
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 {266Please respect copyright.PENANA7lSi5aDZ4D
public int longestPalindrome(String[] words) {266Please respect copyright.PENANADURVVhRVgI
HashMap<String, Integer> count = new HashMap<String, Integer>();266Please respect copyright.PENANALvAvuaSs0G
// Count the number of occurrences of each word using a hashmap266Please respect copyright.PENANAjy3tIGR2kx
for (String word : words) {266Please respect copyright.PENANAz6FKbCh6Dz
Integer countOfTheWord = count.get(word);266Please respect copyright.PENANAijt0a1iZR0
if (countOfTheWord == null) {266Please respect copyright.PENANAuVIhNRjRjX
count.put(word, 1);266Please respect copyright.PENANAPIDFjYNnlX
} else {266Please respect copyright.PENANAnr1z0wAPZg
count.put(word, countOfTheWord + 1);266Please respect copyright.PENANAY29SQjEdNS
}266Please respect copyright.PENANAmOQfh2lxri
}266Please respect copyright.PENANAIKuMkzTqWy
// 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 word266Please respect copyright.PENANAVlO2mj88zK
int answer = 0;266Please respect copyright.PENANAZUsGmVZ6SG
boolean central = false;266Please respect copyright.PENANAYqGxjgww2U
for (Map.Entry<String, Integer> entry : count.entrySet()) {266Please respect copyright.PENANAp104eFGY2C
String word = entry.getKey();266Please respect copyright.PENANAGyknFQyUuJ
int countOfTheWord = entry.getValue();266Please respect copyright.PENANAJPjV0JwQB1
// if the word is a palindrome266Please respect copyright.PENANA0pb3NDuKoO
if (word.charAt(0) == word.charAt(1)) {266Please respect copyright.PENANAwtrvyG0ppD
if (countOfTheWord % 2 == 0) {266Please respect copyright.PENANAXPUOL2P8Md
answer += countOfTheWord;266Please respect copyright.PENANAZGitD7qNjb
} else {266Please respect copyright.PENANAMtLJE8wdiX
answer += countOfTheWord - 1;266Please respect copyright.PENANAk5VxdYwWzD
central = true;266Please respect copyright.PENANASpKsaYR3CP
}266Please respect copyright.PENANAOwAnoRcvg9
// consider a pair of non-palindrome words such that one is the reverse of another266Please respect copyright.PENANA9TAEBPgQV0
} else if (word.charAt(0) < word.charAt(1)) {266Please respect copyright.PENANAStYHoSk2fj
String reversedWord = "" + word.charAt(1) + word.charAt(0);266Please respect copyright.PENANArHN098gSEE
if (count.containsKey(reversedWord)) {266Please respect copyright.PENANAa5014805r7
answer += 2 * Math.min(countOfTheWord, count.get(reversedWord));266Please respect copyright.PENANATxrREgMGO6
}266Please respect copyright.PENANAezomlTpgp4
}266Please respect copyright.PENANA8zck5Z7Gv9
}266Please respect copyright.PENANACvldxUilSB
if (central) {266Please respect copyright.PENANAEETWjhQbDR
answer++;266Please respect copyright.PENANA7jrp8o2Xbb
}266Please respect copyright.PENANA6BbBWjUTuY
return 2 * answer;266Please respect copyright.PENANAkky2EllfBU
}266Please respect copyright.PENANAZJedOw4QcZ
}
2: A Two-Dimensional Array Approach
class Solution {266Please respect copyright.PENANAskxnix8RJX
public int longestPalindrome(String[] words) {266Please respect copyright.PENANAGYAZqSUIQ1
final int alphabetSize = 26;266Please respect copyright.PENANAZYoUilGg4y
int[][] count = new int[alphabetSize][alphabetSize];266Please respect copyright.PENANA99YXf8nIoA
// Count the number of occurrences of each word using a two-dimensional array. 266Please respect copyright.PENANAsM6Nkiu8Xx
for (String word : words) {266Please respect copyright.PENANAW1eGZnocZI
count[word.charAt(0) - 'a'][word.charAt(1) - 'a']++;266Please respect copyright.PENANAg1gDls889n
}266Please respect copyright.PENANABqbKHyAyVC
// The answer will denote the number of words in the final string and the boolean variable central will denote whether we have a central word266Please respect copyright.PENANABPMdGhcGC5
int answer = 0;266Please respect copyright.PENANAPgex54W7cX
boolean central = false;266Please respect copyright.PENANA4Orl1doSTU
for (int i = 0; i < alphabetSize; i++) {266Please respect copyright.PENANAzrbhWaOscb
if (count[i][i] % 2 == 0) {266Please respect copyright.PENANACpBC4fkao5
answer += count[i][i];266Please respect copyright.PENANAmrC7Ki8Ojh
} else {266Please respect copyright.PENANAccIAc68Cjp
answer += count[i][i] - 1;266Please respect copyright.PENANAiNDv5g1b4P
central = true;266Please respect copyright.PENANAK5WrYf23e0
}266Please respect copyright.PENANAe1Lvxa83sI
for (int j = i + 1; j < alphabetSize; j++) {266Please respect copyright.PENANAUTGtD1Xuef
answer += 2 * Math.min(count[i][j], count[j][i]);266Please respect copyright.PENANAB36ETod3a0
}266Please respect copyright.PENANAmBK0gPGrpQ
}266Please respect copyright.PENANAmmdJ30eOJf
if (central) {266Please respect copyright.PENANAWTYZiV38Ng
answer++;266Please respect copyright.PENANA4JP9DjFrBy
}266Please respect copyright.PENANAZQYp8P6dff
return 2 * answer;266Please respect copyright.PENANA4rKfoZwXec
}266Please respect copyright.PENANAvQ2eFuEub8
}
266Please respect copyright.PENANAasC53cqzBC