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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 {204Please respect copyright.ＰＥＮＡＮＡNRN6Mk3Fnu
public int longestPalindrome(String[] words) {204Please respect copyright.ＰＥＮＡＮＡ6KdbOw4RC6
HashMap<String, Integer> count = new HashMap<String, Integer>();204Please respect copyright.ＰＥＮＡＮＡsWqLbNgMNt
// Count the number of occurrences of each word using a hashmap204Please respect copyright.ＰＥＮＡＮＡoXvYoLkyuz
for (String word : words) {204Please respect copyright.ＰＥＮＡＮＡEovXkVaev8
Integer countOfTheWord = count.get(word);204Please respect copyright.ＰＥＮＡＮＡh5PSdAnDUT
if (countOfTheWord == null) {204Please respect copyright.ＰＥＮＡＮＡpe4Jx9v2ey
count.put(word, 1);204Please respect copyright.ＰＥＮＡＮＡGIPoqPYnmp
} else {204Please respect copyright.ＰＥＮＡＮＡvVUz4xHBfA
count.put(word, countOfTheWord + 1);204Please respect copyright.ＰＥＮＡＮＡfCkB69Lvth
}204Please respect copyright.ＰＥＮＡＮＡGl0QoK4AH8
}204Please respect copyright.ＰＥＮＡＮＡA6vvmf11Sh
// 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 word204Please respect copyright.ＰＥＮＡＮＡYmlc3Q76Fj
int answer = 0;204Please respect copyright.ＰＥＮＡＮＡNMY9aA1SP2
boolean central = false;204Please respect copyright.ＰＥＮＡＮＡosLjnlolVM
for (Map.Entry<String, Integer> entry : count.entrySet()) {204Please respect copyright.ＰＥＮＡＮＡOyyZjWFUtZ
String word = entry.getKey();204Please respect copyright.ＰＥＮＡＮＡtkI2VgMghI
int countOfTheWord = entry.getValue();204Please respect copyright.ＰＥＮＡＮＡt1FPaMsnbN
// if the word is a palindrome204Please respect copyright.ＰＥＮＡＮＡsepMnwKq6l
if (word.charAt(0) == word.charAt(1)) {204Please respect copyright.ＰＥＮＡＮＡxzxYSvx1bE
if (countOfTheWord % 2 == 0) {204Please respect copyright.ＰＥＮＡＮＡWoJ97KDCU3
answer += countOfTheWord;204Please respect copyright.ＰＥＮＡＮＡ7dnoYRwpPz
} else {204Please respect copyright.ＰＥＮＡＮＡVYTmossXb7
answer += countOfTheWord - 1;204Please respect copyright.ＰＥＮＡＮＡ8Dxcl29LWI
central = true;204Please respect copyright.ＰＥＮＡＮＡK7sZFICspU
}204Please respect copyright.ＰＥＮＡＮＡZkviM6FRfp
// consider a pair of non-palindrome words such that one is the reverse of another204Please respect copyright.ＰＥＮＡＮＡmDbCuQe405
} else if (word.charAt(0) < word.charAt(1)) {204Please respect copyright.ＰＥＮＡＮＡ9EjkxCIozI
String reversedWord = "" + word.charAt(1) + word.charAt(0);204Please respect copyright.ＰＥＮＡＮＡ2fczfvK1DU
if (count.containsKey(reversedWord)) {204Please respect copyright.ＰＥＮＡＮＡM8gd4Dqztj
answer += 2 * Math.min(countOfTheWord, count.get(reversedWord));204Please respect copyright.ＰＥＮＡＮＡ9XHrul8cif
}204Please respect copyright.ＰＥＮＡＮＡ8SWryYRbHG
}204Please respect copyright.ＰＥＮＡＮＡp7bc0IdtgT
}204Please respect copyright.ＰＥＮＡＮＡSKWBFW4eX5
if (central) {204Please respect copyright.ＰＥＮＡＮＡxc1c9TsALQ
answer++;204Please respect copyright.ＰＥＮＡＮＡD9kTezlMGs
}204Please respect copyright.ＰＥＮＡＮＡOef7Hyrxot
return 2 * answer;204Please respect copyright.ＰＥＮＡＮＡEoViuxtxe3
}204Please respect copyright.ＰＥＮＡＮＡvBWKd62vm2
}
2: A Two-Dimensional Array Approach
class Solution {204Please respect copyright.ＰＥＮＡＮＡmX4A22hGhN
public int longestPalindrome(String[] words) {204Please respect copyright.ＰＥＮＡＮＡfuKLig7iXp
final int alphabetSize = 26;204Please respect copyright.ＰＥＮＡＮＡYyS72AtSdO
int[][] count = new int[alphabetSize][alphabetSize];204Please respect copyright.ＰＥＮＡＮＡFDS3M8aDQH
// Count the number of occurrences of each word using a two-dimensional array. 204Please respect copyright.ＰＥＮＡＮＡz8BULeAJL6
for (String word : words) {204Please respect copyright.ＰＥＮＡＮＡ35n0lkfggg
count[word.charAt(0) - 'a'][word.charAt(1) - 'a']++;204Please respect copyright.ＰＥＮＡＮＡz93RCzgcEj
}204Please respect copyright.ＰＥＮＡＮＡxsPZ5RoBsC
// The answer will denote the number of words in the final string and the boolean variable central will denote whether we have a central word204Please respect copyright.ＰＥＮＡＮＡQMQGzW9LBL
int answer = 0;204Please respect copyright.ＰＥＮＡＮＡ0M852nUXkS
boolean central = false;204Please respect copyright.ＰＥＮＡＮＡS3FWqNe2i1
for (int i = 0; i < alphabetSize; i++) {204Please respect copyright.ＰＥＮＡＮＡSRxI4kX7vP
if (count[i][i] % 2 == 0) {204Please respect copyright.ＰＥＮＡＮＡuX2fxwxPtl
answer += count[i][i];204Please respect copyright.ＰＥＮＡＮＡr5u6Ph1hYj
} else {204Please respect copyright.ＰＥＮＡＮＡGD8W05c2iT
answer += count[i][i] - 1;204Please respect copyright.ＰＥＮＡＮＡeujT4QYlzH
central = true;204Please respect copyright.ＰＥＮＡＮＡShEHeC7b7R
}204Please respect copyright.ＰＥＮＡＮＡ9lHGLsGnDg
for (int j = i + 1; j < alphabetSize; j++) {204Please respect copyright.ＰＥＮＡＮＡlDZ6WevaOs
answer += 2 * Math.min(count[i][j], count[j][i]);204Please respect copyright.ＰＥＮＡＮＡqD2HpnLVDC
}204Please respect copyright.ＰＥＮＡＮＡcefyffoVpR
}204Please respect copyright.ＰＥＮＡＮＡxVvinFeYyI
if (central) {204Please respect copyright.ＰＥＮＡＮＡIMC2bBfNsB
answer++;204Please respect copyright.ＰＥＮＡＮＡcpeU58P7LC
}204Please respect copyright.ＰＥＮＡＮＡjauwMAwiBv
return 2 * answer;204Please respect copyright.ＰＥＮＡＮＡFMV00tlKcw
}204Please respect copyright.ＰＥＮＡＮＡ6wbwecGt2x
}
204Please respect copyright.ＰＥＮＡＮＡGzklDGhGTU
