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 {249Please respect copyright.ＰＥＮＡＮＡTKZPtXBQJI
public int longestPalindrome(String[] words) {249Please respect copyright.ＰＥＮＡＮＡ6Rt5PPlABe
HashMap<String, Integer> count = new HashMap<String, Integer>();249Please respect copyright.ＰＥＮＡＮＡcKTKBJ02TJ
// Count the number of occurrences of each word using a hashmap249Please respect copyright.ＰＥＮＡＮＡu13YGcrEeX
for (String word : words) {249Please respect copyright.ＰＥＮＡＮＡjszaEMYQCr
Integer countOfTheWord = count.get(word);249Please respect copyright.ＰＥＮＡＮＡb86jfGKPb2
if (countOfTheWord == null) {249Please respect copyright.ＰＥＮＡＮＡsHyKqQLJbV
count.put(word, 1);249Please respect copyright.ＰＥＮＡＮＡscQpcdjAQF
} else {249Please respect copyright.ＰＥＮＡＮＡZnKEmLIG23
count.put(word, countOfTheWord + 1);249Please respect copyright.ＰＥＮＡＮＡheGt6liznz
}249Please respect copyright.ＰＥＮＡＮＡomg1JTPyrv
}249Please respect copyright.ＰＥＮＡＮＡqICf0gV7W6
// 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 word249Please respect copyright.ＰＥＮＡＮＡILGHDmNE3A
int answer = 0;249Please respect copyright.ＰＥＮＡＮＡtugLuAPzKB
boolean central = false;249Please respect copyright.ＰＥＮＡＮＡV4LGBy4Jqh
for (Map.Entry<String, Integer> entry : count.entrySet()) {249Please respect copyright.ＰＥＮＡＮＡmDhiHHMU1T
String word = entry.getKey();249Please respect copyright.ＰＥＮＡＮＡGHGSdVVX8o
int countOfTheWord = entry.getValue();249Please respect copyright.ＰＥＮＡＮＡXL6vxwPUqP
// if the word is a palindrome249Please respect copyright.ＰＥＮＡＮＡrzaWMgmA9k
if (word.charAt(0) == word.charAt(1)) {249Please respect copyright.ＰＥＮＡＮＡQvEV9d8Brs
if (countOfTheWord % 2 == 0) {249Please respect copyright.ＰＥＮＡＮＡbGDH9QfOsR
answer += countOfTheWord;249Please respect copyright.ＰＥＮＡＮＡN8bsjzjP9S
} else {249Please respect copyright.ＰＥＮＡＮＡGr9sLJf7vY
answer += countOfTheWord - 1;249Please respect copyright.ＰＥＮＡＮＡlHcfN3r45q
central = true;249Please respect copyright.ＰＥＮＡＮＡddBuxF75DV
}249Please respect copyright.ＰＥＮＡＮＡaVJsCkGt7m
// consider a pair of non-palindrome words such that one is the reverse of another249Please respect copyright.ＰＥＮＡＮＡXdEgSMHMz6
} else if (word.charAt(0) < word.charAt(1)) {249Please respect copyright.ＰＥＮＡＮＡ2jeviK9IwN
String reversedWord = "" + word.charAt(1) + word.charAt(0);249Please respect copyright.ＰＥＮＡＮＡXUNHI9xJXM
if (count.containsKey(reversedWord)) {249Please respect copyright.ＰＥＮＡＮＡrrBqXb7SXA
answer += 2 * Math.min(countOfTheWord, count.get(reversedWord));249Please respect copyright.ＰＥＮＡＮＡXcZI5lBTJy
}249Please respect copyright.ＰＥＮＡＮＡVe6LswfVx6
}249Please respect copyright.ＰＥＮＡＮＡXDL17G2LL7
}249Please respect copyright.ＰＥＮＡＮＡr3jNpDa3gi
if (central) {249Please respect copyright.ＰＥＮＡＮＡRntPn1gH2t
answer++;249Please respect copyright.ＰＥＮＡＮＡumL1LReCYj
}249Please respect copyright.ＰＥＮＡＮＡw8JakFkNNQ
return 2 * answer;249Please respect copyright.ＰＥＮＡＮＡ4iIGa9KcnJ
}249Please respect copyright.ＰＥＮＡＮＡJ2C8ZGkIuJ
}
2: A Two-Dimensional Array Approach
class Solution {249Please respect copyright.ＰＥＮＡＮＡurATOFzl2B
public int longestPalindrome(String[] words) {249Please respect copyright.ＰＥＮＡＮＡgwYU1F22fq
final int alphabetSize = 26;249Please respect copyright.ＰＥＮＡＮＡESO9eYfhrz
int[][] count = new int[alphabetSize][alphabetSize];249Please respect copyright.ＰＥＮＡＮＡFdrog0Kv6j
// Count the number of occurrences of each word using a two-dimensional array. 249Please respect copyright.ＰＥＮＡＮＡo9GznLWOWK
for (String word : words) {249Please respect copyright.ＰＥＮＡＮＡzXHm6Vozmp
count[word.charAt(0) - 'a'][word.charAt(1) - 'a']++;249Please respect copyright.ＰＥＮＡＮＡODvYtTeLBW
}249Please respect copyright.ＰＥＮＡＮＡB4dpGgLedr
// The answer will denote the number of words in the final string and the boolean variable central will denote whether we have a central word249Please respect copyright.ＰＥＮＡＮＡX2GMwcPM4q
int answer = 0;249Please respect copyright.ＰＥＮＡＮＡFxRoX6oJgn
boolean central = false;249Please respect copyright.ＰＥＮＡＮＡ7tO0dCykFs
for (int i = 0; i < alphabetSize; i++) {249Please respect copyright.ＰＥＮＡＮＡP7KE8drHXd
if (count[i][i] % 2 == 0) {249Please respect copyright.ＰＥＮＡＮＡCJ7IXxkbcP
answer += count[i][i];249Please respect copyright.ＰＥＮＡＮＡlOkqrnoHOE
} else {249Please respect copyright.ＰＥＮＡＮＡzaXqtuj5rS
answer += count[i][i] - 1;249Please respect copyright.ＰＥＮＡＮＡtQY1p4xrj3
central = true;249Please respect copyright.ＰＥＮＡＮＡ2hHm4ZPmIR
}249Please respect copyright.ＰＥＮＡＮＡF3Q4W1tees
for (int j = i + 1; j < alphabetSize; j++) {249Please respect copyright.ＰＥＮＡＮＡQ4RN5WMDwR
answer += 2 * Math.min(count[i][j], count[j][i]);249Please respect copyright.ＰＥＮＡＮＡ6jGPcvmJdm
}249Please respect copyright.ＰＥＮＡＮＡb04sNfmsMl
}249Please respect copyright.ＰＥＮＡＮＡZwt7zbHv8D
if (central) {249Please respect copyright.ＰＥＮＡＮＡ0GYmQaAHhv
answer++;249Please respect copyright.ＰＥＮＡＮＡm5UB7epDkE
}249Please respect copyright.ＰＥＮＡＮＡ47EqDeIt6x
return 2 * answer;249Please respect copyright.ＰＥＮＡＮＡXtDDaVeOgX
}249Please respect copyright.ＰＥＮＡＮＡ2edGq9kXzW
}
249Please respect copyright.ＰＥＮＡＮＡpQJ5BAy6vr
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