
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 {287Please respect copyright.PENANAS3ojRxJeEs
public int longestPalindrome(String[] words) {287Please respect copyright.PENANAUU6pBKNwbY
HashMap<String, Integer> count = new HashMap<String, Integer>();287Please respect copyright.PENANAkUSDASs0Jd
// Count the number of occurrences of each word using a hashmap287Please respect copyright.PENANAQjhja8zepK
for (String word : words) {287Please respect copyright.PENANAydMdflCJr6
Integer countOfTheWord = count.get(word);287Please respect copyright.PENANAeJ1YhNRAHe
if (countOfTheWord == null) {287Please respect copyright.PENANAlGrCzs1dIi
count.put(word, 1);287Please respect copyright.PENANAM5ynlzEd67
} else {287Please respect copyright.PENANArO07UpgfGY
count.put(word, countOfTheWord + 1);287Please respect copyright.PENANAJLt0Cmv4sN
}287Please respect copyright.PENANAcfu9scOzW2
}287Please respect copyright.PENANAedgd5GMaIx
// 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 word287Please respect copyright.PENANA32IcUml4JN
int answer = 0;287Please respect copyright.PENANAtilOeO7Cms
boolean central = false;287Please respect copyright.PENANAgWK2zbXW62
for (Map.Entry<String, Integer> entry : count.entrySet()) {287Please respect copyright.PENANAWjiDm8ivBm
String word = entry.getKey();287Please respect copyright.PENANArfdwqPYK7b
int countOfTheWord = entry.getValue();287Please respect copyright.PENANA1ipgsqHAXZ
// if the word is a palindrome287Please respect copyright.PENANAXw3YdglXYz
if (word.charAt(0) == word.charAt(1)) {287Please respect copyright.PENANAy9IpJEf1a2
if (countOfTheWord % 2 == 0) {287Please respect copyright.PENANASqe8Ro8geK
answer += countOfTheWord;287Please respect copyright.PENANAK9iwSzb2DZ
} else {287Please respect copyright.PENANAjf67i4a7zt
answer += countOfTheWord - 1;287Please respect copyright.PENANA682wgyrWkC
central = true;287Please respect copyright.PENANA5NMhgn1Qfd
}287Please respect copyright.PENANAXrctDSDU0a
// consider a pair of non-palindrome words such that one is the reverse of another287Please respect copyright.PENANAnrPtfInFEB
} else if (word.charAt(0) < word.charAt(1)) {287Please respect copyright.PENANANW7AdhTLFm
String reversedWord = "" + word.charAt(1) + word.charAt(0);287Please respect copyright.PENANAiYh3yidAPs
if (count.containsKey(reversedWord)) {287Please respect copyright.PENANACan7hXYxFc
answer += 2 * Math.min(countOfTheWord, count.get(reversedWord));287Please respect copyright.PENANAYGVSYZsd66
}287Please respect copyright.PENANAO5HFv4oN6I
}287Please respect copyright.PENANAU9l55iTUvS
}287Please respect copyright.PENANAQ5f1oUgQoh
if (central) {287Please respect copyright.PENANA5moV5Awwu9
answer++;287Please respect copyright.PENANAMj7FSTLq71
}287Please respect copyright.PENANAKrOkjRBCel
return 2 * answer;287Please respect copyright.PENANAqPt19oExig
}287Please respect copyright.PENANAjdLgkXMLS3
}
2: A Two-Dimensional Array Approach
class Solution {287Please respect copyright.PENANAn5Z5fcl2Zq
public int longestPalindrome(String[] words) {287Please respect copyright.PENANARMOyHfOPKN
final int alphabetSize = 26;287Please respect copyright.PENANAZSai1uBABT
int[][] count = new int[alphabetSize][alphabetSize];287Please respect copyright.PENANAGHvOYBIpDw
// Count the number of occurrences of each word using a two-dimensional array. 287Please respect copyright.PENANA3BQgsfhroE
for (String word : words) {287Please respect copyright.PENANAIIpgtSu2uZ
count[word.charAt(0) - 'a'][word.charAt(1) - 'a']++;287Please respect copyright.PENANA4lHz5bVGh3
}287Please respect copyright.PENANAngZ0YCQKq5
// The answer will denote the number of words in the final string and the boolean variable central will denote whether we have a central word287Please respect copyright.PENANA9AKad4kmYo
int answer = 0;287Please respect copyright.PENANAC29K4Q3omS
boolean central = false;287Please respect copyright.PENANAXnPdkFLzsR
for (int i = 0; i < alphabetSize; i++) {287Please respect copyright.PENANAEtrvoVbKrq
if (count[i][i] % 2 == 0) {287Please respect copyright.PENANACYpxuMJ44E
answer += count[i][i];287Please respect copyright.PENANAIWFEGFr6Ep
} else {287Please respect copyright.PENANApllO5Er3Vk
answer += count[i][i] - 1;287Please respect copyright.PENANAoowA7kGrxj
central = true;287Please respect copyright.PENANA9CkOrm4KYq
}287Please respect copyright.PENANA7idcucAc6l
for (int j = i + 1; j < alphabetSize; j++) {287Please respect copyright.PENANANT12kmEyZ0
answer += 2 * Math.min(count[i][j], count[j][i]);287Please respect copyright.PENANAAOH1IVJenf
}287Please respect copyright.PENANAfQ0tREx4eN
}287Please respect copyright.PENANA4bx41gJqdx
if (central) {287Please respect copyright.PENANA5dPHjVxRRm
answer++;287Please respect copyright.PENANAFgzgaIh2Qf
}287Please respect copyright.PENANAkM19shsm78
return 2 * answer;287Please respect copyright.PENANAjuDgV1XrzN
}287Please respect copyright.PENANAJx8vgSDrY7
}
287Please respect copyright.PENANA53kjCPKd22