Is Valid Palindrome IV easy or hard?

Valid Palindrome IV is typically rated Medium. The implementation is short, but the key insight is realizing that each mismatched mirrored pair requires one change, so the string must have at most two mismatches.

Valid Palindrome IV Python/Java solution

The implementation is identical across languages. Initialize two indices at the start and end of the string, compare characters, and increment a mismatch counter when they differ. Stop early if mismatches exceed two. This logic translates directly to Python, Java, C++, Go, and TypeScript.

How to solve Valid Palindrome IV in O(n)?

Use two pointers starting at the beginning and end of the string. If characters differ, increment a mismatch counter because one change can fix that pair. Continue moving inward until the pointers meet. If mismatches are at most two, the string can be converted into a palindrome.

What is the best approach for Valid Palindrome IV?

The best approach uses the two pointers technique. Place one pointer at the start and one at the end of the string and compare characters while moving inward. Count mismatched pairs and ensure the count does not exceed two. This runs in O(n) time and O(1) space.

Is Valid Palindrome IV asked at Google/Amazon/Meta?

Palindrome and two-pointer string problems frequently appear in interviews at companies like Google, Amazon, and Meta. Variants such as Valid Palindrome II and edit-limited palindrome checks are common because they test pattern recognition and efficient string traversal.

What data structure is used in Valid Palindrome IV?

The problem primarily uses the two pointers technique on a string. No additional data structures like hash maps or stacks are required, which allows the solution to run with constant extra space.

What is the time complexity of Valid Palindrome IV?

The optimal solution runs in O(n) time where n is the length of the string. Each iteration compares a pair of characters using two pointers moving from both ends. Space complexity remains O(1) because only counters and indices are stored.

Description Hints 2 Solution Video 6:47

2330. Valid Palindrome IV

Medium75.8% acceptance2 companiesPremium elsewhere, free here

You are given a 0-indexed string s consisting of only lowercase English letters. In one operation, you can change any character of s to any other character.

Return true if you can make s a palindrome after performing exactly one or two operations, or return false otherwise.

Example 1:

Input: s = "abcdba"
Output: true
Explanation: One way to make s a palindrome using 1 operation is:
- Change s[2] to 'd'. Now, s = "abddba".
One operation could be performed to make s a palindrome so return true.

Example 2:

Input: s = "aa"
Output: true
Explanation: One way to make s a palindrome using 2 operations is:
- Change s[0] to 'b'. Now, s = "ba".
- Change s[1] to 'b'. Now, s = "bb".
Two operations could be performed to make s a palindrome so return true.

Example 3:

Input: s = "abcdef"
Output: false
Explanation: It is not possible to make s a palindrome using one or two operations so return false.

Constraints:

1 <= s.length <= 10⁵
s consists only of lowercase English letters.

Asked at 2 companies

Amazon Meta

Topics

String Two Pointers

Hints open one at a time

Hint 1

Count the number of pairs of characters that should be equal if s were a palindrome, but are not equal.

Hint 2

If the number of pairs is 0, then s is already palindrome. You can pick any pair of characters and change them both to a different character and s will stay a palindrome.

6:47

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