Valid Palindrome Two Pointer in DSA Rust - Time & Space Complexity

Time Complexity: Valid Palindrome Two Pointer

O(n)

Understanding Time Complexity

We want to know how long it takes to check if a string is a palindrome using two pointers.

How does the time needed grow as the string gets longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code checks if a string reads the same forwards and backwards by comparing characters from both ends moving towards the center.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Comparing characters at two positions in the string.
• How many times: Up to half the length of the string (n/2 times).

How Execution Grows With Input

As the string gets longer, the number of comparisons grows roughly half as fast.

Input Size (n)	Approx. Operations
10	5 comparisons
100	50 comparisons
1000	500 comparisons

Pattern observation: The operations grow linearly with input size, but only half as many steps are needed.

Final Time Complexity

Time Complexity: O(n)

This means the time to check grows directly in proportion to the string length.

Common Mistake

[X] Wrong: "The time complexity is O(n²) because we compare pairs of characters."

[OK] Correct: Each comparison is done once per pair, and we only move inward, so total comparisons grow linearly, not quadratically.

Interview Connect

Understanding this pattern helps you quickly analyze string problems and shows you can reason about efficient scanning techniques.

Self-Check

"What if we ignored non-alphanumeric characters and case differences? How would the time complexity change?"

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