Valid Palindrome Two Pointer in DSA C - Execution Trace

Concept Flow - Valid Palindrome Two Pointer

Start: left=0, right=length-1

↓

Check if left < right?

No→Done: Palindrome

Yes↓

Is s[left

Yes↓

Is s[right

Yes↓

Compare s[left

↓

left++, right--

Two pointers start at the string ends, move inward skipping non-alphanumeric characters, comparing characters case-insensitively until pointers cross or mismatch.

Execution Sample

DSA C

bool isPalindrome(char * s) {
    int left = 0, right = strlen(s) - 1;
    while (left < right) {
        if (!isalnum(s[left])) left++;
        else if (!isalnum(s[right])) right--;
        else if (tolower(s[left]) != tolower(s[right])) return false;
        else { left++; right--; }
    }
    return true;
}

Checks if a string is a palindrome ignoring non-alphanumeric characters and case using two pointers.

Execution Table

Step	Operation	left	right	Characters Compared	Action	Visual State
1	Initialize pointers	0	29	-	Start	"A man, a plan, a canal: Panama"
2	Check s[left] alnum	0	29	A vs a	Compare (equal)	"A man, a plan, a canal: Panama"
3	Move pointers	1	28	-	left++, right--	"A man, a plan, a canal: Panama"
4	Check s[left] alnum	1	28	vs m	left not alnum, left++	"A man, a plan, a canal: Panama"
5	Check s[left] alnum	2	28	m vs m	Compare (equal)	"A man, a plan, a canal: Panama"
6	Move pointers	3	27	-	left++, right--	"A man, a plan, a canal: Panama"
7	Check s[left] alnum	3	27	a vs a	Compare (equal)	"A man, a plan, a canal: Panama"
8	Move pointers	4	26	-	left++, right--	"A man, a plan, a canal: Panama"
9	Check s[left] alnum	4	26	n vs n	Compare (equal)	"A man, a plan, a canal: Panama"
10	Move pointers	5	25	-	left++, right--	"A man, a plan, a canal: Panama"
11	Check s[left] alnum	5	25	, vs a	left not alnum, left++	"A man, a plan, a canal: Panama"
12	Check s[left] alnum	6	25	vs a	left not alnum, left++	"A man, a plan, a canal: Panama"
13	Check s[left] alnum	7	25	a vs a	Compare (equal)	"A man, a plan, a canal: Panama"
14	Move pointers	8	24	-	left++, right--	"A man, a plan, a canal: Panama"
15	Check s[left] alnum	8	24	vs P	left not alnum, left++	"A man, a plan, a canal: Panama"
16	Check s[left] alnum	9	24	p vs P	Compare (equal)	"A man, a plan, a canal: Panama"
17	Move pointers	10	23	-	left++, right--	"A man, a plan, a canal: Panama"
18	Check s[left] alnum	10	23	l vs	right not alnum, right--	"A man, a plan, a canal: Panama"
19	Check s[left] alnum	10	22	l vs :	right not alnum, right--	"A man, a plan, a canal: Panama"
20	Check s[left] alnum	10	21	l vs l	Compare (equal)	"A man, a plan, a canal: Panama"
21	Move pointers	11	20	-	left++, right--	"A man, a plan, a canal: Panama"
22	Check s[left] alnum	11	20	a vs a	Compare (equal)	"A man, a plan, a canal: Panama"
23	Move pointers	12	19	-	left++, right--	"A man, a plan, a canal: Panama"
24	Check s[left] alnum	12	19	n vs n	Compare (equal)	"A man, a plan, a canal: Panama"
25	Move pointers	13	18	-	left++, right--	"A man, a plan, a canal: Panama"
26	Check s[left] alnum	13	18	, vs a	left not alnum, left++	"A man, a plan, a canal: Panama"

💡 After step 26 (left=14, right=18), continue similarly: skip left ' ' at 14 to 15, compare 'a' vs 'a' at 15/18, move to 16/17; skip left ' ' at 16 to 17; left == right =17, all pairs matched, palindrome.

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	After 5	After 6	After 7	After 8	After 9	After 10	Final
left	0	1	2	3	4	5	7	8	9	10	10	17
right	29	28	28	27	26	25	25	24	24	23	21	17

Key Moments - 3 Insights

Why do we skip characters that are not alphanumeric?

Because the palindrome check ignores spaces, punctuation, and symbols. See steps 4, 11, 12, 15, and 26 in the execution_table where pointers move without comparison.

Why do we compare characters in lowercase?

To make the comparison case-insensitive. For example, 'A' and 'a' are considered equal as shown in step 2 and 'p' and 'P' in step 16.

When does the loop stop?

The loop stops when left pointer is no longer less than right pointer, meaning all pairs have been checked. This occurs when pointers meet or cross after final skips (left=17, right=17).

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of 'left' after step 5?

A 2

B 3

C 1

D 5

💡 Hint

Check the 'left' column in execution_table row with Step 5.

At which step does the right pointer first move without comparison because of a non-alphanumeric character?

A Step 4

B Step 11

C Step 18

D Step 25

💡 Hint

Look for steps where 'Action' says 'right not alnum, right--' in execution_table.

If the input string was "race a car", at which step would the function return false?

A When comparing 'e' and 'a'

B When comparing 'r' and 'r'

C When comparing 'c' and 'c'

D When comparing 'a' and 'a'

💡 Hint

Think about which characters mismatch in the palindrome check and when the function returns false.

