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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 {257Please respect copyright.ＰＥＮＡＮＡd2vbMWqaCX
public int longestPalindrome(String[] words) {257Please respect copyright.ＰＥＮＡＮＡSfsbwMA0Nj
HashMap<String, Integer> count = new HashMap<String, Integer>();257Please respect copyright.ＰＥＮＡＮＡjZTs6UWhgx
// Count the number of occurrences of each word using a hashmap257Please respect copyright.ＰＥＮＡＮＡO41vubMsDO
for (String word : words) {257Please respect copyright.ＰＥＮＡＮＡPxc3p4hdnb
Integer countOfTheWord = count.get(word);257Please respect copyright.ＰＥＮＡＮＡLnAqepRakQ
if (countOfTheWord == null) {257Please respect copyright.ＰＥＮＡＮＡYa62NLnJbL
count.put(word, 1);257Please respect copyright.ＰＥＮＡＮＡyrIZOIo4IU
} else {257Please respect copyright.ＰＥＮＡＮＡVghIm4iKXg
count.put(word, countOfTheWord + 1);257Please respect copyright.ＰＥＮＡＮＡLzxuzY86ve
}257Please respect copyright.ＰＥＮＡＮＡ108ZzMIH9b
}257Please respect copyright.ＰＥＮＡＮＡiKxn9CA2hE
// 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 word257Please respect copyright.ＰＥＮＡＮＡSOwRkPN4ya
int answer = 0;257Please respect copyright.ＰＥＮＡＮＡD24pV8dVRu
boolean central = false;257Please respect copyright.ＰＥＮＡＮＡ62bQAKf0ym
for (Map.Entry<String, Integer> entry : count.entrySet()) {257Please respect copyright.ＰＥＮＡＮＡCJ7AYH6Bpg
String word = entry.getKey();257Please respect copyright.ＰＥＮＡＮＡ6NdZqzAg0V
int countOfTheWord = entry.getValue();257Please respect copyright.ＰＥＮＡＮＡ0YAKepYcaH
// if the word is a palindrome257Please respect copyright.ＰＥＮＡＮＡrDewqDeAxE
if (word.charAt(0) == word.charAt(1)) {257Please respect copyright.ＰＥＮＡＮＡBtEeZAa3yE
if (countOfTheWord % 2 == 0) {257Please respect copyright.ＰＥＮＡＮＡVgtdLCxTMh
answer += countOfTheWord;257Please respect copyright.ＰＥＮＡＮＡH7aWTPcltm
} else {257Please respect copyright.ＰＥＮＡＮＡ6R47kSJimD
answer += countOfTheWord - 1;257Please respect copyright.ＰＥＮＡＮＡGEFGNYB00C
central = true;257Please respect copyright.ＰＥＮＡＮＡ5IfgEY3WiR
}257Please respect copyright.ＰＥＮＡＮＡJ01XT1q33Z
// consider a pair of non-palindrome words such that one is the reverse of another257Please respect copyright.ＰＥＮＡＮＡXVjFDXUBgh
} else if (word.charAt(0) < word.charAt(1)) {257Please respect copyright.ＰＥＮＡＮＡUTj0tt2vos
String reversedWord = "" + word.charAt(1) + word.charAt(0);257Please respect copyright.ＰＥＮＡＮＡPirLun4SZi
if (count.containsKey(reversedWord)) {257Please respect copyright.ＰＥＮＡＮＡqDqpQt7S82
answer += 2 * Math.min(countOfTheWord, count.get(reversedWord));257Please respect copyright.ＰＥＮＡＮＡ3Vq9XmTzHy
}257Please respect copyright.ＰＥＮＡＮＡAYE5PCCQde
}257Please respect copyright.ＰＥＮＡＮＡLwNnoFJEfJ
}257Please respect copyright.ＰＥＮＡＮＡawrLD6hlDT
if (central) {257Please respect copyright.ＰＥＮＡＮＡtKn3lJxZAY
answer++;257Please respect copyright.ＰＥＮＡＮＡ4oGZ2kVF74
}257Please respect copyright.ＰＥＮＡＮＡuJs8tML4D1
return 2 * answer;257Please respect copyright.ＰＥＮＡＮＡIg5UfCOR6L
}257Please respect copyright.ＰＥＮＡＮＡXcp6UwbEQG
}
2: A Two-Dimensional Array Approach
class Solution {257Please respect copyright.ＰＥＮＡＮＡ60Dk5fhim7
public int longestPalindrome(String[] words) {257Please respect copyright.ＰＥＮＡＮＡJCrreKH0mJ
final int alphabetSize = 26;257Please respect copyright.ＰＥＮＡＮＡAspLw0BYlp
int[][] count = new int[alphabetSize][alphabetSize];257Please respect copyright.ＰＥＮＡＮＡ12j8QMq1D7
// Count the number of occurrences of each word using a two-dimensional array. 257Please respect copyright.ＰＥＮＡＮＡkMS3drmc1n
for (String word : words) {257Please respect copyright.ＰＥＮＡＮＡBAUm6M6Siq
count[word.charAt(0) - 'a'][word.charAt(1) - 'a']++;257Please respect copyright.ＰＥＮＡＮＡOavhnRlUip
}257Please respect copyright.ＰＥＮＡＮＡvBgHQba3mx
// The answer will denote the number of words in the final string and the boolean variable central will denote whether we have a central word257Please respect copyright.ＰＥＮＡＮＡyCSPYA8OKS
int answer = 0;257Please respect copyright.ＰＥＮＡＮＡ7zDIXgJUOK
boolean central = false;257Please respect copyright.ＰＥＮＡＮＡlbttTcMRNL
for (int i = 0; i < alphabetSize; i++) {257Please respect copyright.ＰＥＮＡＮＡPYYTXTsOuJ
if (count[i][i] % 2 == 0) {257Please respect copyright.ＰＥＮＡＮＡIJVxq1NMga
answer += count[i][i];257Please respect copyright.ＰＥＮＡＮＡdJijYTreK6
} else {257Please respect copyright.ＰＥＮＡＮＡ55KZnoP7Ec
answer += count[i][i] - 1;257Please respect copyright.ＰＥＮＡＮＡ7ByE1p0N9G
central = true;257Please respect copyright.ＰＥＮＡＮＡ72p2AB35oZ
}257Please respect copyright.ＰＥＮＡＮＡ60PyK9YKRP
for (int j = i + 1; j < alphabetSize; j++) {257Please respect copyright.ＰＥＮＡＮＡI7QTPiT1Bg
answer += 2 * Math.min(count[i][j], count[j][i]);257Please respect copyright.ＰＥＮＡＮＡ5Xp2gX3QvN
}257Please respect copyright.ＰＥＮＡＮＡ4j8xKfCuzv
}257Please respect copyright.ＰＥＮＡＮＡeB8snkTUDO
if (central) {257Please respect copyright.ＰＥＮＡＮＡrpeM73nae0
answer++;257Please respect copyright.ＰＥＮＡＮＡ02GTKs0fzE
}257Please respect copyright.ＰＥＮＡＮＡ4urbv62feQ
return 2 * answer;257Please respect copyright.ＰＥＮＡＮＡJmuHHWlzrS
}257Please respect copyright.ＰＥＮＡＮＡnyRPtDGsVC
}
257Please respect copyright.ＰＥＮＡＮＡInBVfZV8gG
