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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 {266Please respect copyright.ＰＥＮＡＮＡ7lSi5aDZ4D
public int longestPalindrome(String[] words) {266Please respect copyright.ＰＥＮＡＮＡDURVVhRVgI
HashMap<String, Integer> count = new HashMap<String, Integer>();266Please respect copyright.ＰＥＮＡＮＡLvAvuaSs0G
// Count the number of occurrences of each word using a hashmap266Please respect copyright.ＰＥＮＡＮＡjy3tIGR2kx
for (String word : words) {266Please respect copyright.ＰＥＮＡＮＡz6FKbCh6Dz
Integer countOfTheWord = count.get(word);266Please respect copyright.ＰＥＮＡＮＡijt0a1iZR0
if (countOfTheWord == null) {266Please respect copyright.ＰＥＮＡＮＡuVIhNRjRjX
count.put(word, 1);266Please respect copyright.ＰＥＮＡＮＡPIDFjYNnlX
} else {266Please respect copyright.ＰＥＮＡＮＡnr1z0wAPZg
count.put(word, countOfTheWord + 1);266Please respect copyright.ＰＥＮＡＮＡY29SQjEdNS
}266Please respect copyright.ＰＥＮＡＮＡmOQfh2lxri
}266Please respect copyright.ＰＥＮＡＮＡIKuMkzTqWy
// 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 word266Please respect copyright.ＰＥＮＡＮＡVlO2mj88zK
int answer = 0;266Please respect copyright.ＰＥＮＡＮＡZUsGmVZ6SG
boolean central = false;266Please respect copyright.ＰＥＮＡＮＡYqGxjgww2U
for (Map.Entry<String, Integer> entry : count.entrySet()) {266Please respect copyright.ＰＥＮＡＮＡp104eFGY2C
String word = entry.getKey();266Please respect copyright.ＰＥＮＡＮＡGyknFQyUuJ
int countOfTheWord = entry.getValue();266Please respect copyright.ＰＥＮＡＮＡJPjV0JwQB1
// if the word is a palindrome266Please respect copyright.ＰＥＮＡＮＡ0pb3NDuKoO
if (word.charAt(0) == word.charAt(1)) {266Please respect copyright.ＰＥＮＡＮＡwtrvyG0ppD
if (countOfTheWord % 2 == 0) {266Please respect copyright.ＰＥＮＡＮＡXPUOL2P8Md
answer += countOfTheWord;266Please respect copyright.ＰＥＮＡＮＡZGitD7qNjb
} else {266Please respect copyright.ＰＥＮＡＮＡMtLJE8wdiX
answer += countOfTheWord - 1;266Please respect copyright.ＰＥＮＡＮＡk5VxdYwWzD
central = true;266Please respect copyright.ＰＥＮＡＮＡSpKsaYR3CP
}266Please respect copyright.ＰＥＮＡＮＡOwAnoRcvg9
// consider a pair of non-palindrome words such that one is the reverse of another266Please respect copyright.ＰＥＮＡＮＡ9TAEBPgQV0
} else if (word.charAt(0) < word.charAt(1)) {266Please respect copyright.ＰＥＮＡＮＡStYHoSk2fj
String reversedWord = "" + word.charAt(1) + word.charAt(0);266Please respect copyright.ＰＥＮＡＮＡrHN098gSEE
if (count.containsKey(reversedWord)) {266Please respect copyright.ＰＥＮＡＮＡa5014805r7
answer += 2 * Math.min(countOfTheWord, count.get(reversedWord));266Please respect copyright.ＰＥＮＡＮＡTxrREgMGO6
}266Please respect copyright.ＰＥＮＡＮＡezomlTpgp4
}266Please respect copyright.ＰＥＮＡＮＡ8zck5Z7Gv9
}266Please respect copyright.ＰＥＮＡＮＡCvldxUilSB
if (central) {266Please respect copyright.ＰＥＮＡＮＡEETWjhQbDR
answer++;266Please respect copyright.ＰＥＮＡＮＡ7jrp8o2Xbb
}266Please respect copyright.ＰＥＮＡＮＡ6BbBWjUTuY
return 2 * answer;266Please respect copyright.ＰＥＮＡＮＡkky2EllfBU
}266Please respect copyright.ＰＥＮＡＮＡZJedOw4QcZ
}
2: A Two-Dimensional Array Approach
class Solution {266Please respect copyright.ＰＥＮＡＮＡskxnix8RJX
public int longestPalindrome(String[] words) {266Please respect copyright.ＰＥＮＡＮＡGYAZqSUIQ1
final int alphabetSize = 26;266Please respect copyright.ＰＥＮＡＮＡZYoUilGg4y
int[][] count = new int[alphabetSize][alphabetSize];266Please respect copyright.ＰＥＮＡＮＡ99YXf8nIoA
// Count the number of occurrences of each word using a two-dimensional array. 266Please respect copyright.ＰＥＮＡＮＡsM6Nkiu8Xx
for (String word : words) {266Please respect copyright.ＰＥＮＡＮＡW1eGZnocZI
count[word.charAt(0) - 'a'][word.charAt(1) - 'a']++;266Please respect copyright.ＰＥＮＡＮＡg1gDls889n
}266Please respect copyright.ＰＥＮＡＮＡBqbKHyAyVC
// The answer will denote the number of words in the final string and the boolean variable central will denote whether we have a central word266Please respect copyright.ＰＥＮＡＮＡBPMdGhcGC5
int answer = 0;266Please respect copyright.ＰＥＮＡＮＡPgex54W7cX
boolean central = false;266Please respect copyright.ＰＥＮＡＮＡ4Orl1doSTU
for (int i = 0; i < alphabetSize; i++) {266Please respect copyright.ＰＥＮＡＮＡzrbhWaOscb
if (count[i][i] % 2 == 0) {266Please respect copyright.ＰＥＮＡＮＡCpBC4fkao5
answer += count[i][i];266Please respect copyright.ＰＥＮＡＮＡmrC7Ki8Ojh
} else {266Please respect copyright.ＰＥＮＡＮＡccIAc68Cjp
answer += count[i][i] - 1;266Please respect copyright.ＰＥＮＡＮＡiNDv5g1b4P
central = true;266Please respect copyright.ＰＥＮＡＮＡK5WrYf23e0
}266Please respect copyright.ＰＥＮＡＮＡe1Lvxa83sI
for (int j = i + 1; j < alphabetSize; j++) {266Please respect copyright.ＰＥＮＡＮＡUTGtD1Xuef
answer += 2 * Math.min(count[i][j], count[j][i]);266Please respect copyright.ＰＥＮＡＮＡB36ETod3a0
}266Please respect copyright.ＰＥＮＡＮＡmBK0gPGrpQ
}266Please respect copyright.ＰＥＮＡＮＡmmdJ30eOJf
if (central) {266Please respect copyright.ＰＥＮＡＮＡWTYZiV38Ng
answer++;266Please respect copyright.ＰＥＮＡＮＡ4JP9DjFrBy
}266Please respect copyright.ＰＥＮＡＮＡZQYp8P6dff
return 2 * answer;266Please respect copyright.ＰＥＮＡＮＡ4rKfoZwXec
}266Please respect copyright.ＰＥＮＡＮＡvQ2eFuEub8
}
266Please respect copyright.ＰＥＮＡＮＡasC53cqzBC
