Valid Palindrome Two Pointer in DSA Rust - 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 = left+1, right=right-1

↩Loop back to check left < right

Two pointers start at the string ends, move inward skipping non-alphanumeric, compare characters ignoring case, stop if mismatch or pointers cross.

Execution Sample

DSA Rust

fn is_palindrome(s: &str) -> bool {
    let (mut left, mut right) = (0, s.len().saturating_sub(1));
    while left < right {
        // skip non-alphanumeric
        // compare characters
    }
    true
}

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

Execution Table

Step	Operation	left	right	s[left]	s[right]	Action	Visual State
1	Initialize pointers	0	10	A	a	Compare 'A' and 'a'	A b, 3, a, B, a, 3, b, A
2	Characters equal (case-insensitive)	1	9			Skip left (space)	A b, 3, a, B, a, 3, b, A
3	Skip non-alphanumeric left	2	9	b		Skip right (space)	A b, 3, a, B, a, 3, b, A
4	Skip non-alphanumeric right	2	8	b	b	Compare 'b' and 'b'	A b, 3, a, B, a, 3, b, A
5	Characters equal	3	7	3	3	Compare '3' and '3'	A b, 3, a, B, a, 3, b, A
6	Characters equal	4	6	a	a	Compare 'a' and 'a'	A b, 3, a, B, a, 3, b, A
7	Characters equal	5	5	B	B	Pointers met, palindrome confirmed	A b, 3, a, B, a, 3, b, A
8	End	5	5	-	-	Return True	Palindrome confirmed

💡 Pointers crossed or met without mismatch, string is palindrome

Variable Tracker

Variable	Step 1	Step 2	Step 3	Step 4	Step 5	Step 6	Step 7	Step 8
left	0	1	2	2	3	4	5	5
right	10	9	9	8	7	6	5	5
s[left]	A		b	b	3	a	B	-
s[right]	a			b	3	a	B	-

Key Moments - 3 Insights

Why do we skip characters that are not letters or numbers?

Because the palindrome check ignores spaces and punctuation. See steps 2 and 3 in execution_table where left and right pointers skip spaces.

Why do we compare characters ignoring case?

Because 'A' and 'a' are considered the same letter in palindrome check. Step 1 shows comparing 'A' and 'a' as equal.

What happens when left and right pointers meet or cross?

It means all characters matched so far, so the string is palindrome. Step 7 shows pointers meeting and confirming palindrome.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of 'left' after step 4?

A 1

B 3

C 2

D 4

💡 Hint

Check the 'left' column in execution_table row for step 4

At which step do the pointers first skip a non-alphanumeric character?

A Step 1

B Step 2

C Step 4

D Step 5

💡 Hint

Look at the 'Action' column in execution_table for skipping characters

If the string had a mismatch at step 5, what would happen?

A Return False immediately

B Return True immediately

C Skip characters and continue

D Move pointers inward and continue

💡 Hint

Refer to the concept_flow where mismatch leads to returning False

Concept Snapshot

Two pointers start at string ends
Skip non-alphanumeric characters
Compare characters ignoring case
If mismatch, return false
If pointers meet/cross, return true

Full Transcript

This concept uses two pointers starting at the beginning and end of the string. They move inward, skipping any characters that are not letters or numbers. At each step, the characters at the pointers are compared ignoring case. If they differ, the function returns false immediately. If the pointers meet or cross without mismatches, the string is a palindrome and the function returns true.

Practice

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

easy

A. Reverse the string and compare it with the original string

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

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

D. Sort the characters and check if the sorted string equals the original

Solution

1. Step 1: Understand two-pointer approach

   The two-pointer technique uses one pointer at the start and one at the end of the string.

2. Step 2: Compare characters moving inward

   Both pointers move towards the center, comparing characters at each step to check equality.

3. Final Answer:

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

4. Quick Check:

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

Hint: Compare characters from both ends moving inward [OK]

Common Mistakes:

• Thinking reversing string is the two-pointer method
• Checking character counts instead of positions
• Sorting string changes order, not palindrome check

2. Which Rust code snippet correctly initializes two pointers for palindrome checking on a string s?

easy

A. let (mut left, mut right) = (0, s.len() - 1);

B. let (left, right) = (0, s.len());

C. let (mut left, mut right) = (1, s.len() - 1);

D. let (left, right) = (1, s.len());

Solution

1. Step 1: Identify pointer initialization for indices

   Left pointer starts at 0 (first char), right pointer at last index (length - 1).

