String Reversal Approaches in DSA C

Mental Model

Reversing a string means flipping its characters so the last becomes first and the first becomes last.

Analogy: Imagine reading a sentence backward, starting from the end and moving to the beginning, like rewinding a tape.

Original: h -> e -> l -> l -> o -> \nReversed: o -> l -> l -> e -> h -> \n↑start ↑end

Dry Run Walkthrough

Input: string: "hello"

Goal: Reverse the string so it reads "olleh"

Step 1: Swap first and last characters: 'h' and 'o'

o -> e -> l -> l -> h

Why: First and last characters must switch places to start reversal

Step 2: Swap second and second-last characters: 'e' and 'l'

o -> l -> l -> e -> h

Why: Continue swapping inward to reverse the string

Step 3: Middle character 'l' remains as is (odd length string)

o -> l -> l -> e -> h

Why: Middle character does not move in odd length strings

Result:

o -> l -> l -> e -> h
Reversed string: "olleh"

Annotated Code

DSA C

#include <stdio.h>
#include <string.h>

void reverseString(char *str) {
    int left = 0;
    int right = strlen(str) - 1;
    while (left < right) {
        // Swap characters at left and right
        char temp = str[left];
        str[left] = str[right];
        str[right] = temp;
        // Move pointers inward
        left++;
        right--;
    }
}

int main() {
    char str[] = "hello";
    reverseString(str);
    printf("%s\n", str);
    return 0;
}

while (left < right) {

loop until pointers meet or cross to cover all swaps

char temp = str[left]; str[left] = str[right]; str[right] = temp;

swap characters at left and right positions

left++; right--;

move pointers inward to next pair

Output

olleh

Complexity Analysis

Time: O(n) because each character is visited once during swapping

Space: O(1) because reversal is done in place without extra storage

vs Alternative: Compared to creating a new reversed string (O(n) space), this approach saves memory by swapping in place

Edge Cases

empty string

function returns immediately with no changes

DSA C

while (left < right) {

single character string

no swaps occur, string remains same

DSA C

while (left < right) {

string with all same characters

swapping identical characters keeps string unchanged

DSA C

char temp = str[left];

When to Use This Pattern

When asked to reverse a string or array in place, use two pointers moving inward swapping elements to achieve reversal efficiently.

Common Mistakes

Mistake: Swapping characters without moving pointers inward causes infinite loop
Fix: Increment left and decrement right pointers after each swap

Mistake: Using a new string without freeing memory in languages like C
Fix: Prefer in-place reversal to avoid extra memory management

Summary

Reverses a string by swapping characters from ends moving inward.

Use when you need to reverse strings efficiently without extra space.

The key is two pointers swapping characters until they meet in the middle.

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