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easyBasicsSometimes asked

Palindrome String

Basics · 6 min

Given a string s, return whether it reads the same forwards and backwards (case-sensitive, comparing all characters).

In plain English

Compare characters from both ends moving inward; any mismatch means it's not a palindrome.

Examples

Example 1

Input: s = "madam"
Output: true
Note: Same both ways.

Example 2

Input: s = "hello"
Output: false
Note: h ≠ o.

Example 3

Input: s = "a"
Output: true
Note: Single char.

Constraints

0 ≤ s.length ≤ 10^4

Clarifying questions1First question: Case-sensitive? Ignore punctuation?Explore questions

Q1.Case-sensitive? Ignore punctuation?
This version compares all chars literally; LC125 normalises first.

Quick interview revision

Understand

30-second explanation

Build the interview-ready mental model before comparing implementations.

Two pointers from the ends: if `s[left] != s[right]` return false; otherwise converge. If they cross, it's a palindrome. O(n) time, O(1) space.

Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview context

What's being tested

Whether you use the O(1)-space two-pointer symmetry check.

Where candidates slip

Reverse-and-compare with extra space, or overshooting pointers.

How to stand out

Mention the case/alphanumeric-insensitive variant (LC125).

Choose

Compare the available approaches

Use constraints and trade-offs to select the route you can defend in an interview.

Learn & trace

Build each solution step by step

Move from the core idea to a visual trace, implementation, proof, and edge cases.

03.1

Approach 1 of 1 · Optimal

Two-pointer compare

TimeO(n)SpaceO(1)

This is the standard, expected solution — lead with it.

Stuck? Progressive hintsReveal one at a time

1Hint 1Explore hint

What relationship must the i-th front and i-th back characters have?

2Hint 2Explore hint

Two pointers inward; mismatch ⇒ not a palindrome.

3Hint 3Explore hint

Avoid building a reversed copy — two pointers are O(1) space.

A · Core idea

A palindrome is symmetric about its centre, so the i-th character from the front must equal the i-th from the back. Compare inward with two pointers and bail on the first mismatch — O(n) time, O(1) space.

Why not build a reversed copy. Reversing and comparing works but uses O(n) extra space and always scans the whole string. The two-pointer check is O(1) space and can exit early on the first mismatched pair.

Why the pointers meet, not cross. With left < right, the middle character of an odd-length string is never compared against itself (it's trivially equal), and even-length strings compare every pair exactly once.

Common real-world variant. 'Valid palindrome' often ignores case and non-alphanumeric characters — there you skip non-letters and lower-case before comparing. This basic version compares all characters as-is.

B · Visual dry run

Interactive dry run

Select a frame to connect the decision with its visual state.

s = "madam"

l=0,r=4

m == m

Stateok

Final resulttrue

C · Code with teaching comments

solution.javaJava

Two-pointer compare — optimal

1 teaching note

1
public boolean isPalindrome(String s) {
2
    int left = 0, right = s.length() - 1;
3
    while (left < right) {
4
        // First mismatched pair disproves the palindrome.
5
        if (s.charAt(left) != s.charAt(right)) return false;
6
        left++;
7
        right--;
8
    }
9
    return true;
10
}

Yellow row = the algorithm's core logic. Amber `//` lines above it = what you'd say out loud.10 lines

D · Why it works and complexityO(n) timeO(1) spaceOpen the invariant and proofExplore the proof

ComplexityO(n) time, O(1) space — at most n/2 comparisons, with early exit on the first mismatch, and only two index variables.

E · Edge cases and pitfalls5s = "a" — Single char → true.Explore edge cases

Edge cases handled

s = "a"Single char → true.
s = ""Empty → true by convention.
s = "ab"a ≠ b → false.

Pitfalls

Building a reversed string (O(n) space) instead of two pointers.
Off-by-one letting the pointers overshoot.

Practice

Try the solution yourself

Edit the optimal implementation, test another input, and observe the result.

Editable code previewpublic boolean isPalindrome(String s) { int left = 0, right = s.length() - 1; while (left < right) { if (s.charAt(left) != s.charAt(right)) return false; left++;Explore the playground

Try it — Java· run a complete program; method snippets need a driver

Avoid mistakes

Recognize the interview traps

Understand why common attempts fail before reviewing the corrected implementation.

Building a reversed string (O(n) space) when two pointers are O(1).
Off-by-one letting the pointers overshoot.

Revise

Carry the pattern into the interview

Quick interview revision

Pattern to remember

Two-pointer symmetry check

Code pattern to memorize

l,r=0,n-1; while l<r: if s[l]!=s[r] return false; l++; r--

Rules to remember

1Compare ends inward.
2Mismatch ⟹ false.
3O(1) space.

Use this pattern when

Symmetry checks
Two-pointer string scans

Interview takeaway

Palindrome checks are symmetry — compare from both ends.

