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 {73Please respect copyright.PENANANjBgNDwj7l
public int longestPalindrome(String[] words) {73Please respect copyright.PENANASlDach2w27
HashMap<String, Integer> count = new HashMap<String, Integer>();73Please respect copyright.PENANA1G1qfY9FEM
// Count the number of occurrences of each word using a hashmap73Please respect copyright.PENANA33QxDdWtuR
for (String word : words) {73Please respect copyright.PENANA3cgfzo2YiR
Integer countOfTheWord = count.get(word);73Please respect copyright.PENANA80VjG0DvYX
if (countOfTheWord == null) {73Please respect copyright.PENANAGdIEmNJhRN
count.put(word, 1);73Please respect copyright.PENANAenKxNuKoMn
} else {73Please respect copyright.PENANAdqxi6mD0NR
count.put(word, countOfTheWord + 1);73Please respect copyright.PENANAa9hKs8FTaO
}73Please respect copyright.PENANA1kqb6RI3MN
}73Please respect copyright.PENANAT3BWEesYcq
// 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 word73Please respect copyright.PENANAtRIsWve6iR
int answer = 0;73Please respect copyright.PENANAUO5OZnHbzk
boolean central = false;73Please respect copyright.PENANAXRb6KPHHbZ
for (Map.Entry<String, Integer> entry : count.entrySet()) {73Please respect copyright.PENANA6VOp5wr2bK
String word = entry.getKey();73Please respect copyright.PENANAoT7MqsLnGH
int countOfTheWord = entry.getValue();73Please respect copyright.PENANAveFlz7PEZ1
// if the word is a palindrome73Please respect copyright.PENANAvAXRArn6uw
if (word.charAt(0) == word.charAt(1)) {73Please respect copyright.PENANABXZP3VdAg0
if (countOfTheWord % 2 == 0) {73Please respect copyright.PENANAX7BYtBg3aQ
answer += countOfTheWord;73Please respect copyright.PENANA62B7ySLcCV
} else {73Please respect copyright.PENANAA0ZadlC2gu
answer += countOfTheWord - 1;73Please respect copyright.PENANAlpXqSRy5p3
central = true;73Please respect copyright.PENANAdvjO5gzUkV
}73Please respect copyright.PENANAcK2QaXkbMA
// consider a pair of non-palindrome words such that one is the reverse of another73Please respect copyright.PENANAQEmwvAxNUk
} else if (word.charAt(0) < word.charAt(1)) {73Please respect copyright.PENANAY7ESwnttYM
String reversedWord = "" + word.charAt(1) + word.charAt(0);73Please respect copyright.PENANATpIPMEgvkU
if (count.containsKey(reversedWord)) {73Please respect copyright.PENANAGL5bNI9pNk
answer += 2 * Math.min(countOfTheWord, count.get(reversedWord));73Please respect copyright.PENANAcTfoNHQOyR
}73Please respect copyright.PENANA6WCr2E5M4A
}73Please respect copyright.PENANAQ1iAzg85hy
}73Please respect copyright.PENANAxEVsLyzJs3
if (central) {73Please respect copyright.PENANAcA4zo1zGvw
answer++;73Please respect copyright.PENANAp4TAUHZMsp
}73Please respect copyright.PENANAVadQAQCPpn
return 2 * answer;73Please respect copyright.PENANAdZCODtYOUG
}73Please respect copyright.PENANAA2T8r37bKs
}
2: A Two-Dimensional Array Approach
class Solution {73Please respect copyright.PENANAxqZnTKv1h9
public int longestPalindrome(String[] words) {73Please respect copyright.PENANAZpnABTdTB7
final int alphabetSize = 26;73Please respect copyright.PENANA20SKaTRInc
int[][] count = new int[alphabetSize][alphabetSize];73Please respect copyright.PENANA5tGvbZuS6n
// Count the number of occurrences of each word using a two-dimensional array. 73Please respect copyright.PENANAsyReImfCns
for (String word : words) {73Please respect copyright.PENANAoBpz0ajuma
count[word.charAt(0) - 'a'][word.charAt(1) - 'a']++;73Please respect copyright.PENANAEkRyNynPL4
}73Please respect copyright.PENANAwPoebcjEEA
// The answer will denote the number of words in the final string and the boolean variable central will denote whether we have a central word73Please respect copyright.PENANAdu5PSYKNKP
int answer = 0;73Please respect copyright.PENANAed79UFhOjL
boolean central = false;73Please respect copyright.PENANACjqYqPtJIK
for (int i = 0; i < alphabetSize; i++) {73Please respect copyright.PENANA5WCrsokRbv
if (count[i][i] % 2 == 0) {73Please respect copyright.PENANAUz0f3350ND
answer += count[i][i];73Please respect copyright.PENANASsJIPk3kom
} else {73Please respect copyright.PENANASzt8ICkQy0
answer += count[i][i] - 1;73Please respect copyright.PENANA5GAanLAwzO
central = true;73Please respect copyright.PENANAtetAhskQzU
}73Please respect copyright.PENANAuRqumksHyE
for (int j = i + 1; j < alphabetSize; j++) {73Please respect copyright.PENANADfwXaZAEAb
answer += 2 * Math.min(count[i][j], count[j][i]);73Please respect copyright.PENANA9Il8kvoXpV
}73Please respect copyright.PENANAx71PIPkPN5
}73Please respect copyright.PENANA1IeepvDzXC
if (central) {73Please respect copyright.PENANAQUAi9iTTyb
answer++;73Please respect copyright.PENANALwJJPGm96E
}73Please respect copyright.PENANATsbdbVKiUn
return 2 * answer;73Please respect copyright.PENANAeCk63IiNTH
}73Please respect copyright.PENANATEiu5P8g9P
}
73Please respect copyright.PENANAv6CkovaOJU