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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 {270Please respect copyright.ＰＥＮＡＮＡaEOTiu48DG
public int longestPalindrome(String[] words) {270Please respect copyright.ＰＥＮＡＮＡMbHkwjX40n
HashMap<String, Integer> count = new HashMap<String, Integer>();270Please respect copyright.ＰＥＮＡＮＡM5TDr0pAiL
// Count the number of occurrences of each word using a hashmap270Please respect copyright.ＰＥＮＡＮＡQxkfC4JUSV
for (String word : words) {270Please respect copyright.ＰＥＮＡＮＡXET4q8t9Wc
Integer countOfTheWord = count.get(word);270Please respect copyright.ＰＥＮＡＮＡY1lCz4wsc5
if (countOfTheWord == null) {270Please respect copyright.ＰＥＮＡＮＡb2I1CNR2MQ
count.put(word, 1);270Please respect copyright.ＰＥＮＡＮＡLtSBJpV8rl
} else {270Please respect copyright.ＰＥＮＡＮＡZPp7pKODJU
count.put(word, countOfTheWord + 1);270Please respect copyright.ＰＥＮＡＮＡBAs16iqWuL
}270Please respect copyright.ＰＥＮＡＮＡFe3CqnwDcs
}270Please respect copyright.ＰＥＮＡＮＡqywQLPKKyD
// 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 word270Please respect copyright.ＰＥＮＡＮＡupUi1nBdq9
int answer = 0;270Please respect copyright.ＰＥＮＡＮＡlStA1owYoQ
boolean central = false;270Please respect copyright.ＰＥＮＡＮＡ54kWYtFUw6
for (Map.Entry<String, Integer> entry : count.entrySet()) {270Please respect copyright.ＰＥＮＡＮＡWAv9x9iv1I
String word = entry.getKey();270Please respect copyright.ＰＥＮＡＮＡ9tRxZyfhNi
int countOfTheWord = entry.getValue();270Please respect copyright.ＰＥＮＡＮＡseGGDwNfyx
// if the word is a palindrome270Please respect copyright.ＰＥＮＡＮＡqzNfjwVUNU
if (word.charAt(0) == word.charAt(1)) {270Please respect copyright.ＰＥＮＡＮＡQnLWHPbZZI
if (countOfTheWord % 2 == 0) {270Please respect copyright.ＰＥＮＡＮＡLmi0Q27IXX
answer += countOfTheWord;270Please respect copyright.ＰＥＮＡＮＡjN4BJppDJi
} else {270Please respect copyright.ＰＥＮＡＮＡdL18kzhXNX
answer += countOfTheWord - 1;270Please respect copyright.ＰＥＮＡＮＡFVeGjuoOcI
central = true;270Please respect copyright.ＰＥＮＡＮＡSxaHKJKYUo
}270Please respect copyright.ＰＥＮＡＮＡMHEcTtahcb
// consider a pair of non-palindrome words such that one is the reverse of another270Please respect copyright.ＰＥＮＡＮＡdN0gEU0XAA
} else if (word.charAt(0) < word.charAt(1)) {270Please respect copyright.ＰＥＮＡＮＡdSErEur35F
String reversedWord = "" + word.charAt(1) + word.charAt(0);270Please respect copyright.ＰＥＮＡＮＡpKtIUZ7eMp
if (count.containsKey(reversedWord)) {270Please respect copyright.ＰＥＮＡＮＡfGnmX3eAhE
answer += 2 * Math.min(countOfTheWord, count.get(reversedWord));270Please respect copyright.ＰＥＮＡＮＡTS25Sr8I1B
}270Please respect copyright.ＰＥＮＡＮＡDGjExFbdYv
}270Please respect copyright.ＰＥＮＡＮＡ8NFQO9VZ2Z
}270Please respect copyright.ＰＥＮＡＮＡCtFcgY7P4O
if (central) {270Please respect copyright.ＰＥＮＡＮＡJcFgmBOX6o
answer++;270Please respect copyright.ＰＥＮＡＮＡ72YA52u6kg
}270Please respect copyright.ＰＥＮＡＮＡGWM3en2TzS
return 2 * answer;270Please respect copyright.ＰＥＮＡＮＡJHTifyvSBr
}270Please respect copyright.ＰＥＮＡＮＡ93bPwEMXzq
}
2: A Two-Dimensional Array Approach
class Solution {270Please respect copyright.ＰＥＮＡＮＡ7osZBzChIG
public int longestPalindrome(String[] words) {270Please respect copyright.ＰＥＮＡＮＡPvFQRCDLcm
final int alphabetSize = 26;270Please respect copyright.ＰＥＮＡＮＡrztSUkQxGD
int[][] count = new int[alphabetSize][alphabetSize];270Please respect copyright.ＰＥＮＡＮＡDZGZSy3olj
// Count the number of occurrences of each word using a two-dimensional array. 270Please respect copyright.ＰＥＮＡＮＡLLR5lHMLDR
for (String word : words) {270Please respect copyright.ＰＥＮＡＮＡgBJoSH2V3M
count[word.charAt(0) - 'a'][word.charAt(1) - 'a']++;270Please respect copyright.ＰＥＮＡＮＡGLjqxS00qW
}270Please respect copyright.ＰＥＮＡＮＡ9GCfph0XCw
// The answer will denote the number of words in the final string and the boolean variable central will denote whether we have a central word270Please respect copyright.ＰＥＮＡＮＡ9mdNbaDqQm
int answer = 0;270Please respect copyright.ＰＥＮＡＮＡGqMKpAPbHI
boolean central = false;270Please respect copyright.ＰＥＮＡＮＡTZmCJuZyIt
for (int i = 0; i < alphabetSize; i++) {270Please respect copyright.ＰＥＮＡＮＡYphlt6ENAq
if (count[i][i] % 2 == 0) {270Please respect copyright.ＰＥＮＡＮＡNcj7pyu0dx
answer += count[i][i];270Please respect copyright.ＰＥＮＡＮＡpYDQPEjXeg
} else {270Please respect copyright.ＰＥＮＡＮＡRMUkParzSP
answer += count[i][i] - 1;270Please respect copyright.ＰＥＮＡＮＡ7u2oBGgfCk
central = true;270Please respect copyright.ＰＥＮＡＮＡw4Z2GebLdB
}270Please respect copyright.ＰＥＮＡＮＡyqPo0F0VvP
for (int j = i + 1; j < alphabetSize; j++) {270Please respect copyright.ＰＥＮＡＮＡH4CIalunwR
answer += 2 * Math.min(count[i][j], count[j][i]);270Please respect copyright.ＰＥＮＡＮＡCaF57apalJ
}270Please respect copyright.ＰＥＮＡＮＡFyhFZPyXXz
}270Please respect copyright.ＰＥＮＡＮＡAnfu10wpv3
if (central) {270Please respect copyright.ＰＥＮＡＮＡkuQkyLnHMQ
answer++;270Please respect copyright.ＰＥＮＡＮＡBY2XV6WiYi
}270Please respect copyright.ＰＥＮＡＮＡzBgSoJaFEP
return 2 * answer;270Please respect copyright.ＰＥＮＡＮＡxA3SCgH84Q
}270Please respect copyright.ＰＥＮＡＮＡuPYGLtazPx
}
270Please respect copyright.ＰＥＮＡＮＡwtvwQaFs2u
