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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 {202Please respect copyright.ＰＥＮＡＮＡqHMbMEyJkV
public int longestPalindrome(String[] words) {202Please respect copyright.ＰＥＮＡＮＡeMvAmXZGLH
HashMap<String, Integer> count = new HashMap<String, Integer>();202Please respect copyright.ＰＥＮＡＮＡAxiMckLgXZ
// Count the number of occurrences of each word using a hashmap202Please respect copyright.ＰＥＮＡＮＡojcLdGMA7l
for (String word : words) {202Please respect copyright.ＰＥＮＡＮＡ7wCOYYlVZn
Integer countOfTheWord = count.get(word);202Please respect copyright.ＰＥＮＡＮＡ1mjApI6lEw
if (countOfTheWord == null) {202Please respect copyright.ＰＥＮＡＮＡ7zYvm6eyXq
count.put(word, 1);202Please respect copyright.ＰＥＮＡＮＡdZPrA8G7jM
} else {202Please respect copyright.ＰＥＮＡＮＡlAsAwciifI
count.put(word, countOfTheWord + 1);202Please respect copyright.ＰＥＮＡＮＡci9cQj9DQz
}202Please respect copyright.ＰＥＮＡＮＡfz4ZSlKWes
}202Please respect copyright.ＰＥＮＡＮＡIPEWRBNs2w
// 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 word202Please respect copyright.ＰＥＮＡＮＡRzCjRaTcvu
int answer = 0;202Please respect copyright.ＰＥＮＡＮＡ6Earls97Mg
boolean central = false;202Please respect copyright.ＰＥＮＡＮＡxCDmRQXLl9
for (Map.Entry<String, Integer> entry : count.entrySet()) {202Please respect copyright.ＰＥＮＡＮＡhfqT5FGMz2
String word = entry.getKey();202Please respect copyright.ＰＥＮＡＮＡFZdNrfpAoy
int countOfTheWord = entry.getValue();202Please respect copyright.ＰＥＮＡＮＡgNPJwPnejg
// if the word is a palindrome202Please respect copyright.ＰＥＮＡＮＡIRuSSod2Is
if (word.charAt(0) == word.charAt(1)) {202Please respect copyright.ＰＥＮＡＮＡgvdw3uly5n
if (countOfTheWord % 2 == 0) {202Please respect copyright.ＰＥＮＡＮＡLkHL5Y2h6B
answer += countOfTheWord;202Please respect copyright.ＰＥＮＡＮＡLS3FLbzHQY
} else {202Please respect copyright.ＰＥＮＡＮＡk24hq6ZSoG
answer += countOfTheWord - 1;202Please respect copyright.ＰＥＮＡＮＡi1lsTFyRWA
central = true;202Please respect copyright.ＰＥＮＡＮＡ7jhQafiFBh
}202Please respect copyright.ＰＥＮＡＮＡp3Hkis9zQ4
// consider a pair of non-palindrome words such that one is the reverse of another202Please respect copyright.ＰＥＮＡＮＡRMWKqYsJ1Y
} else if (word.charAt(0) < word.charAt(1)) {202Please respect copyright.ＰＥＮＡＮＡRF9uKpjsIR
String reversedWord = "" + word.charAt(1) + word.charAt(0);202Please respect copyright.ＰＥＮＡＮＡVTn42LbUAv
if (count.containsKey(reversedWord)) {202Please respect copyright.ＰＥＮＡＮＡfvu5WGwF5t
answer += 2 * Math.min(countOfTheWord, count.get(reversedWord));202Please respect copyright.ＰＥＮＡＮＡOLT1VFwB4p
}202Please respect copyright.ＰＥＮＡＮＡUwEY3LToza
}202Please respect copyright.ＰＥＮＡＮＡtPP0nHp6MA
}202Please respect copyright.ＰＥＮＡＮＡqE8nlTOzOv
if (central) {202Please respect copyright.ＰＥＮＡＮＡlKwlWh7jlM
answer++;202Please respect copyright.ＰＥＮＡＮＡ4652Ebb9e5
}202Please respect copyright.ＰＥＮＡＮＡda92ZqeKY0
return 2 * answer;202Please respect copyright.ＰＥＮＡＮＡrfVtopqeNa
}202Please respect copyright.ＰＥＮＡＮＡmCe8c5y5UJ
}
2: A Two-Dimensional Array Approach
class Solution {202Please respect copyright.ＰＥＮＡＮＡT5yPCgxH0P
public int longestPalindrome(String[] words) {202Please respect copyright.ＰＥＮＡＮＡF6WoFrajjc
final int alphabetSize = 26;202Please respect copyright.ＰＥＮＡＮＡ43lLdwG2XU
int[][] count = new int[alphabetSize][alphabetSize];202Please respect copyright.ＰＥＮＡＮＡxFHLAHWSxc
// Count the number of occurrences of each word using a two-dimensional array. 202Please respect copyright.ＰＥＮＡＮＡd2UMSYdwzB
for (String word : words) {202Please respect copyright.ＰＥＮＡＮＡa3nFQbnxq3
count[word.charAt(0) - 'a'][word.charAt(1) - 'a']++;202Please respect copyright.ＰＥＮＡＮＡkij90Whm7G
}202Please respect copyright.ＰＥＮＡＮＡqV1O4qnd3L
// The answer will denote the number of words in the final string and the boolean variable central will denote whether we have a central word202Please respect copyright.ＰＥＮＡＮＡWDZX2yUUsS
int answer = 0;202Please respect copyright.ＰＥＮＡＮＡkhwI1WmBLG
boolean central = false;202Please respect copyright.ＰＥＮＡＮＡcbnWceTz0m
for (int i = 0; i < alphabetSize; i++) {202Please respect copyright.ＰＥＮＡＮＡATHcm4OL9t
if (count[i][i] % 2 == 0) {202Please respect copyright.ＰＥＮＡＮＡlDmCakYorM
answer += count[i][i];202Please respect copyright.ＰＥＮＡＮＡs8V5Bd4aoL
} else {202Please respect copyright.ＰＥＮＡＮＡ58BAn6tB83
answer += count[i][i] - 1;202Please respect copyright.ＰＥＮＡＮＡxDWnyzEldP
central = true;202Please respect copyright.ＰＥＮＡＮＡ4wsENNEmGJ
}202Please respect copyright.ＰＥＮＡＮＡOKrcQgYdxO
for (int j = i + 1; j < alphabetSize; j++) {202Please respect copyright.ＰＥＮＡＮＡa7UPWYrBOJ
answer += 2 * Math.min(count[i][j], count[j][i]);202Please respect copyright.ＰＥＮＡＮＡtht93H6RFl
}202Please respect copyright.ＰＥＮＡＮＡUhRQX7rc59
}202Please respect copyright.ＰＥＮＡＮＡwGe1h9Ve3J
if (central) {202Please respect copyright.ＰＥＮＡＮＡlVcgL0iAER
answer++;202Please respect copyright.ＰＥＮＡＮＡbS94Upjwkl
}202Please respect copyright.ＰＥＮＡＮＡq9xkC4Cy7V
return 2 * answer;202Please respect copyright.ＰＥＮＡＮＡW6gVFl5Q0g
}202Please respect copyright.ＰＥＮＡＮＡ8ypkdzfx9l
}
202Please respect copyright.ＰＥＮＡＮＡgsOjLuAjAh
