The solution to challenge #11 of AdventJS 2023
The solution to the previous challenge
The solution to the next challenge
Challenge Description
In Santa's workshop, the elves love puzzles 🧠. This year, they have created a special one: a challenge to form a Christmas palindrome.
A palindrome is a word that reads the same forwards and backward. The elves want to know if it is possible to form a palindrome by making, at most, one exchange of letters.
Create a function getIndexsForPalindrome that receives a string and returns:
If it is already a palindrome, an empty array.
If it is not possible, null.
If a palindrome can be formed with one change, an array with the two positions (indexes) must be swapped to create it.
For example:
getIndexsForPalindrome('anna') // []
getIndexsForPalindrome('abab') // [0, 1]
getIndexsForPalindrome('abac') // null
getIndexsForPalindrome('aaaaaaaa') // []
getIndexsForPalindrome('aaababa') // [1, 3]
getIndexsForPalindrome('caababa') // null
If the palindrome can be formed with different swaps, always return the first one found.
Analysis
In this challenge, our goal is to find a palindrome by swapping two elements of the string and return the indices of those two elements. If a palindrome cannot be found, return an empty array.
Input
- Word (
word): A string with the word to evaluate.
Output
An array of two elements with the indices to change to find the palindrome.
If it's already a palindrome, return an empty array.
If none of the above conditions are met, return
null.
Considerations
- If multiple palindromes can be formed, always return the first one found.
Solution
It can be solved by iterating over each element of the string, and within that iteration, iterating again and swapping elements to compare.
Code
/**
* Returns the indices of characters that need to be reversed in order to make a word a palindrome.
*
* @param {string} word - The word to check for palindromic properties.
* @return {number[] | null} - An array of two indices representing the characters to be reversed, or null if no such indices exist.
*/
function getIndexsForPalindrome(word) {
// Check if the given word is already a palindrome
const isPalindrome = (s) => s === s.split("").reverse().join("");
// If the word is already a palindrome, return an empty array
if (isPalindrome(word)) {
return [];
}
// Iterate through all possible pairs of indices
for (let i = 0; i < word.length; i += 1) {
for (let j = i + 1; j < word.length; j += 1) {
// Create a new word by swapping the characters at indices i and j
const newWord =
word.substring(0, i) +
word[j] +
word.substring(i + 1, j) +
word[i] +
word.substring(j + 1);
// If the new word is a palindrome, return the indices i and j
if (isPalindrome(newWord)) {
return [i, j];
}
}
}
// If no such indices exist, return null
return null;
}
Community Solutions
Solution by iswilljr:
function getIndexsForPalindrome(word: string) {
const _letters = [...word]
const palindrome = word === [..._letters].reverse().join('')
let initial: number[] | null = [null, []][+palindrome]
let index = 0
let aux = 1
const letters = [_letters, []][+palindrome]
let auxLetters = letters.slice(1)
for (const letter of letters) {
for (const auxLetter of auxLetters) {
const w = [...letters]
w[index] = auxLetter
w[aux] = letter
const isPalindrome = +(w.join('') === w.reverse().join(''))
const isInitialNull = +(initial == null)
const isDifferentIndex = +(index !== aux)
const values = [initial, initial, initial, [index, aux]]
initial = values[isInitialNull + isDifferentIndex + isPalindrome]
aux++
}
index++
aux = 1
auxLetters = [[], auxLetters][+(initial == null)]
}
return initial
}
Solution by Achalogy:
function getIndexsForPalindrome(word) {
let res: any = null
for (const a of Array.from({ length: word.length }).keys()) {
for (const b of Array.from({ length: word.length }).keys()) {
let swapped = [...word]
let aux = word[a]
swapped[a] = word[b]
swapped[b] = aux
let left = swapped.slice(0, Math.floor(word.length / 2)).join("")
let right = swapped.slice(Math.ceil(word.length / 2)).reverse().join("")
res = [
[
null
, [
[]
, [a, b]
][+((a + b) > 0)]
][+(left == right)]
, res
][+!!res]
}
}
return res
}
And that was the challenge for December 11th and its solutions. Give it a like if you enjoyed the challenge or the solution! Do you have an alternative solution? Leave it in the comments!
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