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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 {276Please respect copyright.ＰＥＮＡＮＡFbU81tiNYD
public int longestPalindrome(String[] words) {276Please respect copyright.ＰＥＮＡＮＡ74xyP4Z2lR
HashMap<String, Integer> count = new HashMap<String, Integer>();276Please respect copyright.ＰＥＮＡＮＡdcrJDocHkn
// Count the number of occurrences of each word using a hashmap276Please respect copyright.ＰＥＮＡＮＡYPSogNojFq
for (String word : words) {276Please respect copyright.ＰＥＮＡＮＡseuex093g9
Integer countOfTheWord = count.get(word);276Please respect copyright.ＰＥＮＡＮＡb4uUJEYhvL
if (countOfTheWord == null) {276Please respect copyright.ＰＥＮＡＮＡqjTwN0rnMS
count.put(word, 1);276Please respect copyright.ＰＥＮＡＮＡekVnc87QtW
} else {276Please respect copyright.ＰＥＮＡＮＡDskY5S8ro2
count.put(word, countOfTheWord + 1);276Please respect copyright.ＰＥＮＡＮＡvnzw5f0zyA
}276Please respect copyright.ＰＥＮＡＮＡH4ZVZsd3Q7
}276Please respect copyright.ＰＥＮＡＮＡw27icMfdvA
// 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 word276Please respect copyright.ＰＥＮＡＮＡkB195R99af
int answer = 0;276Please respect copyright.ＰＥＮＡＮＡfxeLFhdMvK
boolean central = false;276Please respect copyright.ＰＥＮＡＮＡ4YWz1gh0Z8
for (Map.Entry<String, Integer> entry : count.entrySet()) {276Please respect copyright.ＰＥＮＡＮＡExuHH83HLk
String word = entry.getKey();276Please respect copyright.ＰＥＮＡＮＡ1qOgdReBB1
int countOfTheWord = entry.getValue();276Please respect copyright.ＰＥＮＡＮＡcnAum4MyZg
// if the word is a palindrome276Please respect copyright.ＰＥＮＡＮＡ8xl0pVYMLt
if (word.charAt(0) == word.charAt(1)) {276Please respect copyright.ＰＥＮＡＮＡcZVoaKdp9L
if (countOfTheWord % 2 == 0) {276Please respect copyright.ＰＥＮＡＮＡDUkptrTbrT
answer += countOfTheWord;276Please respect copyright.ＰＥＮＡＮＡ5zVZZKniiI
} else {276Please respect copyright.ＰＥＮＡＮＡiL2B9PaTah
answer += countOfTheWord - 1;276Please respect copyright.ＰＥＮＡＮＡQFANXrEhjR
central = true;276Please respect copyright.ＰＥＮＡＮＡgcCdRd8KZg
}276Please respect copyright.ＰＥＮＡＮＡRJOd3luepC
// consider a pair of non-palindrome words such that one is the reverse of another276Please respect copyright.ＰＥＮＡＮＡX4JAc5dlbV
} else if (word.charAt(0) < word.charAt(1)) {276Please respect copyright.ＰＥＮＡＮＡsF1NOseWb9
String reversedWord = "" + word.charAt(1) + word.charAt(0);276Please respect copyright.ＰＥＮＡＮＡPyPP6JVu5y
if (count.containsKey(reversedWord)) {276Please respect copyright.ＰＥＮＡＮＡiC73uYh4IO
answer += 2 * Math.min(countOfTheWord, count.get(reversedWord));276Please respect copyright.ＰＥＮＡＮＡp3tgGXoA39
}276Please respect copyright.ＰＥＮＡＮＡR0EyrJAClL
}276Please respect copyright.ＰＥＮＡＮＡw20gbMr5ng
}276Please respect copyright.ＰＥＮＡＮＡCFxoK5XJnU
if (central) {276Please respect copyright.ＰＥＮＡＮＡs4M8kShgoW
answer++;276Please respect copyright.ＰＥＮＡＮＡIWPctZqfWO
}276Please respect copyright.ＰＥＮＡＮＡVMfMbCRzYk
return 2 * answer;276Please respect copyright.ＰＥＮＡＮＡPJWCFukih3
}276Please respect copyright.ＰＥＮＡＮＡMCViFH9tSx
}
2: A Two-Dimensional Array Approach
class Solution {276Please respect copyright.ＰＥＮＡＮＡnBShPcdYuI
public int longestPalindrome(String[] words) {276Please respect copyright.ＰＥＮＡＮＡICLULr04LI
final int alphabetSize = 26;276Please respect copyright.ＰＥＮＡＮＡ2wydSdV9po
int[][] count = new int[alphabetSize][alphabetSize];276Please respect copyright.ＰＥＮＡＮＡtYQINCd49x
// Count the number of occurrences of each word using a two-dimensional array. 276Please respect copyright.ＰＥＮＡＮＡSyeLuFGans
for (String word : words) {276Please respect copyright.ＰＥＮＡＮＡFfkRIhDrbg
count[word.charAt(0) - 'a'][word.charAt(1) - 'a']++;276Please respect copyright.ＰＥＮＡＮＡIQQLxOxSXM
}276Please respect copyright.ＰＥＮＡＮＡpcMXKdhFHJ
// The answer will denote the number of words in the final string and the boolean variable central will denote whether we have a central word276Please respect copyright.ＰＥＮＡＮＡ74AnCI9rpj
int answer = 0;276Please respect copyright.ＰＥＮＡＮＡp80lfoqpMH
boolean central = false;276Please respect copyright.ＰＥＮＡＮＡvROjH2z9S3
for (int i = 0; i < alphabetSize; i++) {276Please respect copyright.ＰＥＮＡＮＡ9IUOLSbIxE
if (count[i][i] % 2 == 0) {276Please respect copyright.ＰＥＮＡＮＡjX9UWXOOTZ
answer += count[i][i];276Please respect copyright.ＰＥＮＡＮＡkocBvDUjWr
} else {276Please respect copyright.ＰＥＮＡＮＡjhKx0tzKZQ
answer += count[i][i] - 1;276Please respect copyright.ＰＥＮＡＮＡymHgCRizwA
central = true;276Please respect copyright.ＰＥＮＡＮＡKap5taDbIt
}276Please respect copyright.ＰＥＮＡＮＡn5kE7TBbsF
for (int j = i + 1; j < alphabetSize; j++) {276Please respect copyright.ＰＥＮＡＮＡYwkmjdqJa5
answer += 2 * Math.min(count[i][j], count[j][i]);276Please respect copyright.ＰＥＮＡＮＡacZGWbYn8B
}276Please respect copyright.ＰＥＮＡＮＡ1pDM0cP6HJ
}276Please respect copyright.ＰＥＮＡＮＡh5FEZkbG5Z
if (central) {276Please respect copyright.ＰＥＮＡＮＡhfbXhpZkSn
answer++;276Please respect copyright.ＰＥＮＡＮＡZInXLeIjII
}276Please respect copyright.ＰＥＮＡＮＡT9RailJkul
return 2 * answer;276Please respect copyright.ＰＥＮＡＮＡRRkhZguJ1r
}276Please respect copyright.ＰＥＮＡＮＡ09XRo4ck7R
}
276Please respect copyright.ＰＥＮＡＮＡN8yaBX2R4e