Continue

Transfer the pattern to another problem

Use a direct variation or pattern sibling to prove the idea is reusable.

Transferable patternA core fresher pattern: Two-pointer symmetry check. Master it on small inputs and it transfers to harder problems.

Practice queueDirect variations and pattern siblings

Direct follow-ups

Valid Palindrome
Coming soon
Ignore case and non-alphanumerics while two-pointer scanning.
Palindrome Number
Same symmetry idea on digits.

Same pattern

easyBasicsSometimes asked

Palindrome String

Basics · 6 min

Given a string s, return whether it reads the same forwards and backwards (case-sensitive, comparing all characters).

In plain English

Compare characters from both ends moving inward; any mismatch means it's not a palindrome.

Examples

Example 1

Input: s = "madam"
Output: true
Note: Same both ways.

Example 2

Input: s = "hello"
Output: false
Note: h ≠ o.

Example 3

Input: s = "a"
Output: true
Note: Single char.

Constraints

0 ≤ s.length ≤ 10^4

Clarifying questions1First question: Case-sensitive? Ignore punctuation?Explore questions

Q1.Case-sensitive? Ignore punctuation?
This version compares all chars literally; LC125 normalises first.

Quick interview revision

Understand

30-second explanation

Build the interview-ready mental model before comparing implementations.

Two pointers from the ends: if `s[left] != s[right]` return false; otherwise converge. If they cross, it's a palindrome. O(n) time, O(1) space.

Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview context

What's being tested

Whether you use the O(1)-space two-pointer symmetry check.

Where candidates slip

Reverse-and-compare with extra space, or overshooting pointers.

How to stand out

Mention the case/alphanumeric-insensitive variant (LC125).

Choose

Compare the available approaches

Use constraints and trade-offs to select the route you can defend in an interview.

Learn & trace

Build each solution step by step

Move from the core idea to a visual trace, implementation, proof, and edge cases.

03.1

Approach 1 of 1 · Optimal

Two-pointer compare

TimeO(n)SpaceO(1)

This is the standard, expected solution — lead with it.

Stuck? Progressive hintsReveal one at a time

1Hint 1Explore hint

What relationship must the i-th front and i-th back characters have?

2Hint 2Explore hint

Two pointers inward; mismatch ⇒ not a palindrome.

3Hint 3Explore hint

Avoid building a reversed copy — two pointers are O(1) space.

A · Core idea

B · Visual dry run

Interactive dry run

Select a frame to connect the decision with its visual state.

s = "madam"

l=0,r=4

m == m

Stateok

Final resulttrue

C · Code with teaching comments

solution.javaJava

Two-pointer compare — optimal

1 teaching note

1
public boolean isPalindrome(String s) {
2
    int left = 0, right = s.length() - 1;
3
    while (left < right) {
4
        // First mismatched pair disproves the palindrome.
5
        if (s.charAt(left) != s.charAt(right)) return false;
6
        left++;
7
        right--;
8
    }
9
    return true;
10
}

Yellow row = the algorithm's core logic. Amber `//` lines above it = what you'd say out loud.10 lines

D · Why it works and complexityO(n) timeO(1) spaceOpen the invariant and proofExplore the proof

ComplexityO(n) time, O(1) space — at most n/2 comparisons, with early exit on the first mismatch, and only two index variables.

E · Edge cases and pitfalls5s = "a" — Single char → true.Explore edge cases

Edge cases handled

s = "a"Single char → true.
s = ""Empty → true by convention.
s = "ab"a ≠ b → false.

Pitfalls

Building a reversed string (O(n) space) instead of two pointers.
Off-by-one letting the pointers overshoot.

Practice

Try the solution yourself

Edit the optimal implementation, test another input, and observe the result.

Try it — Java· run a complete program; method snippets need a driver

Avoid mistakes

Recognize the interview traps

Understand why common attempts fail before reviewing the corrected implementation.

Building a reversed string (O(n) space) when two pointers are O(1).
Off-by-one letting the pointers overshoot.

Revise

Carry the pattern into the interview

Quick interview revision

Pattern to remember

Two-pointer symmetry check

Code pattern to memorize

l,r=0,n-1; while l<r: if s[l]!=s[r] return false; l++; r--

Rules to remember

1Compare ends inward.
2Mismatch ⟹ false.
3O(1) space.

Use this pattern when

Symmetry checks
Two-pointer string scans

Interview takeaway

Palindrome checks are symmetry — compare from both ends.

Continue

Transfer the pattern to another problem

Use a direct variation or pattern sibling to prove the idea is reusable.

Transferable patternA core fresher pattern: Two-pointer symmetry check. Master it on small inputs and it transfers to harder problems.

Practice queueDirect variations and pattern siblings

Direct follow-ups

Valid Palindrome
Coming soon
Ignore case and non-alphanumerics while two-pointer scanning.
Palindrome Number
Same symmetry idea on digits.

Same pattern