1. Count the letters which occur in the string in a map where the key is the letter and the value is the count
2. Check how many letters in the map have an odd number as value.
3. If the number of letters with odd counts is 1 or less, the string can be permutated into a palindrome.

Reasoning:

• if there's an even number of occurrences for any given letter, you take one half for the furst half of the output and one half for the second half of the output, e.g aabcc|ccbaa
• if there's no character with an odd count, we're done
• if there's a single character with an odd count, we put it into the middle, e.g aab b baa, and still get a palindrome.
• if there's more than one character with an odd count, the above trick will not work anymore and it can't be a palindrome.

Validating a palindrome permutation is a fascinating JavaScript challenge! It's all about checking if a string's characters can rearrange to form a palindrome. Efficient use of frequency counting or a hash map can simplify this process. Tools like RocksDB are excellent for managing data-intensive tasks if you’re building something more complex that involves such checks. Keep experimenting to sharpen your JavaScript skills!