
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 {232Please respect copyright.PENANAdaA24B0Abt
public int longestPalindrome(String[] words) {232Please respect copyright.PENANAH1Tob86Kyz
HashMap<String, Integer> count = new HashMap<String, Integer>();232Please respect copyright.PENANAAVJzActCNY
// Count the number of occurrences of each word using a hashmap232Please respect copyright.PENANAtXsUXOwxPH
for (String word : words) {232Please respect copyright.PENANAsXGUPZR5uJ
Integer countOfTheWord = count.get(word);232Please respect copyright.PENANAU8u0cYZSuL
if (countOfTheWord == null) {232Please respect copyright.PENANAYSA2qsMw70
count.put(word, 1);232Please respect copyright.PENANAQdJPnvcloK
} else {232Please respect copyright.PENANA3kukYCmP7i
count.put(word, countOfTheWord + 1);232Please respect copyright.PENANAk6il7aLq0k
}232Please respect copyright.PENANAN5ieREbsTn
}232Please respect copyright.PENANAoP649GPhzM
// 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 word232Please respect copyright.PENANAWFJCdpY6jv
int answer = 0;232Please respect copyright.PENANAep8PVzJ9iI
boolean central = false;232Please respect copyright.PENANA0cni8pRkI4
for (Map.Entry<String, Integer> entry : count.entrySet()) {232Please respect copyright.PENANAufGaiwcVBL
String word = entry.getKey();232Please respect copyright.PENANAqd3ncRbjNq
int countOfTheWord = entry.getValue();232Please respect copyright.PENANAW6VnXmFVUI
// if the word is a palindrome232Please respect copyright.PENANAuunlKktUjR
if (word.charAt(0) == word.charAt(1)) {232Please respect copyright.PENANAUeHGJCHlDk
if (countOfTheWord % 2 == 0) {232Please respect copyright.PENANA9h8IcV5FMe
answer += countOfTheWord;232Please respect copyright.PENANAq6FgILMF3u
} else {232Please respect copyright.PENANAzy577AaOAe
answer += countOfTheWord - 1;232Please respect copyright.PENANAOk6ICOIblT
central = true;232Please respect copyright.PENANAk6lpsugUJM
}232Please respect copyright.PENANAAuEaFPG8AX
// consider a pair of non-palindrome words such that one is the reverse of another232Please respect copyright.PENANAZrIDNVnB2P
} else if (word.charAt(0) < word.charAt(1)) {232Please respect copyright.PENANAVHApF8LtqS
String reversedWord = "" + word.charAt(1) + word.charAt(0);232Please respect copyright.PENANAHiwrRuwOrI
if (count.containsKey(reversedWord)) {232Please respect copyright.PENANAHSQ2iaf7NM
answer += 2 * Math.min(countOfTheWord, count.get(reversedWord));232Please respect copyright.PENANAYdlUTrFrRz
}232Please respect copyright.PENANAYVuSyUZexB
}232Please respect copyright.PENANAHySciWDsLR
}232Please respect copyright.PENANAPMIcthNudI
if (central) {232Please respect copyright.PENANAaVFSK1VmtO
answer++;232Please respect copyright.PENANAdiWdI7PvMr
}232Please respect copyright.PENANAdYIwBep9VT
return 2 * answer;232Please respect copyright.PENANAyrgcAtOtnF
}232Please respect copyright.PENANAFVvTVXjuiZ
}
2: A Two-Dimensional Array Approach
class Solution {232Please respect copyright.PENANA5eG7obzBr3
public int longestPalindrome(String[] words) {232Please respect copyright.PENANAn97hiSxNMj
final int alphabetSize = 26;232Please respect copyright.PENANAVJA0wrR7J6
int[][] count = new int[alphabetSize][alphabetSize];232Please respect copyright.PENANAW26OBPKeyw
// Count the number of occurrences of each word using a two-dimensional array. 232Please respect copyright.PENANAhFryjotYxA
for (String word : words) {232Please respect copyright.PENANAUQCPn6330K
count[word.charAt(0) - 'a'][word.charAt(1) - 'a']++;232Please respect copyright.PENANAdPwEA24F7C
}232Please respect copyright.PENANAKloiX7A6ea
// The answer will denote the number of words in the final string and the boolean variable central will denote whether we have a central word232Please respect copyright.PENANAlfmR2K6pM9
int answer = 0;232Please respect copyright.PENANA7P5el42BTl
boolean central = false;232Please respect copyright.PENANAkL6JmSClII
for (int i = 0; i < alphabetSize; i++) {232Please respect copyright.PENANAHU05HO0gZK
if (count[i][i] % 2 == 0) {232Please respect copyright.PENANAcN1OPDsisX
answer += count[i][i];232Please respect copyright.PENANA2V5pLmnEAS
} else {232Please respect copyright.PENANABD5eTqIdOJ
answer += count[i][i] - 1;232Please respect copyright.PENANAxPA2bdkC5f
central = true;232Please respect copyright.PENANAsJPTsCS9IM
}232Please respect copyright.PENANAfBYvzp9IPi
for (int j = i + 1; j < alphabetSize; j++) {232Please respect copyright.PENANAPor5zRZ8iR
answer += 2 * Math.min(count[i][j], count[j][i]);232Please respect copyright.PENANAdt0ur2RdvW
}232Please respect copyright.PENANAvYRdgKLA9y
}232Please respect copyright.PENANAVLKfxR3AbW
if (central) {232Please respect copyright.PENANApo9sxu5DCt
answer++;232Please respect copyright.PENANATJ7bp9jxPd
}232Please respect copyright.PENANAEDDzhD8RYy
return 2 * answer;232Please respect copyright.PENANAAW0aDwNOQ4
}232Please respect copyright.PENANAgn638WF4ZG
}
232Please respect copyright.PENANAK4OOgKhzpN