
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 {265Please respect copyright.PENANAaNjJ9NPXfv
public int longestPalindrome(String[] words) {265Please respect copyright.PENANAHs3UEctH3z
HashMap<String, Integer> count = new HashMap<String, Integer>();265Please respect copyright.PENANAfyvrf42CIQ
// Count the number of occurrences of each word using a hashmap265Please respect copyright.PENANAiAedWTKWpS
for (String word : words) {265Please respect copyright.PENANA7yiMpO8aTy
Integer countOfTheWord = count.get(word);265Please respect copyright.PENANAYLwlMBHYO1
if (countOfTheWord == null) {265Please respect copyright.PENANAX0PRzpZiHu
count.put(word, 1);265Please respect copyright.PENANAjokuvicuCa
} else {265Please respect copyright.PENANAsUAvsRzWK9
count.put(word, countOfTheWord + 1);265Please respect copyright.PENANAxQ0ulbMhXt
}265Please respect copyright.PENANAiWfwUGPkTn
}265Please respect copyright.PENANAv40SKaH500
// 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 word265Please respect copyright.PENANAtaElZX9QVS
int answer = 0;265Please respect copyright.PENANAkbUuSHfMZN
boolean central = false;265Please respect copyright.PENANA9SEOKHWM47
for (Map.Entry<String, Integer> entry : count.entrySet()) {265Please respect copyright.PENANAxj1UO5Aai8
String word = entry.getKey();265Please respect copyright.PENANAr2ry9wrkib
int countOfTheWord = entry.getValue();265Please respect copyright.PENANAGoH6j6ozBK
// if the word is a palindrome265Please respect copyright.PENANARnjFdIOdwp
if (word.charAt(0) == word.charAt(1)) {265Please respect copyright.PENANAyedFToBpCM
if (countOfTheWord % 2 == 0) {265Please respect copyright.PENANAMVtjWnZAdV
answer += countOfTheWord;265Please respect copyright.PENANABfJlycjnPd
} else {265Please respect copyright.PENANAe3bweFxqRj
answer += countOfTheWord - 1;265Please respect copyright.PENANAQWsjZrqUpM
central = true;265Please respect copyright.PENANAkmA4RvaqZS
}265Please respect copyright.PENANANsxu68bjv7
// consider a pair of non-palindrome words such that one is the reverse of another265Please respect copyright.PENANAsI1tCuRFzv
} else if (word.charAt(0) < word.charAt(1)) {265Please respect copyright.PENANAIei9nbCn5i
String reversedWord = "" + word.charAt(1) + word.charAt(0);265Please respect copyright.PENANA74JFFasP1Z
if (count.containsKey(reversedWord)) {265Please respect copyright.PENANArCty3tcZFg
answer += 2 * Math.min(countOfTheWord, count.get(reversedWord));265Please respect copyright.PENANAHXhxRCm5bY
}265Please respect copyright.PENANAKddutEzwYu
}265Please respect copyright.PENANAaFsaFeBZ3c
}265Please respect copyright.PENANADV3CJX7NsL
if (central) {265Please respect copyright.PENANACYIvzO07NN
answer++;265Please respect copyright.PENANANzVdx3P13e
}265Please respect copyright.PENANAH6mi9jCyix
return 2 * answer;265Please respect copyright.PENANAqmuHHYXHAg
}265Please respect copyright.PENANA2V9YHaG1ux
}
2: A Two-Dimensional Array Approach
class Solution {265Please respect copyright.PENANAc7jkKqv119
public int longestPalindrome(String[] words) {265Please respect copyright.PENANAGr9camqOCT
final int alphabetSize = 26;265Please respect copyright.PENANAr8naEj2YU4
int[][] count = new int[alphabetSize][alphabetSize];265Please respect copyright.PENANA0lvDdsdrMR
// Count the number of occurrences of each word using a two-dimensional array. 265Please respect copyright.PENANAe3QajgUSCO
for (String word : words) {265Please respect copyright.PENANA5UK046MY8S
count[word.charAt(0) - 'a'][word.charAt(1) - 'a']++;265Please respect copyright.PENANAqPGMeg8doa
}265Please respect copyright.PENANAY0Rykv1d8J
// The answer will denote the number of words in the final string and the boolean variable central will denote whether we have a central word265Please respect copyright.PENANA47hsbgE5du
int answer = 0;265Please respect copyright.PENANAJZv6VFqSk5
boolean central = false;265Please respect copyright.PENANAu23gj1GqW0
for (int i = 0; i < alphabetSize; i++) {265Please respect copyright.PENANAZYT0BOu2it
if (count[i][i] % 2 == 0) {265Please respect copyright.PENANAQju1FeWAFU
answer += count[i][i];265Please respect copyright.PENANASBUvPN1Onk
} else {265Please respect copyright.PENANA5lmUN6Kror
answer += count[i][i] - 1;265Please respect copyright.PENANAwAxC6c0lG7
central = true;265Please respect copyright.PENANAWLuIQfRrp9
}265Please respect copyright.PENANAYmabBaoYGW
for (int j = i + 1; j < alphabetSize; j++) {265Please respect copyright.PENANAXONBFOEyTP
answer += 2 * Math.min(count[i][j], count[j][i]);265Please respect copyright.PENANAe5uXRshYvJ
}265Please respect copyright.PENANAwYRnFd95dD
}265Please respect copyright.PENANAnCB97axODy
if (central) {265Please respect copyright.PENANABuRwmY244u
answer++;265Please respect copyright.PENANAIul9sC08Cc
}265Please respect copyright.PENANA0e2R5hEA37
return 2 * answer;265Please respect copyright.PENANAwz4Y2v24vP
}265Please respect copyright.PENANAa7SAmQq2qz
}
265Please respect copyright.PENANAtUqyGcZKQQ