String Reversal Approaches in DSA C - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the simplest way to reverse a string in C using a loop?

Use two pointers: one at the start and one at the end of the string. Swap characters at these pointers and move them towards the center until they meet or cross.

intermediate

How does recursion help in reversing a string?

Recursion reverses a string by swapping the first and last characters and then calling itself on the substring excluding these characters until the base case (empty or single character) is reached.

beginner

What is the role of the null character '\0' in C strings during reversal?

The null character marks the end of the string. It should not be moved or swapped during reversal to avoid corrupting the string termination.

beginner

Why is it important to handle string length correctly before reversal?

Knowing the string length helps set the correct end pointer for swapping characters. Incorrect length can cause out-of-bound errors or incomplete reversal.

intermediate

Can you reverse a string in C without using extra memory? How?

Yes, by swapping characters in place using two pointers moving from start and end towards the center, no extra memory is needed.

Which method uses two pointers to reverse a string in C?

A Using string concatenation

B Using a temporary array to store reversed string

C Using recursion without swapping

D Swapping characters from start and end moving inward

The two-pointer method swaps characters from the start and end moving inward until the string is reversed.

What is the base case in recursive string reversal?

A When the string is longer than 10 characters

B When the string contains only vowels

C When the string length is zero or one

D When the string contains spaces

The recursion stops when the string is empty or has a single character, as it is already reversed.

Why should the null character '\0' not be swapped during reversal?

A It marks the end of the string and swapping it breaks the string

B It is a visible character

C It is a space character

D It is a number

The null character marks the string end; moving it corrupts the string termination.

Which function is commonly used to find string length before reversal in C?

A strlen()

B strrev()

C strcpy()

D strcmp()

strlen() returns the length of the string, needed to set pointers for reversal.

What is the time complexity of reversing a string using the two-pointer approach?

A O(n^2)

B O(n)

C O(log n)

D O(1)

Each character is swapped once, so the time complexity is linear, O(n).

Explain how to reverse a string in C using the two-pointer approach.

Think about swapping characters from both ends moving towards the center.

Describe the recursive method to reverse a string and its base case.

Recursion breaks the problem into smaller parts until a simple case is reached.

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