
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 {270Please respect copyright.PENANAaEOTiu48DG
public int longestPalindrome(String[] words) {270Please respect copyright.PENANAMbHkwjX40n
HashMap<String, Integer> count = new HashMap<String, Integer>();270Please respect copyright.PENANAM5TDr0pAiL
// Count the number of occurrences of each word using a hashmap270Please respect copyright.PENANAQxkfC4JUSV
for (String word : words) {270Please respect copyright.PENANAXET4q8t9Wc
Integer countOfTheWord = count.get(word);270Please respect copyright.PENANAY1lCz4wsc5
if (countOfTheWord == null) {270Please respect copyright.PENANAb2I1CNR2MQ
count.put(word, 1);270Please respect copyright.PENANALtSBJpV8rl
} else {270Please respect copyright.PENANAZPp7pKODJU
count.put(word, countOfTheWord + 1);270Please respect copyright.PENANABAs16iqWuL
}270Please respect copyright.PENANAFe3CqnwDcs
}270Please respect copyright.PENANAqywQLPKKyD
// 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 word270Please respect copyright.PENANAupUi1nBdq9
int answer = 0;270Please respect copyright.PENANAlStA1owYoQ
boolean central = false;270Please respect copyright.PENANA54kWYtFUw6
for (Map.Entry<String, Integer> entry : count.entrySet()) {270Please respect copyright.PENANAWAv9x9iv1I
String word = entry.getKey();270Please respect copyright.PENANA9tRxZyfhNi
int countOfTheWord = entry.getValue();270Please respect copyright.PENANAseGGDwNfyx
// if the word is a palindrome270Please respect copyright.PENANAqzNfjwVUNU
if (word.charAt(0) == word.charAt(1)) {270Please respect copyright.PENANAQnLWHPbZZI
if (countOfTheWord % 2 == 0) {270Please respect copyright.PENANALmi0Q27IXX
answer += countOfTheWord;270Please respect copyright.PENANAjN4BJppDJi
} else {270Please respect copyright.PENANAdL18kzhXNX
answer += countOfTheWord - 1;270Please respect copyright.PENANAFVeGjuoOcI
central = true;270Please respect copyright.PENANASxaHKJKYUo
}270Please respect copyright.PENANAMHEcTtahcb
// consider a pair of non-palindrome words such that one is the reverse of another270Please respect copyright.PENANAdN0gEU0XAA
} else if (word.charAt(0) < word.charAt(1)) {270Please respect copyright.PENANAdSErEur35F
String reversedWord = "" + word.charAt(1) + word.charAt(0);270Please respect copyright.PENANApKtIUZ7eMp
if (count.containsKey(reversedWord)) {270Please respect copyright.PENANAfGnmX3eAhE
answer += 2 * Math.min(countOfTheWord, count.get(reversedWord));270Please respect copyright.PENANATS25Sr8I1B
}270Please respect copyright.PENANADGjExFbdYv
}270Please respect copyright.PENANA8NFQO9VZ2Z
}270Please respect copyright.PENANACtFcgY7P4O
if (central) {270Please respect copyright.PENANAJcFgmBOX6o
answer++;270Please respect copyright.PENANA72YA52u6kg
}270Please respect copyright.PENANAGWM3en2TzS
return 2 * answer;270Please respect copyright.PENANAJHTifyvSBr
}270Please respect copyright.PENANA93bPwEMXzq
}
2: A Two-Dimensional Array Approach
class Solution {270Please respect copyright.PENANA7osZBzChIG
public int longestPalindrome(String[] words) {270Please respect copyright.PENANAPvFQRCDLcm
final int alphabetSize = 26;270Please respect copyright.PENANArztSUkQxGD
int[][] count = new int[alphabetSize][alphabetSize];270Please respect copyright.PENANADZGZSy3olj
// Count the number of occurrences of each word using a two-dimensional array. 270Please respect copyright.PENANALLR5lHMLDR
for (String word : words) {270Please respect copyright.PENANAgBJoSH2V3M
count[word.charAt(0) - 'a'][word.charAt(1) - 'a']++;270Please respect copyright.PENANAGLjqxS00qW
}270Please respect copyright.PENANA9GCfph0XCw
// The answer will denote the number of words in the final string and the boolean variable central will denote whether we have a central word270Please respect copyright.PENANA9mdNbaDqQm
int answer = 0;270Please respect copyright.PENANAGqMKpAPbHI
boolean central = false;270Please respect copyright.PENANATZmCJuZyIt
for (int i = 0; i < alphabetSize; i++) {270Please respect copyright.PENANAYphlt6ENAq
if (count[i][i] % 2 == 0) {270Please respect copyright.PENANANcj7pyu0dx
answer += count[i][i];270Please respect copyright.PENANApYDQPEjXeg
} else {270Please respect copyright.PENANARMUkParzSP
answer += count[i][i] - 1;270Please respect copyright.PENANA7u2oBGgfCk
central = true;270Please respect copyright.PENANAw4Z2GebLdB
}270Please respect copyright.PENANAyqPo0F0VvP
for (int j = i + 1; j < alphabetSize; j++) {270Please respect copyright.PENANAH4CIalunwR
answer += 2 * Math.min(count[i][j], count[j][i]);270Please respect copyright.PENANACaF57apalJ
}270Please respect copyright.PENANAFyhFZPyXXz
}270Please respect copyright.PENANAAnfu10wpv3
if (central) {270Please respect copyright.PENANAkuQkyLnHMQ
answer++;270Please respect copyright.PENANABY2XV6WiYi
}270Please respect copyright.PENANAzBgSoJaFEP
return 2 * answer;270Please respect copyright.PENANAxA3SCgH84Q
}270Please respect copyright.PENANAuPYGLtazPx
}
270Please respect copyright.PENANAwtvwQaFs2u