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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.
1#include <iostream>
2#include <unordered_map>
3using namespace std;
4
5int longestPalindrome(vector<string>& words) {
6 unordered_map<string, int> freq;
7 int length = 0, middleUsed = 0;
8
9 // Fill the map with word counts
10 for (const auto& word : words) {
11 freq[word]++;
12 }
13
14 for (auto& [word, count] : freq) {
15 string rev = string(1, word[1]) + word[0];
16 if (word == rev) { // Self-mirrored like "gg"
17 length += (count / 2) * 4;
18 if (count % 2 == 1 && !middleUsed) {
19 length += 2;
20 middleUsed = 1;
21 }
22 } else if (freq.find(rev) != freq.end()) {
23 int pairs = min(count, freq[rev]);
24 length += pairs * 4;
25 freq[rev] = 0; // Mark used
26 }
27 }
28
29 return length;
30}
31
32int main() {
33 vector<string> words = {"lc", "cl", "gg"};
34 cout << longestPalindrome(words) << endl;
35 return 0;
36}
37Using C++, we organize the problem by hashing word counts in an unordered_map, leveraging reversals to find palindrome-forming pairs. Self-mirrored words use a specific mechanism to determine if a middle can be formed in a longer palindrome.
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.