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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.
1using System;
2using System.Collections.Generic;
3
4public class Solution {
5 public int LongestPalindrome(string[] words) {
6 var dict = new Dictionary<string, int>();
7 int length = 0;
8 bool middleUsed = false;
9
10 foreach (var word in words) {
11 if (!dict.ContainsKey(word)) {
12 dict[word] = 0;
13 }
14 dict[word]++;
15 }
16
17 foreach (var kv in dict) {
18 string word = kv.Key;
19 int count = kv.Value;
20 string revWord = new string(new char[] { word[1], word[0] });
21
22 if (word == revWord) { // Symmetric word
23 length += (count / 2) * 4;
24 if (count % 2 == 1 && !middleUsed) {
25 length += 2;
26 middleUsed = true;
27 }
28 } else if (dict.ContainsKey(revWord)) {
29 int pairs = Math.Min(count, dict[revWord]);
30 length += pairs * 4;
31 dict[revWord] = 0; // Mark reverse used
32 }
33 }
34 return length;
35 }
36
37 public static void Main(string[] args) {
38 Solution sol = new Solution();
39 string[] words = new string[] { "lc", "cl", "gg" };
40 Console.WriteLine(sol.LongestPalindrome(words));
41 }
42}
43The C# solution uses a Dictionary to manage word counts and efficiently checks for reverse pairs and self-reversible words for palindrome creation. It optimizes palindrome construction by pairing and using extra 'odd' occurrences of mirrored symmetrical words.
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.