What is the time complexity of Valid Palindrome II?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Valid Palindrome II?

This problem uses Two Pointers, String, Greedy. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Valid Palindrome II?

Insight 1: The problem can be solved using the two-pointer approach to check if a string is a palindrome. If a mismatch is found, we explore two possibilities: deleting the left character or deleting the right character. Insight 2: The core idea is to identify the first mismatch when checking for a palindrome and then recursively check if the string becomes a palindrome after deleting either the left or the right character that caused the mismatch.

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Valid Palindrome II - LeetCode 680 Solution

Q: How to solve Valid Palindrome II?

The solution uses a two-pointer approach with a helper function. The main function iterates through the string using two pointers, `l` and `r`, starting from the beginning and end respectively. If the characters at these pointers match, they are incremented/decremented. If they don't match, it means we need to delete one of these characters. The solution then calls the `isPalindrome` helper function twice: once with `l + 1, r` (deleting the left character) and once with `l, r - 1` (deleting the right character). If either of these calls returns `true`, the original string can be made a palindrome by deleting at most one character. If the initial loop completes without finding any mismatches, the string is already a palindrome.

Q: Which companies ask LeetCode 680. Valid Palindrome II in interviews?

This problem is frequently asked at Attentive, Fortinet, Roku, Walmart Labs, Whatnot and 2 more companies.

Valid Palindrome II - Complete Solution Guide

Valid Palindrome II is LeetCode problem 680, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given a string s , return true if the s can be palindrome after deleting at most one character from it . Example 1: Input: s = "aba" Output: true Example 2: Input: s = "abca" Output: true Explanation: You could delete the character 'c'. Example 3: Input: s = "abc" Output: false Constraints: 1 <= s.length <= 10 5 s consists of lowercase English letters.

Detailed Explanation

The problem asks us to determine if a given string `s` can be made a palindrome by deleting at most one character. A palindrome is a string that reads the same forwards and backward. The input is a string `s` consisting of lowercase English letters. The output is a boolean value: `true` if the string can become a palindrome after deleting at most one character, and `false` otherwise. The constraints are that the string length is between 1 and 10^5.

Solution Approach

The solution uses a two-pointer approach with a helper function. The main function iterates through the string using two pointers, `l` and `r`, starting from the beginning and end respectively. If the characters at these pointers match, they are incremented/decremented. If they don't match, it means we need to delete one of these characters. The solution then calls the `isPalindrome` helper function twice: once with `l + 1, r` (deleting the left character) and once with `l, r - 1` (deleting the right character). If either of these calls returns `true`, the original string can be made a palindrome by deleting at most one character. If the initial loop completes without finding any mismatches, the string is already a palindrome.

Step-by-Step Algorithm

Step 1: Initialize two pointers, `l` and `r`, to the start and end of the string, respectively.
Step 2: While `l` is less than `r`, compare the characters at `s[l]` and `s[r]`.
Step 3: If `s[l]` and `s[r]` are not equal, call the `isPalindrome` helper function twice: once with `l + 1, r` and once with `l, r - 1`. Return `true` if either of these calls returns `true`.
Step 4: If `s[l]` and `s[r]` are equal, increment `l` and decrement `r`.
Step 5: If the loop completes without finding any mismatches, return `true` (the string is already a palindrome).
Step 6: The `isPalindrome` helper function takes a string and two pointers, `l` and `r`, and checks if the substring between `l` and `r` is a palindrome. It returns `true` if it is, and `false` otherwise.

Key Insights

Insight 1: The problem can be solved using the two-pointer approach to check if a string is a palindrome. If a mismatch is found, we explore two possibilities: deleting the left character or deleting the right character.
Insight 2: The core idea is to identify the first mismatch when checking for a palindrome and then recursively check if the string becomes a palindrome after deleting either the left or the right character that caused the mismatch.
Insight 3: The `isPalindrome` helper function helps to abstract out the palindrome checking logic, making the code cleaner and easier to understand.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Two Pointers, String, Greedy.

Companies

Asked at: Attentive, Fortinet, Roku, Walmart Labs, Whatnot, Yandex, eBay.

Valid Palindrome II | LeetCode 680

Easy

Two Pointers String Greedy

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public boolean validPalindrome(String s) {
        int left = 0;
        int right = s.length() - 1;
        while (left < right) {
            if (s.charAt(left) != s.charAt(right)) {
                return isPalindrome(s, left + 1, right) || isPalindrome(s, left, right - 1);
            }
            left++;
            right--;
        }
        return true;
    }

    private boolean isPalindrome(String s, int left, int right) {
        while (left < right) {
            if (s.charAt(left) != s.charAt(right)) {
                return false;
            }
            left++;
            right--;
        }
        return true;
    }
}

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Problem Info

Description

Problem 680: Valid Palindrome II - Detailed explanation and solution approaches.

