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Daily JavaScript Challenge #JS-73: Validate Palindrome Permutation

Daily JavaScript Challenge: Validate Palindrome Permutation

Hey fellow developers! 👋 Welcome to today's JavaScript coding challenge. Let's keep those programming skills sharp!

The Challenge

Difficulty: Medium

Topic: String Manipulation

Description

Given a string, determine if any permutation of the string is a palindrome. A palindrome is a word that reads the same backward as forward.

Ready to Begin?

https://www.dpcdev.com/

  1. Fork this challenge
  2. Write your solution
  3. Test it against the provided test cases
  4. Share your approach in the comments below!

Want to Learn More?

Check out the documentation about this topic here: https://en.wikipedia.org/wiki/Palindrome

Join the Discussion!

  • How did you approach this problem?
  • Did you find any interesting edge cases?
  • What was your biggest learning from this challenge?

Let's learn together! Drop your thoughts and questions in the comments below. 👇


This is part of our Daily JavaScript Challenge series. Follow me for daily programming challenges and let's grow together! 🚀

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Top comments (3)

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martinhaeusler profile image
Martin Häusler •
  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.
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rockksdb profile image
RocksDB •

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!

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