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 {450Please respect copyright.ＰＥＮＡＮＡNzUth6NozU
public int longestPalindrome(String[] words) {450Please respect copyright.ＰＥＮＡＮＡNJWFtMlVrm
HashMap<String, Integer> count = new HashMap<String, Integer>();450Please respect copyright.ＰＥＮＡＮＡFgiS5B32Dj
// Count the number of occurrences of each word using a hashmap450Please respect copyright.ＰＥＮＡＮＡ6W1FJdED6a
for (String word : words) {450Please respect copyright.ＰＥＮＡＮＡFyVyH7tfaM
Integer countOfTheWord = count.get(word);450Please respect copyright.ＰＥＮＡＮＡKvMcSjXm2N
if (countOfTheWord == null) {450Please respect copyright.ＰＥＮＡＮＡ7LGMTSs0aV
count.put(word, 1);450Please respect copyright.ＰＥＮＡＮＡ2dNh4cSfbw
} else {450Please respect copyright.ＰＥＮＡＮＡisKpvCDcYG
count.put(word, countOfTheWord + 1);450Please respect copyright.ＰＥＮＡＮＡDQ0xhH696K
}450Please respect copyright.ＰＥＮＡＮＡ19cyjQCPRI
}450Please respect copyright.ＰＥＮＡＮＡZqIhI3DHHB
// 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 word450Please respect copyright.ＰＥＮＡＮＡPXqBdrUrF8
int answer = 0;450Please respect copyright.ＰＥＮＡＮＡE9nsr2gxOw
boolean central = false;450Please respect copyright.ＰＥＮＡＮＡqX8mAzWJeP
for (Map.Entry<String, Integer> entry : count.entrySet()) {450Please respect copyright.ＰＥＮＡＮＡleQrSTDpvL
String word = entry.getKey();450Please respect copyright.ＰＥＮＡＮＡUF1rBFvvKj
int countOfTheWord = entry.getValue();450Please respect copyright.ＰＥＮＡＮＡmABukNpxKY
// if the word is a palindrome450Please respect copyright.ＰＥＮＡＮＡBdO3zYYMYK
if (word.charAt(0) == word.charAt(1)) {450Please respect copyright.ＰＥＮＡＮＡeOjV1JEeoU
if (countOfTheWord % 2 == 0) {450Please respect copyright.ＰＥＮＡＮＡFueREB9TIE
answer += countOfTheWord;450Please respect copyright.ＰＥＮＡＮＡOO591W8MZ7
} else {450Please respect copyright.ＰＥＮＡＮＡRn3mpykPwN
answer += countOfTheWord - 1;450Please respect copyright.ＰＥＮＡＮＡN7N2MUgfLF
central = true;450Please respect copyright.ＰＥＮＡＮＡZBCgvzo67O
}450Please respect copyright.ＰＥＮＡＮＡlAE4zyJjB9
// consider a pair of non-palindrome words such that one is the reverse of another450Please respect copyright.ＰＥＮＡＮＡMFEqXtqxJw
} else if (word.charAt(0) < word.charAt(1)) {450Please respect copyright.ＰＥＮＡＮＡmsZLtGMHfW
String reversedWord = "" + word.charAt(1) + word.charAt(0);450Please respect copyright.ＰＥＮＡＮＡn7TlJstNxL
if (count.containsKey(reversedWord)) {450Please respect copyright.ＰＥＮＡＮＡeEHqRcPDbS
answer += 2 * Math.min(countOfTheWord, count.get(reversedWord));450Please respect copyright.ＰＥＮＡＮＡAScmbqHDn1
}450Please respect copyright.ＰＥＮＡＮＡzF0pN7DhBQ
}450Please respect copyright.ＰＥＮＡＮＡYmuFkTJ4rD
}450Please respect copyright.ＰＥＮＡＮＡGUZRhGqLAQ
if (central) {450Please respect copyright.ＰＥＮＡＮＡWbcUQ13cMt
answer++;450Please respect copyright.ＰＥＮＡＮＡdrIxI2LOAt
}450Please respect copyright.ＰＥＮＡＮＡSTeIWB3uGu
return 2 * answer;450Please respect copyright.ＰＥＮＡＮＡ8DnJa3fom6
}450Please respect copyright.ＰＥＮＡＮＡKXgb7M8aTN
}
2: A Two-Dimensional Array Approach
class Solution {450Please respect copyright.ＰＥＮＡＮＡaXuWYk5ZVr
public int longestPalindrome(String[] words) {450Please respect copyright.ＰＥＮＡＮＡ2rUyy1uI9w
final int alphabetSize = 26;450Please respect copyright.ＰＥＮＡＮＡIEBfqqa8RE
int[][] count = new int[alphabetSize][alphabetSize];450Please respect copyright.ＰＥＮＡＮＡwcMyLnzWvW
// Count the number of occurrences of each word using a two-dimensional array. 450Please respect copyright.ＰＥＮＡＮＡrO4MvGNVIs
for (String word : words) {450Please respect copyright.ＰＥＮＡＮＡXVo1qXkeFy
count[word.charAt(0) - 'a'][word.charAt(1) - 'a']++;450Please respect copyright.ＰＥＮＡＮＡJ88rS5Cuno
}450Please respect copyright.ＰＥＮＡＮＡMJW4oCv086
// The answer will denote the number of words in the final string and the boolean variable central will denote whether we have a central word450Please respect copyright.ＰＥＮＡＮＡa6TEyRydrF
int answer = 0;450Please respect copyright.ＰＥＮＡＮＡY5Ama5vTb2
boolean central = false;450Please respect copyright.ＰＥＮＡＮＡqtBPTpy6V1
for (int i = 0; i < alphabetSize; i++) {450Please respect copyright.ＰＥＮＡＮＡWOV3gsX4ip
if (count[i][i] % 2 == 0) {450Please respect copyright.ＰＥＮＡＮＡ66hRhHdB1n
answer += count[i][i];450Please respect copyright.ＰＥＮＡＮＡiBHNtTrGcC
} else {450Please respect copyright.ＰＥＮＡＮＡYnVwr8Jkbh
answer += count[i][i] - 1;450Please respect copyright.ＰＥＮＡＮＡmomNdpp61E
central = true;450Please respect copyright.ＰＥＮＡＮＡJfvc555Cga
}450Please respect copyright.ＰＥＮＡＮＡTP3fRfXH08
for (int j = i + 1; j < alphabetSize; j++) {450Please respect copyright.ＰＥＮＡＮＡuKX88v89I1
answer += 2 * Math.min(count[i][j], count[j][i]);450Please respect copyright.ＰＥＮＡＮＡyK7HDQwu4C
}450Please respect copyright.ＰＥＮＡＮＡ0GISbKQ3PL
}450Please respect copyright.ＰＥＮＡＮＡ5xOGcB0ZTk
if (central) {450Please respect copyright.ＰＥＮＡＮＡ4U2yxVJAn9
answer++;450Please respect copyright.ＰＥＮＡＮＡqXWE8byTWw
}450Please respect copyright.ＰＥＮＡＮＡwNJRFtUuKF
return 2 * answer;450Please respect copyright.ＰＥＮＡＮＡdqpkClgdoM
}450Please respect copyright.ＰＥＮＡＮＡIXCWa3526V
}
450Please respect copyright.ＰＥＮＡＮＡ4D6Xel4Cxa
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