Is Palindrome Permutation easy or hard?

Palindrome Permutation is classified as an Easy problem. The core idea relies on a simple observation about palindrome character frequencies, making it a common early interview question for practicing hash table and string manipulation concepts.

Palindrome Permutation Python/Java solution

In Python or Java, iterate through the string and maintain a frequency map using a dictionary or HashMap. After counting characters, verify that at most one character has an odd count. This approach is concise, readable, and runs in O(n) time.

How to solve Palindrome Permutation in O(n)?

Iterate through the string and count occurrences of each character using a hash map or array. After counting, check how many characters have odd frequencies. If the number of odd counts is greater than one, a palindrome permutation is impossible; otherwise it is valid.

What is the best approach for Palindrome Permutation?

The best approach uses a hash table to count character frequencies. A string can form a palindrome if at most one character has an odd frequency. This method runs in O(n) time and O(1) space since the character set is bounded.

Is Palindrome Permutation asked at Google/Amazon/Meta?

Palindrome Permutation is a classic string and hash table interview question. Variants of this problem frequently appear in interviews at companies like Amazon, Google, and Meta because it tests understanding of character frequency patterns and basic data structures.

What data structure is used in Palindrome Permutation?

The most common data structure is a hash table (or fixed-size array) to track character frequencies. Some optimized solutions use a bitmask to toggle bits representing characters, which relies on bit manipulation instead of storing explicit counts.

What is the time complexity of Palindrome Permutation?

The optimal solution runs in O(n) time where n is the length of the string. You scan the string once to count characters and then check the number of odd frequencies. Space complexity is O(1) because only a fixed-size frequency structure is required.