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The first approach involves counting occurrences of each word and its reverse. If a word appears with its reverse, they can be directly paired to contribute to the palindrome length. Symmetrical words like 'gg' can be used optimally to form palindromes, with one potentially serving as a center if their count is odd.
Time Complexity: O(n log n) due to sorting.
Space Complexity: O(n) for storage of words and their counts.
1import java.util.HashMap;
2
3class Solution {
4 public int longestPalindrome(String[] words) {
5 HashMap<String, Integer> map = new HashMap<>();
6 int length = 0;
7 boolean midUsed = false;
8
9 for (String word : words) {
10 map.put(word, map.getOrDefault(word, 0) + 1);
11 }
12
13 for (String word : map.keySet()) {
14 int count = map.get(word);
15 String reverse = new StringBuilder(word).reverse().toString();
16
17 if (word.equals(reverse)) { // Symmetrical
18 length += (count / 2) * 4;
19 if (count % 2 == 1 && !midUsed) {
20 length += 2;
21 midUsed = true;
22 }
23 } else if (map.containsKey(reverse)) {
24 int minPairs = Math.min(count, map.get(reverse));
25 length += minPairs * 4;
26 map.put(reverse, 0); // Mark reverse used
27 }
28 }
29
30 return length;
31 }
32
33 public static void main(String[] args) {
34 Solution sol = new Solution();
35 String[] words = {"lc", "cl", "gg"};
36 System.out.println(sol.longestPalindrome(words));
37 }
38}
39In Java, the solution uses a HashMap to count occurrences of each word. It then checks for reverse pairs or uses words consisting of identical letters to maximize palindrome length.
This approach involves creating a hash map to quickly check for the existence of a word's reverse in the input, allowing us to efficiently pair words or use symmetric words optimally. We calculate pairs and handle middle contributions by taking account of unsused symmetric words.
Time Complexity: O(n) with a constant factor given by alphabets.
Space Complexity: O(1) as the 26x26 map is constant in size.
This C solution employs a two-dimensional array approach (map) to store the frequency of character pairs. It counts how many of each type of word pair exist and checks for their reversal, adding contributions to the length based on pair formation. It separately checks for symmetrical pairs to add potential middle characters.