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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 <stdio.h>
2#include <string.h>
3#include <stdlib.h>
4#define MAX_WORDS 100000
5
6typedef struct {
7 char word[3];
8 int count;
9} WordCount;
10
11// Comparator for qsort
12int cmp(const void *a, const void *b) {
13 return strcmp(((WordCount *)a)->word, ((WordCount *)b)->word);
14}
15
16int longestPalindrome(char **words, int wordsSize) {
17 WordCount map[MAX_WORDS];
18 int mapSize = 0;
19 int length = 0, middleUsed = 0;
20
21 // Fill the map with word counts
22 for (int i = 0; i < wordsSize; ++i) {
23 int j;
24 for (j = 0; j < mapSize; ++j) {
25 if (strcmp(map[j].word, words[i]) == 0) {
26 map[j].count++;
27 break;
28 }
29 }
30 if (j == mapSize) {
31 strcpy(map[mapSize].word, words[i]);
32 map[mapSize++].count = 1;
33 }
34 }
35
36 qsort(map, mapSize, sizeof(WordCount), cmp);
37
38 for (int i = 0; i < mapSize; ++i) {
39 char reversed[3];
40 reversed[0] = map[i].word[1];
41 reversed[1] = map[i].word[0];
42 reversed[2] = '\0';
43
44 if (strcmp(map[i].word, reversed) == 0) { // cases like "gg"
45 length += (map[i].count / 2) * 4;
46 if (map[i].count % 2 != 0 && !middleUsed) {
47 length += 2;
48 middleUsed = 1;
49 }
50 } else {
51 for (int j = i + 1; j < mapSize; ++j) {
52 if (strcmp(map[j].word, reversed) == 0) {
53 length += 4 * (map[i].count < map[j].count ? map[i].count : map[j].count);
54 map[j].count = 0; // Mark used
55 break;
56 }
57 }
58 }
59 }
60
61 return length;
62}
63
64int main() {
65 char *words[] = {"lc", "cl", "gg"};
66 int wordsSize = 3;
67 printf("%d", longestPalindrome(words, wordsSize));
68 return 0;
69}
70The C solution uses a manual structure to count word occurrences and then utilizes sorting and dual iterations to pair words and compute maximum palindrome length. Reversals are computed for each word, ensuring that all possible pairs contribute optimally to the 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.
1#include <vector>
using namespace std;
int longestPalindrome(vector<string>& words) {
int map[26][26] = {0};
int length = 0;
int symmetricCount = 0;
for (const string& word : words) {
int first = word[0] - 'a';
int second = word[1] - 'a';
if (map[second][first] > 0) {
length += 4;
map[second][first]--;
} else {
map[first][second]++;
}
}
for (int i = 0; i < 26; ++i) {
if (map[i][i] > 0) {
symmetricCount = 1;
break;
}
}
return length + symmetricCount * 2;
}
int main() {
vector<string> words = {"lc", "cl", "gg"};
cout << longestPalindrome(words) << endl;
return 0;
}
The C++ solution also uses a 2D array to track counts of specific first-second character combinations and their reverses, efficiently managing pair and middle character use. It organizes operations into element retrieval and comparison for palindromic construction.