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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 {264Please respect copyright.ＰＥＮＡＮＡE5L4T5PXji
public int longestPalindrome(String[] words) {264Please respect copyright.ＰＥＮＡＮＡAj24na4rUZ
HashMap<String, Integer> count = new HashMap<String, Integer>();264Please respect copyright.ＰＥＮＡＮＡWkSsB9uxOX
// Count the number of occurrences of each word using a hashmap264Please respect copyright.ＰＥＮＡＮＡ7C5KMowWEL
for (String word : words) {264Please respect copyright.ＰＥＮＡＮＡs3mh5Yv6iY
Integer countOfTheWord = count.get(word);264Please respect copyright.ＰＥＮＡＮＡ1nMWDJ0KBF
if (countOfTheWord == null) {264Please respect copyright.ＰＥＮＡＮＡeJ2FzjDsA9
count.put(word, 1);264Please respect copyright.ＰＥＮＡＮＡPnhCXyvDEq
} else {264Please respect copyright.ＰＥＮＡＮＡ2RKOQYADt8
count.put(word, countOfTheWord + 1);264Please respect copyright.ＰＥＮＡＮＡGzUx5htdAV
}264Please respect copyright.ＰＥＮＡＮＡoP0AbPThZn
}264Please respect copyright.ＰＥＮＡＮＡMFTIviiS2i
// 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 word264Please respect copyright.ＰＥＮＡＮＡspHrhY8Ntz
int answer = 0;264Please respect copyright.ＰＥＮＡＮＡl9zJQqbQid
boolean central = false;264Please respect copyright.ＰＥＮＡＮＡecRWcwmBGX
for (Map.Entry<String, Integer> entry : count.entrySet()) {264Please respect copyright.ＰＥＮＡＮＡcZzLBwtNlO
String word = entry.getKey();264Please respect copyright.ＰＥＮＡＮＡD8eK54Zk0a
int countOfTheWord = entry.getValue();264Please respect copyright.ＰＥＮＡＮＡB2cOx6knZd
// if the word is a palindrome264Please respect copyright.ＰＥＮＡＮＡcip1zMjSi9
if (word.charAt(0) == word.charAt(1)) {264Please respect copyright.ＰＥＮＡＮＡUetg7okgBl
if (countOfTheWord % 2 == 0) {264Please respect copyright.ＰＥＮＡＮＡP9fUcT97Hp
answer += countOfTheWord;264Please respect copyright.ＰＥＮＡＮＡ2jjSyagG0a
} else {264Please respect copyright.ＰＥＮＡＮＡtKl5JkAaUL
answer += countOfTheWord - 1;264Please respect copyright.ＰＥＮＡＮＡyf7WoEu7O5
central = true;264Please respect copyright.ＰＥＮＡＮＡvoiyCI1XVw
}264Please respect copyright.ＰＥＮＡＮＡMNxWh4Curl
// consider a pair of non-palindrome words such that one is the reverse of another264Please respect copyright.ＰＥＮＡＮＡ5ABNUgETTi
} else if (word.charAt(0) < word.charAt(1)) {264Please respect copyright.ＰＥＮＡＮＡaZ4SgtFzn5
String reversedWord = "" + word.charAt(1) + word.charAt(0);264Please respect copyright.ＰＥＮＡＮＡiFI8cOy3YL
if (count.containsKey(reversedWord)) {264Please respect copyright.ＰＥＮＡＮＡC4SGI8Vf41
answer += 2 * Math.min(countOfTheWord, count.get(reversedWord));264Please respect copyright.ＰＥＮＡＮＡeqRljAKt8v
}264Please respect copyright.ＰＥＮＡＮＡKFzCXyuZhD
}264Please respect copyright.ＰＥＮＡＮＡqaydAGZ0Cf
}264Please respect copyright.ＰＥＮＡＮＡCstwbhn5Nl
if (central) {264Please respect copyright.ＰＥＮＡＮＡfFWcXcBZRn
answer++;264Please respect copyright.ＰＥＮＡＮＡT6vz4YvtEt
}264Please respect copyright.ＰＥＮＡＮＡhDg6bC8eXB
return 2 * answer;264Please respect copyright.ＰＥＮＡＮＡZMcIhc3PEy
}264Please respect copyright.ＰＥＮＡＮＡmWrPdiyInj
}
2: A Two-Dimensional Array Approach
class Solution {264Please respect copyright.ＰＥＮＡＮＡO7n7y1Csyr
public int longestPalindrome(String[] words) {264Please respect copyright.ＰＥＮＡＮＡMdMrcGXGru
final int alphabetSize = 26;264Please respect copyright.ＰＥＮＡＮＡ23yx0iKEOn
int[][] count = new int[alphabetSize][alphabetSize];264Please respect copyright.ＰＥＮＡＮＡ1eU8R7pszc
// Count the number of occurrences of each word using a two-dimensional array. 264Please respect copyright.ＰＥＮＡＮＡJktlOLh0Oa
for (String word : words) {264Please respect copyright.ＰＥＮＡＮＡxGVwqDU4XK
count[word.charAt(0) - 'a'][word.charAt(1) - 'a']++;264Please respect copyright.ＰＥＮＡＮＡF3eRRgijVB
}264Please respect copyright.ＰＥＮＡＮＡgYjRjDp8xF
// The answer will denote the number of words in the final string and the boolean variable central will denote whether we have a central word264Please respect copyright.ＰＥＮＡＮＡKM5YtTQ793
int answer = 0;264Please respect copyright.ＰＥＮＡＮＡIyQjkGj7sR
boolean central = false;264Please respect copyright.ＰＥＮＡＮＡkW0oQLnfQ1
for (int i = 0; i < alphabetSize; i++) {264Please respect copyright.ＰＥＮＡＮＡkFiLdy06Mk
if (count[i][i] % 2 == 0) {264Please respect copyright.ＰＥＮＡＮＡimeLXU8ixY
answer += count[i][i];264Please respect copyright.ＰＥＮＡＮＡewQEOekP3F
} else {264Please respect copyright.ＰＥＮＡＮＡczkn8soAlS
answer += count[i][i] - 1;264Please respect copyright.ＰＥＮＡＮＡkt40xxpCtu
central = true;264Please respect copyright.ＰＥＮＡＮＡmBiSwFWIwO
}264Please respect copyright.ＰＥＮＡＮＡE79Pfuc6nD
for (int j = i + 1; j < alphabetSize; j++) {264Please respect copyright.ＰＥＮＡＮＡOvpRdQaDdI
answer += 2 * Math.min(count[i][j], count[j][i]);264Please respect copyright.ＰＥＮＡＮＡ0ZepPTQ65H
}264Please respect copyright.ＰＥＮＡＮＡxwQebQHA0d
}264Please respect copyright.ＰＥＮＡＮＡlVAK2CDF2O
if (central) {264Please respect copyright.ＰＥＮＡＮＡhDPJS8nJqV
answer++;264Please respect copyright.ＰＥＮＡＮＡPdzNxHDBHn
}264Please respect copyright.ＰＥＮＡＮＡGNuwGbIVHP
return 2 * answer;264Please respect copyright.ＰＥＮＡＮＡ1xRF3V5b5v
}264Please respect copyright.ＰＥＮＡＮＡgvt1flIQfx
}
264Please respect copyright.ＰＥＮＡＮＡOUNFc0uiU0
