinite loop
• Printing original string without reversal

4. Identify the error in this C code that attempts to reverse a string str:

int len = strlen(str);
for(int i = 0; i <= len/2; i++) {
  char temp = str[i];
  str[i] = str[len - i];
  str[len - i] = temp;
}

medium

A. Loop condition should be i < len/2 not <=.

B. Missing null terminator assignment after reversal.

C. Index len - i is out of bounds; should be len - i - 1.

D. Using strlen inside loop causes performance issue.

Solution

1. Step 1: Check index usage in swapping

   String indices go from 0 to len-1. Using str[len - i] accesses one past last character (out of bounds).

2. Step 2: Correct index to len - i - 1

   Adjusting index fixes out-of-bounds error and correctly swaps characters.

3. Final Answer:

   Index len - i is out of bounds; should be len - i - 1 -> Option C

4. Quick Check:

   String last index = len-1, not len [OK]

Hint: String last index = length - 1 [OK]

Common Mistakes:

• Using length as index directly
• Loop running one extra iteration
• Ignoring null terminator position

5. You want to reverse only the vowels in a string "education" using C. Which approach correctly achieves this?

hard

A. Reverse entire string and then remove consonants.

B. Use two pointers from start and end, swap vowels only, skip consonants.

C. Copy vowels to a new string and print it reversed.

D. Replace vowels with uppercase letters in place.

Solution

1. Step 1: Understand vowel-only reversal

   We must swap vowels at start and end positions, ignoring consonants.

2. Step 2: Use two pointers approach

   Initialize pointers at string start and end, move inward skipping consonants, swap vowels when both pointers point to vowels.

3. Step 3: Confirm other options are incorrect

   Reverse entire string and then remove consonants. reverses whole string, not vowels only. Copy vowels to a new string and print it reversed. prints vowels only, losing consonants. Replace vowels with uppercase letters in place. changes vowels but does not reverse.

4. Final Answer:

   Two pointers swapping vowels only -> Option B

5. Quick Check:

   Two-pointer vowel swap reverses vowels [OK]

Hint: Swap vowels with two pointers, skip consonants [OK]

Common Mistakes:

• Reversing entire string instead of vowels only
• Removing consonants accidentally
• Changing vowels without reversing