2. Step 2: Check mutability and correct indices

   Both pointers must be mutable to move; right pointer must be s.len() - 1 to avoid out-of-bounds.

3. Final Answer:

   let (mut left, mut right) = (0, s.len() - 1); -> Option A

4. Quick Check:

   Correct pointer init = let (mut left, mut right) = (0, s.len() - 1); [OK]

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

Common Mistakes:

• Using s.len() as right index causes out-of-bounds
• Not making pointers mutable
• Starting left pointer at 1 skips first character

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

fn is_palindrome(s: &str) -> bool {
    let bytes = s.as_bytes();
    let (mut left, mut right) = (0, bytes.len() - 1);
    while left < right {
        if bytes[left] != bytes[right] {
            return false;
        }
        left += 1;
        right -= 1;
    }
    true
}

fn main() {
    println!("{}", is_palindrome("racecar"));
    println!("{}", is_palindrome("hello"));
}

medium

A. false\nfalse

B. false\ntrue

C. true\nfalse

D. true\ntrue

Solution

1. Step 1: Check palindrome for "racecar"

   Characters match from both ends moving inward, so returns true.

2. Step 2: Check palindrome for "hello"

   Characters differ at first comparison ('h' vs 'o'), so returns false.

3. Final Answer:

   true\nfalse -> Option C

4. Quick Check:

   racecar = true, hello = false [OK]

Hint: Check first and last chars; mismatch means false [OK]

Common Mistakes:

• Assuming hello is palindrome
• Confusing output order
• Ignoring early return on mismatch

4. Identify the bug in this Rust function that checks palindrome and select the fix:

fn is_palindrome(s: &str) -> bool {
    let bytes = s.as_bytes();
    let (mut left, mut right) = (0, bytes.len());
    while left < right {
        if bytes[left] != bytes[right] {
            return false;
        }
        left += 1;
        right -= 1;
    }
    true
}

medium

A. Change right initialization to bytes.len() - 1

B. Change while condition to left <= right

C. Make left and right immutable

D. Use s.chars() instead of bytes

Solution

1. Step 1: Identify out-of-bounds error

   Right pointer is set to bytes.len(), which is out of valid index range (0 to len-1).

2. Step 2: Fix pointer initialization

   Set right to bytes.len() - 1 to point to last valid character index.

3. Final Answer:

   Change right initialization to bytes.len() - 1 -> Option A

4. Quick Check:

   Right pointer must be last index = Change right initialization to bytes.len() - 1 [OK]

Hint: Right pointer = length minus one to avoid out-of-bounds [OK]

Common Mistakes:

• Using length as index causes panic
• Changing while condition without fixing index
• Making pointers immutable prevents movement

5. Given the string s = "A man, a plan, a canal: Panama", which Rust code modification correctly checks if s is a palindrome ignoring non-alphanumeric characters and case using two pointers?

hard

A. Convert string to uppercase and compare characters including punctuation

B. Reverse the string and compare directly without filtering

C. Count characters ignoring case but include spaces and punctuation

D. Use two pointers skipping non-alphanumeric chars and compare lowercase characters

Solution

1. Step 1: Understand problem requirements

   We must ignore non-alphanumeric characters and case differences when checking palindrome.

2. Step 2: Apply two-pointer with skipping and case normalization

   Move pointers inward, skip chars that are not letters or digits, compare lowercase versions of characters.

3. Final Answer:

   Use two pointers skipping non-alphanumeric chars and compare lowercase characters -> Option D

4. Quick Check:

   Skip non-alphanumeric + lowercase compare = Use two pointers skipping non-alphanumeric chars and compare lowercase characters [OK]

Hint: Skip non-letters/digits and compare lowercase chars [OK]

Common Mistakes:

• Comparing with punctuation included
• Ignoring case differences
• Reversing string without filtering