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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 {254Please respect copyright.ＰＥＮＡＮＡB1LylCozIW
public int longestPalindrome(String[] words) {254Please respect copyright.ＰＥＮＡＮＡdwQvrRlD0H
HashMap<String, Integer> count = new HashMap<String, Integer>();254Please respect copyright.ＰＥＮＡＮＡjJsrclVFhB
// Count the number of occurrences of each word using a hashmap254Please respect copyright.ＰＥＮＡＮＡX4J9qVpxy4
for (String word : words) {254Please respect copyright.ＰＥＮＡＮＡTIGLWatweR
Integer countOfTheWord = count.get(word);254Please respect copyright.ＰＥＮＡＮＡbJql9AGlc3
if (countOfTheWord == null) {254Please respect copyright.ＰＥＮＡＮＡQnVJ1h8Bx0
count.put(word, 1);254Please respect copyright.ＰＥＮＡＮＡRBFDXU0bVC
} else {254Please respect copyright.ＰＥＮＡＮＡPtJNIkHmiS
count.put(word, countOfTheWord + 1);254Please respect copyright.ＰＥＮＡＮＡ9teArVhXP6
}254Please respect copyright.ＰＥＮＡＮＡKhBIQpiICW
}254Please respect copyright.ＰＥＮＡＮＡU78Il3rz7G
// 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 word254Please respect copyright.ＰＥＮＡＮＡTyJc0hZMyD
int answer = 0;254Please respect copyright.ＰＥＮＡＮＡ2XKp9srP0T
boolean central = false;254Please respect copyright.ＰＥＮＡＮＡhMtzJxIEdG
for (Map.Entry<String, Integer> entry : count.entrySet()) {254Please respect copyright.ＰＥＮＡＮＡuIZHTGqnRs
String word = entry.getKey();254Please respect copyright.ＰＥＮＡＮＡfVsq2toJFN
int countOfTheWord = entry.getValue();254Please respect copyright.ＰＥＮＡＮＡv3CPnE9YAC
// if the word is a palindrome254Please respect copyright.ＰＥＮＡＮＡm2tRK6pswq
if (word.charAt(0) == word.charAt(1)) {254Please respect copyright.ＰＥＮＡＮＡZe8Z8LpjvS
if (countOfTheWord % 2 == 0) {254Please respect copyright.ＰＥＮＡＮＡS0huxSXJ2e
answer += countOfTheWord;254Please respect copyright.ＰＥＮＡＮＡ9gU1Bvnwic
} else {254Please respect copyright.ＰＥＮＡＮＡko1uAcIwmR
answer += countOfTheWord - 1;254Please respect copyright.ＰＥＮＡＮＡF4FT9EFHBJ
central = true;254Please respect copyright.ＰＥＮＡＮＡAMPCC4KM2a
}254Please respect copyright.ＰＥＮＡＮＡl3vCe1w5KI
// consider a pair of non-palindrome words such that one is the reverse of another254Please respect copyright.ＰＥＮＡＮＡqX26j1Jofa
} else if (word.charAt(0) < word.charAt(1)) {254Please respect copyright.ＰＥＮＡＮＡV0LKC1ouxT
String reversedWord = "" + word.charAt(1) + word.charAt(0);254Please respect copyright.ＰＥＮＡＮＡN0XvKJsSHf
if (count.containsKey(reversedWord)) {254Please respect copyright.ＰＥＮＡＮＡCBtYUcOJg3
answer += 2 * Math.min(countOfTheWord, count.get(reversedWord));254Please respect copyright.ＰＥＮＡＮＡjV0r8qNqVs
}254Please respect copyright.ＰＥＮＡＮＡMzsOBeaLUa
}254Please respect copyright.ＰＥＮＡＮＡA8bNaxsl4g
}254Please respect copyright.ＰＥＮＡＮＡRtKqn9GZa5
if (central) {254Please respect copyright.ＰＥＮＡＮＡlMmiYzDlhG
answer++;254Please respect copyright.ＰＥＮＡＮＡuq4gxb6byN
}254Please respect copyright.ＰＥＮＡＮＡUs1nSMIvi0
return 2 * answer;254Please respect copyright.ＰＥＮＡＮＡOxO9TkBA23
}254Please respect copyright.ＰＥＮＡＮＡPx3cKcOeBV
}
2: A Two-Dimensional Array Approach
class Solution {254Please respect copyright.ＰＥＮＡＮＡ94xvoytp7k
public int longestPalindrome(String[] words) {254Please respect copyright.ＰＥＮＡＮＡGxE2g28PTh
final int alphabetSize = 26;254Please respect copyright.ＰＥＮＡＮＡoxaxbSpc6Z
int[][] count = new int[alphabetSize][alphabetSize];254Please respect copyright.ＰＥＮＡＮＡllTu0XRZks
// Count the number of occurrences of each word using a two-dimensional array. 254Please respect copyright.ＰＥＮＡＮＡ4alyfwUFrS
for (String word : words) {254Please respect copyright.ＰＥＮＡＮＡXbXx1WfjId
count[word.charAt(0) - 'a'][word.charAt(1) - 'a']++;254Please respect copyright.ＰＥＮＡＮＡvJRI3eV1qH
}254Please respect copyright.ＰＥＮＡＮＡFwFVFMIezA
// The answer will denote the number of words in the final string and the boolean variable central will denote whether we have a central word254Please respect copyright.ＰＥＮＡＮＡcbG9bZ3Pw8
int answer = 0;254Please respect copyright.ＰＥＮＡＮＡwzziD0mIRY
boolean central = false;254Please respect copyright.ＰＥＮＡＮＡhARqpTDLe3
for (int i = 0; i < alphabetSize; i++) {254Please respect copyright.ＰＥＮＡＮＡgDrWK9F6tT
if (count[i][i] % 2 == 0) {254Please respect copyright.ＰＥＮＡＮＡ0P3qq6Rp3g
answer += count[i][i];254Please respect copyright.ＰＥＮＡＮＡijPsbgpQZD
} else {254Please respect copyright.ＰＥＮＡＮＡJiKYsr2Zva
answer += count[i][i] - 1;254Please respect copyright.ＰＥＮＡＮＡcLWndeUVcU
central = true;254Please respect copyright.ＰＥＮＡＮＡN8NZKDwtwF
}254Please respect copyright.ＰＥＮＡＮＡQWjcI09hIq
for (int j = i + 1; j < alphabetSize; j++) {254Please respect copyright.ＰＥＮＡＮＡAg5NR3qEv1
answer += 2 * Math.min(count[i][j], count[j][i]);254Please respect copyright.ＰＥＮＡＮＡAhIqcyhP4Q
}254Please respect copyright.ＰＥＮＡＮＡfOg1sU8wxD
}254Please respect copyright.ＰＥＮＡＮＡndIW6VXq61
if (central) {254Please respect copyright.ＰＥＮＡＮＡykw9wrAr4F
answer++;254Please respect copyright.ＰＥＮＡＮＡdU34L8ITco
}254Please respect copyright.ＰＥＮＡＮＡa6kZ6UyMqR
return 2 * answer;254Please respect copyright.ＰＥＮＡＮＡJT7TWm3HDO
}254Please respect copyright.ＰＥＮＡＮＡ4FkAXbmdV3
}
254Please respect copyright.ＰＥＮＡＮＡpH34chIsPc
