
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 {286Please respect copyright.PENANAt60wu7gNYS
public int longestPalindrome(String[] words) {286Please respect copyright.PENANAIIDBhq77mL
HashMap<String, Integer> count = new HashMap<String, Integer>();286Please respect copyright.PENANAQfOzqbkzvi
// Count the number of occurrences of each word using a hashmap286Please respect copyright.PENANAEuhM8spuQ7
for (String word : words) {286Please respect copyright.PENANAM7hJJJNQDu
Integer countOfTheWord = count.get(word);286Please respect copyright.PENANA1mFoeW3fXr
if (countOfTheWord == null) {286Please respect copyright.PENANA1xSN3dqbpX
count.put(word, 1);286Please respect copyright.PENANAR00J5xSWQG
} else {286Please respect copyright.PENANAT6kvNXrd6z
count.put(word, countOfTheWord + 1);286Please respect copyright.PENANAs75snp6Ykc
}286Please respect copyright.PENANALNYZo5adcP
}286Please respect copyright.PENANAihTGPr8bXP
// 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 word286Please respect copyright.PENANA2oNuRggv6C
int answer = 0;286Please respect copyright.PENANAxmfIFcAKWh
boolean central = false;286Please respect copyright.PENANAsZVdpCtnq5
for (Map.Entry<String, Integer> entry : count.entrySet()) {286Please respect copyright.PENANAJaxE6YgEn0
String word = entry.getKey();286Please respect copyright.PENANAoxiF46Lr2n
int countOfTheWord = entry.getValue();286Please respect copyright.PENANASGp2bJ3Fli
// if the word is a palindrome286Please respect copyright.PENANA4qFZkq8kaw
if (word.charAt(0) == word.charAt(1)) {286Please respect copyright.PENANANKB5KVDXsV
if (countOfTheWord % 2 == 0) {286Please respect copyright.PENANAP1xLdTIDqG
answer += countOfTheWord;286Please respect copyright.PENANAPc5oPXLkMV
} else {286Please respect copyright.PENANATTp2MyaReR
answer += countOfTheWord - 1;286Please respect copyright.PENANApMj7nPTcyx
central = true;286Please respect copyright.PENANAz0SsB6uTLb
}286Please respect copyright.PENANAL3kgECwVMY
// consider a pair of non-palindrome words such that one is the reverse of another286Please respect copyright.PENANARa17hKmlAY
} else if (word.charAt(0) < word.charAt(1)) {286Please respect copyright.PENANA7QPL2B6JBD
String reversedWord = "" + word.charAt(1) + word.charAt(0);286Please respect copyright.PENANAks8QmkL17A
if (count.containsKey(reversedWord)) {286Please respect copyright.PENANAmBa5ir1GW3
answer += 2 * Math.min(countOfTheWord, count.get(reversedWord));286Please respect copyright.PENANAeJhCZcvjUA
}286Please respect copyright.PENANAbEL1pzLdYh
}286Please respect copyright.PENANAqO9pSugfTK
}286Please respect copyright.PENANAMoNjVOGBU8
if (central) {286Please respect copyright.PENANAd5my2srj1I
answer++;286Please respect copyright.PENANAEOJyFK7Afo
}286Please respect copyright.PENANAY9W4EwZQtd
return 2 * answer;286Please respect copyright.PENANArfD55PFURM
}286Please respect copyright.PENANAPObe0LXWUV
}
2: A Two-Dimensional Array Approach
class Solution {286Please respect copyright.PENANAWEdiXMZd9v
public int longestPalindrome(String[] words) {286Please respect copyright.PENANA5klHe2wHj2
final int alphabetSize = 26;286Please respect copyright.PENANAj6OOGPoYNt
int[][] count = new int[alphabetSize][alphabetSize];286Please respect copyright.PENANA3ec9Z7Sre4
// Count the number of occurrences of each word using a two-dimensional array. 286Please respect copyright.PENANAJc05hER67S
for (String word : words) {286Please respect copyright.PENANAsm7jiPF8E7
count[word.charAt(0) - 'a'][word.charAt(1) - 'a']++;286Please respect copyright.PENANAlzYt1cmYPU
}286Please respect copyright.PENANARxoBDSdxBa
// The answer will denote the number of words in the final string and the boolean variable central will denote whether we have a central word286Please respect copyright.PENANAj2yh5iC7Wx
int answer = 0;286Please respect copyright.PENANAg9paVCQQ9I
boolean central = false;286Please respect copyright.PENANAKOmD6LY1Mc
for (int i = 0; i < alphabetSize; i++) {286Please respect copyright.PENANARFWyulqEyG
if (count[i][i] % 2 == 0) {286Please respect copyright.PENANAw6JLWvQNv0
answer += count[i][i];286Please respect copyright.PENANAguUPLI0JJf
} else {286Please respect copyright.PENANARnjcABmyn5
answer += count[i][i] - 1;286Please respect copyright.PENANAIXRbKIYr0s
central = true;286Please respect copyright.PENANAyaDNLjsrWb
}286Please respect copyright.PENANANHUMmwWPiE
for (int j = i + 1; j < alphabetSize; j++) {286Please respect copyright.PENANAL1Es4H0Ymq
answer += 2 * Math.min(count[i][j], count[j][i]);286Please respect copyright.PENANAgg9eBOm8J8
}286Please respect copyright.PENANAbbjgpqZ2R1
}286Please respect copyright.PENANAPyu4yO41ua
if (central) {286Please respect copyright.PENANAZZOhXClkiE
answer++;286Please respect copyright.PENANA12Ctj0vPP3
}286Please respect copyright.PENANAiTwqwF2Ef9
return 2 * answer;286Please respect copyright.PENANA6OaHXEFjTW
}286Please respect copyright.PENANA0faeUcGARG
}
286Please respect copyright.PENANATD73bW6ojs