Topics

Two Pointers String Greedy

Companies

AttentiveFortinetRokuWalmart LabsWhatnotYandexeBay

Stats

Difficulty:Easy

Languages:4

Companies:7

Problem #:680

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Initialize two pointers, `l` and `r`, to the start and end of the string, respectively.
2
Step 2: While `l` is less than `r`, compare the characters at `s[l]` and `s[r]`.
3
Step 3: If `s[l]` and `s[r]` are not equal, call the `isPalindrome` helper function twice: once with `l + 1, r` and once with `l, r - 1`. Return `true` if either of these calls returns `true`.
4
Step 4: If `s[l]` and `s[r]` are equal, increment `l` and decrement `r`.
5
Step 5: If the loop completes without finding any mismatches, return `true` (the string is already a palindrome).
6
Step 6: The `isPalindrome` helper function takes a string and two pointers, `l` and `r`, and checks if the substring between `l` and `r` is a palindrome. It returns `true` if it is, and `false` otherwise.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The two-pointer approach iterates through the string at most once in the `validPalindrome` function. If a mismatch is found, the `isPalindrome` function is called twice, each iterating through the string at most once. Thus, the time complexity is linear. No extra space is used that scales with the size of the input string; only constant space is used for variables and function calls.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursive Approach Without Helper

A recursive approach could be implemented without the `isPalindrome` helper function, directly within the `validPalindrome` function. This approach would involve modifying the main function to call itself recursively when a mismatch is found.

Time: O(n)Space: O(n) in worst case due to recursion depth

Trade-offs: The helper function makes the code clearer. Without it, the recursive calls would be harder to read and understand. The space complexity would be increased due to the recursive call stack potentially reaching a depth of n in the worst-case scenario. Using the two pointer approach with a helper function allows to have O(1) space complexity.

✨

Key Insights

1Insight 1: The problem can be solved using the two-pointer approach to check if a string is a palindrome. If a mismatch is found, we explore two possibilities: deleting the left character or deleting the right character.
2Insight 2: The core idea is to identify the first mismatch when checking for a palindrome and then recursively check if the string becomes a palindrome after deleting either the left or the right character that caused the mismatch.
3Insight 3: The `isPalindrome` helper function helps to abstract out the palindrome checking logic, making the code cleaner and easier to understand.

📚

Key Terms Explained

For Students

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Greedy Algorithm

Making the best choice at each step without looking ahead.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty string or single-character string. The code correctly handles these cases as they are already palindromes, and the initial `while` loop condition `l < r` will be false, immediately returning `true`.
Edge case 2: String that is already a palindrome (e.g., "aba"). The `while` loop in `validPalindrome` will complete without finding any mismatches, and the function will correctly return `true`.
Edge case 3: String where deleting the character at the beginning or end makes it a palindrome (e.g., "abca"). The code correctly explores both possibilities by calling `isPalindrome(l + 1, r)` and `isPalindrome(l, r - 1)`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not checking both possibilities when a mismatch is found. It's important to consider both deleting the left character and deleting the right character, as only one of these deletions might result in a palindrome.

✗Common mistake 2: Incorrectly implementing the `isPalindrome` helper function. Make sure the loop condition is `l < r` and that the pointers are incremented/decremented correctly.

✗Common mistake 3: Forgetting to handle the case where the string is already a palindrome. The code should return `true` in this case.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Text editors and word processors often use palindrome detection (or near-palindrome detection) for features like autocorrect or spell checking, to suggest corrections that are close to the user's intended input.

🏦Real-world scenario 2: Bioinformatics. Palindrome sequences are significant in DNA analysis. Slight variations or near-palindrome sequences can indicate mutations or genetic markers.

🎮Real-world scenario 3: Data validation and sanitization. This logic can be adapted to check if data entered by users is 'almost' valid, allowing for minor errors without rejecting the entire input.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

The two-pointer technique is powerful for sorted arrays. The key insight is recognizing when elements can be processed from both ends simultaneously.

🎯 Interview Tip

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

The most common mistake is not handling the case when pointers meet. Always consider the termination condition.

👨‍💻 From Experience

This is one of the first problems I solved when learning Two Pointers. It's a great foundation for more complex problems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 19, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

❓

Frequently Asked Questions

When is the two-pointer technique most effective?

Two pointers work best on sorted arrays or when searching for pairs/triplets that satisfy a condition. Common patterns include: moving pointers inward from both ends (like in two-sum variants), fast/slow pointers for cycle detection, or left/right pointers for partitioning.

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Valid Palindrome II"?

For this easy-level Two Pointers problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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