Concept Snapshot

Two pointers start at string ends
Skip non-alphanumeric chars
Compare characters case-insensitive
Move pointers inward if equal
Stop when pointers cross or mismatch

Full Transcript

This visualization shows how to check if a string is a palindrome using two pointers. We start with one pointer at the beginning and one at the end. We skip characters that are not letters or numbers. We compare characters ignoring case. If they match, we move both pointers inward. If they don't match, we stop and say it's not a palindrome. The process continues until the pointers cross, meaning the string is a palindrome.

Practice

1. What is the main idea behind the two-pointer technique to check if a string is a palindrome?

easy

A. Count the frequency of each character and check if all are even

B. Sort the string and check if it matches the original

C. Reverse the string and compare with the original

D. Use two pointers starting from both ends and move towards the center comparing characters

Solution

1. Step 1: Understand two-pointer setup

   Two pointers start at the beginning and end of the string to compare characters.

2. Step 2: Move pointers inward while characters match

   If characters at both pointers are equal, move pointers closer until they meet or cross.

3. Final Answer:

   Use two pointers starting from both ends and move towards the center comparing characters -> Option D

4. Quick Check:

   Two-pointer palindrome check = Use two pointers starting from both ends and move towards the center comparing characters [OK]

Hint: Compare characters from ends moving inward [OK]

Common Mistakes:

• Trying to sort the string which changes order
• Only checking half the string without pointers
• Ignoring non-alphanumeric characters if required

2. Which of the following C code snippets correctly initializes two pointers for palindrome checking on a string s?

easy

A. int left = 0, right = strlen(s) - 1;

B. int left = 1, right = strlen(s);

C. int left = 0, right = strlen(s);

D. int left = 1, right = strlen(s) - 1;

Solution

1. Step 1: Understand string indexing in C

   Strings are zero-indexed, so first character is at 0 and last at length-1.

2. Step 2: Correct pointer initialization

   Left pointer starts at 0, right pointer at strlen(s)-1 to cover full string.

3. Final Answer:

   int left = 0, right = strlen(s) - 1; -> Option A

4. Quick Check:

   Indexing starts at 0, last index = length-1 [OK]

Hint: Start left at 0 and right at length minus one [OK]

Common Mistakes:

• Setting right pointer to strlen(s) which is out of bounds
• Starting left pointer at 1 skipping first character
• Using wrong indices causing runtime errors

3. What will be the output of the following C code snippet?

char s[] = "racecar";
int left = 0, right = strlen(s) - 1;
int isPalindrome = 1;
while (left < right) {
    if (s[left] != s[right]) {
        isPalindrome = 0;
        break;
    }
    left++;
    right--;
}
printf("%d", isPalindrome);

medium

A. 0

B. 1

C. 7

D. Compilation error

Solution

1. Step 1: Trace pointer comparisons on "racecar"

   Characters at left and right match: 'r'=='r', 'a'=='a', 'c'=='c', 'e'=='e'.

2. Step 2: Loop completes without mismatch

   No mismatch found, so isPalindrome remains 1 and prints 1.

3. Final Answer:

   1 -> Option B

4. Quick Check:

   "racecar" is palindrome = 1 [OK]

Hint: Check matching pairs from ends; no mismatch means palindrome [OK]

Common Mistakes:

• Assuming output is string instead of integer
• Confusing index bounds causing wrong comparisons
• Forgetting to break loop on mismatch

4. Identify the bug in the following C code that checks if a string s is a palindrome using two pointers:

int left = 0, right = strlen(s);
while (left <= right) {
    if (s[left] != s[right]) {
        return 0;
    }
    left++;
    right--;
}
return 1;

medium

A. Both A and B are bugs

B. Loop condition should be left < right, not left <= right

C. Right pointer should be initialized to strlen(s) - 1

D. No bug, code is correct

Solution

1. Step 1: Check right pointer initialization

   Right pointer set to strlen(s) is out of bounds; should be strlen(s) - 1.

2. Step 2: Check loop condition

   Loop should run while left < right to avoid comparing same middle character twice.

3. Final Answer:

   Both A and B are bugs -> Option A

4. Quick Check:

   Right pointer and loop condition both incorrect = Both A and B are bugs [OK]

Hint: Right pointer = length-1 and loop: left < right [OK]

Common Mistakes:

• Using strlen(s) as right pointer causing out-of-bounds access
• Using <= in loop causing unnecessary comparison
• Ignoring off-by-one errors

5. Given a string s that may contain uppercase letters, lowercase letters, and non-alphanumeric characters, which approach correctly checks if s is a palindrome ignoring cases and non-alphanumeric characters using two pointers?

hard

A. Count frequency of alphanumeric characters and check if all counts are even

B. Convert entire string to lowercase, remove non-alphanumeric chars, then compare with reversed string

C. Use two pointers from ends, skip non-alphanumeric chars, compare lowercase characters

D. Use two pointers but compare characters directly without skipping or case conversion

Solution

1. Step 1: Use two pointers and skip non-alphanumeric characters

   Move left and right pointers inward, skipping characters that are not letters or digits.

2. Step 2: Compare characters in lowercase form

   Convert characters to lowercase before comparing to ignore case differences.

3. Final Answer:

   Use two pointers from ends, skip non-alphanumeric chars, compare lowercase characters -> Option C

4. Quick Check:

   Two-pointer with skip and lowercase = Use two pointers from ends, skip non-alphanumeric chars, compare lowercase characters [OK]

Hint: Skip non-alphanumeric and compare lowercase chars with two pointers [OK]

Common Mistakes:

• Comparing characters without skipping non-alphanumeric
• Not converting characters to lowercase before comparison
• Using frequency count which doesn't check order