
Q: You are given an array of strings words. Each element of words consists of two lowercase English letters.
Create the longest possible palindrome by selecting some elements from words and concatenating them in any order. Each element can be selected at most once.
Return the length of the longest palindrome that you can create. If it is impossible to create any palindrome, return 0.
A palindrome is a string that reads the same forward and backward.
A: 1. A Hash Map Approach
class Solution {235Please respect copyright.PENANAU9dDmzLkM3
public int longestPalindrome(String[] words) {235Please respect copyright.PENANASHCfPUSMmF
HashMap<String, Integer> count = new HashMap<String, Integer>();235Please respect copyright.PENANANJBFhMAnU0
// Count the number of occurrences of each word using a hashmap235Please respect copyright.PENANAMn5gk6WIX5
for (String word : words) {235Please respect copyright.PENANApAuHzlZtzj
Integer countOfTheWord = count.get(word);235Please respect copyright.PENANAlNRWEVCpB5
if (countOfTheWord == null) {235Please respect copyright.PENANAIg0sRwxqpS
count.put(word, 1);235Please respect copyright.PENANADx8OvKcHZm
} else {235Please respect copyright.PENANAehIhV4NnrD
count.put(word, countOfTheWord + 1);235Please respect copyright.PENANAlbqcHJbUZ7
}235Please respect copyright.PENANAADCCad5ojP
}235Please respect copyright.PENANAVSxgjqUTm4
// Initialize answer=0, central = false. The answer will denote the number of words in the final string and the boolean variable central will denote whether we have a central word235Please respect copyright.PENANAHCKbrPrY8I
int answer = 0;235Please respect copyright.PENANA4KFTtkjZ2a
boolean central = false;235Please respect copyright.PENANAtNhZZmwNgR
for (Map.Entry<String, Integer> entry : count.entrySet()) {235Please respect copyright.PENANAIYpEIFF8LE
String word = entry.getKey();235Please respect copyright.PENANAv9rL9zrNih
int countOfTheWord = entry.getValue();235Please respect copyright.PENANAlKKUsbKq4u
// if the word is a palindrome235Please respect copyright.PENANAQmS7CrB18t
if (word.charAt(0) == word.charAt(1)) {235Please respect copyright.PENANAdErDuTnagl
if (countOfTheWord % 2 == 0) {235Please respect copyright.PENANAOAihWLQ0PG
answer += countOfTheWord;235Please respect copyright.PENANAOTPe6MKILw
} else {235Please respect copyright.PENANAYMEImtTiH6
answer += countOfTheWord - 1;235Please respect copyright.PENANAACrSsGXaZw
central = true;235Please respect copyright.PENANAQzwrXEx76g
}235Please respect copyright.PENANAK6gENbNJZm
// consider a pair of non-palindrome words such that one is the reverse of another235Please respect copyright.PENANAf6A9kiOneX
} else if (word.charAt(0) < word.charAt(1)) {235Please respect copyright.PENANAkHEgWQLKLD
String reversedWord = "" + word.charAt(1) + word.charAt(0);235Please respect copyright.PENANA9gSbL9u1La
if (count.containsKey(reversedWord)) {235Please respect copyright.PENANA1FxAn4ZrLB
answer += 2 * Math.min(countOfTheWord, count.get(reversedWord));235Please respect copyright.PENANAPY15hPATQu
}235Please respect copyright.PENANAeSatwTW6Yb
}235Please respect copyright.PENANAuKedqW7mNt
}235Please respect copyright.PENANAZeUKWcrT5Q
if (central) {235Please respect copyright.PENANAcydC3GLB2G
answer++;235Please respect copyright.PENANA23bOG9G1n7
}235Please respect copyright.PENANALLIKO4ejbR
return 2 * answer;235Please respect copyright.PENANA29faJJfqZb
}235Please respect copyright.PENANADyfapGZr0T
}
2: A Two-Dimensional Array Approach
class Solution {235Please respect copyright.PENANAkexJZ3olKn
public int longestPalindrome(String[] words) {235Please respect copyright.PENANAlXyT5UGV4D
final int alphabetSize = 26;235Please respect copyright.PENANATtuiqSoqYL
int[][] count = new int[alphabetSize][alphabetSize];235Please respect copyright.PENANAVc8tglzqKD
// Count the number of occurrences of each word using a two-dimensional array. 235Please respect copyright.PENANAwr65JG6yJQ
for (String word : words) {235Please respect copyright.PENANAOnDaV9ffPt
count[word.charAt(0) - 'a'][word.charAt(1) - 'a']++;235Please respect copyright.PENANAPB5zSVWmc3
}235Please respect copyright.PENANAIuLkyfiOPj
// The answer will denote the number of words in the final string and the boolean variable central will denote whether we have a central word235Please respect copyright.PENANA86MfWSOEaG
int answer = 0;235Please respect copyright.PENANANBhNIqC8Z3
boolean central = false;235Please respect copyright.PENANADiYgtq8ICA
for (int i = 0; i < alphabetSize; i++) {235Please respect copyright.PENANAP7VIwEAYVO
if (count[i][i] % 2 == 0) {235Please respect copyright.PENANAjkvM2NL4Z2
answer += count[i][i];235Please respect copyright.PENANAh2fu8aiGAB
} else {235Please respect copyright.PENANAMUxjAHH4hK
answer += count[i][i] - 1;235Please respect copyright.PENANA49EHzJgpVT
central = true;235Please respect copyright.PENANAreTklguPBk
}235Please respect copyright.PENANArdGxu52J2y
for (int j = i + 1; j < alphabetSize; j++) {235Please respect copyright.PENANA3rw87tPd3p
answer += 2 * Math.min(count[i][j], count[j][i]);235Please respect copyright.PENANAhPieEWJc5f
}235Please respect copyright.PENANAJeh7oVLSfm
}235Please respect copyright.PENANA6gh8lrfVNc
if (central) {235Please respect copyright.PENANAQuzztjFewH
answer++;235Please respect copyright.PENANA8MMNl9YHCF
}235Please respect copyright.PENANAv8fQSJ2WK5
return 2 * answer;235Please respect copyright.PENANAJX5XzhoYAt
}235Please respect copyright.PENANA2t7nEvsbJC
}
235Please respect copyright.PENANAJ7hthRAH7e