
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 {237Please respect copyright.PENANAYywTPkjVoF
public int longestPalindrome(String[] words) {237Please respect copyright.PENANAhKwBHChNy7
HashMap<String, Integer> count = new HashMap<String, Integer>();237Please respect copyright.PENANANp8EVUF9Yy
// Count the number of occurrences of each word using a hashmap237Please respect copyright.PENANA389vzTcG34
for (String word : words) {237Please respect copyright.PENANAZBjMMBWQ30
Integer countOfTheWord = count.get(word);237Please respect copyright.PENANAgSrIQwUdqW
if (countOfTheWord == null) {237Please respect copyright.PENANAWqnIoiFJHL
count.put(word, 1);237Please respect copyright.PENANADdi0Y945Sw
} else {237Please respect copyright.PENANAR3gsFWf5rX
count.put(word, countOfTheWord + 1);237Please respect copyright.PENANAsKXZvN9rZP
}237Please respect copyright.PENANAEW64SMlcDx
}237Please respect copyright.PENANA3FCTz4ScuS
// 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 word237Please respect copyright.PENANAhxCd9BSqum
int answer = 0;237Please respect copyright.PENANAiGiAVEEUwy
boolean central = false;237Please respect copyright.PENANACSOhaVaBbW
for (Map.Entry<String, Integer> entry : count.entrySet()) {237Please respect copyright.PENANAZIj8rkXiMT
String word = entry.getKey();237Please respect copyright.PENANAOkUxz4ZAcQ
int countOfTheWord = entry.getValue();237Please respect copyright.PENANAqj9fr9HIQu
// if the word is a palindrome237Please respect copyright.PENANAUkJhStS86E
if (word.charAt(0) == word.charAt(1)) {237Please respect copyright.PENANAb6xTmd1Aoy
if (countOfTheWord % 2 == 0) {237Please respect copyright.PENANAGNeEmWiJSE
answer += countOfTheWord;237Please respect copyright.PENANAvWnFMIojrV
} else {237Please respect copyright.PENANATMEILYU1O9
answer += countOfTheWord - 1;237Please respect copyright.PENANA6WHRe8MZFA
central = true;237Please respect copyright.PENANAbqYmjbkbyX
}237Please respect copyright.PENANAcl8Xo2iGTf
// consider a pair of non-palindrome words such that one is the reverse of another237Please respect copyright.PENANAYlPWH4vmWj
} else if (word.charAt(0) < word.charAt(1)) {237Please respect copyright.PENANAXSZShF0JEW
String reversedWord = "" + word.charAt(1) + word.charAt(0);237Please respect copyright.PENANAwnOwGvsLb1
if (count.containsKey(reversedWord)) {237Please respect copyright.PENANADZkFLWzFoY
answer += 2 * Math.min(countOfTheWord, count.get(reversedWord));237Please respect copyright.PENANAcI0Jsz6G8N
}237Please respect copyright.PENANAgOL5TNgK8D
}237Please respect copyright.PENANARXsVof40Kq
}237Please respect copyright.PENANAaiGNY8CZq0
if (central) {237Please respect copyright.PENANALurZxMdhup
answer++;237Please respect copyright.PENANAyzh0ne9xa5
}237Please respect copyright.PENANAmNMcEM8YcO
return 2 * answer;237Please respect copyright.PENANAclckYZRMln
}237Please respect copyright.PENANAUh3AKGH4Xe
}
2: A Two-Dimensional Array Approach
class Solution {237Please respect copyright.PENANAbvqwXqmlmK
public int longestPalindrome(String[] words) {237Please respect copyright.PENANAu500wyaQvp
final int alphabetSize = 26;237Please respect copyright.PENANAG1QD9AJY62
int[][] count = new int[alphabetSize][alphabetSize];237Please respect copyright.PENANAXOnWN294JS
// Count the number of occurrences of each word using a two-dimensional array. 237Please respect copyright.PENANAT5LPB0vQqw
for (String word : words) {237Please respect copyright.PENANABJcrZAeb8f
count[word.charAt(0) - 'a'][word.charAt(1) - 'a']++;237Please respect copyright.PENANApP9SKoGgqt
}237Please respect copyright.PENANA6Rnpppp3Xl
// The answer will denote the number of words in the final string and the boolean variable central will denote whether we have a central word237Please respect copyright.PENANANg0KzRcWuW
int answer = 0;237Please respect copyright.PENANA57VegD5qN0
boolean central = false;237Please respect copyright.PENANAyXH3IZunhv
for (int i = 0; i < alphabetSize; i++) {237Please respect copyright.PENANAiOESqhU4i1
if (count[i][i] % 2 == 0) {237Please respect copyright.PENANAMjPn4iiSWB
answer += count[i][i];237Please respect copyright.PENANAKKBNOi9Qj7
} else {237Please respect copyright.PENANAWMYCYrLCqW
answer += count[i][i] - 1;237Please respect copyright.PENANAMjm0nlnS6b
central = true;237Please respect copyright.PENANArNGmViEbj0
}237Please respect copyright.PENANAAybZM1sD48
for (int j = i + 1; j < alphabetSize; j++) {237Please respect copyright.PENANA444T21CdIi
answer += 2 * Math.min(count[i][j], count[j][i]);237Please respect copyright.PENANA2nPH9CqhUc
}237Please respect copyright.PENANAVzzFERLSAR
}237Please respect copyright.PENANAeObrA9i39j
if (central) {237Please respect copyright.PENANAZhheSpgYn8
answer++;237Please respect copyright.PENANAHYWSYL80M9
}237Please respect copyright.PENANAXCnWGpmtVP
return 2 * answer;237Please respect copyright.PENANA0dXNeutO3D
}237Please respect copyright.PENANAvFI3XYpwBS
}
237Please respect copyright.PENANA6ABVCZxSth