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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
public class Solution {
public int LongestPalindrome(string[] words) {
int[,] map = new int[26, 26];
int length = 0;
bool hasMiddle = false;
foreach (var word in words) {
int a = word[0] - 'a';
int b = word[1] - 'a';
if (map[b, a] > 0) {
length += 4;
map[b, a]--;
} else {
map[a, b]++;
}
}
for (int i = 0; i < 26; ++i) {
if (map[i, i] > 0) {
hasMiddle = true;
break;
}
}
return length + (hasMiddle ? 2 : 0);
}
public static void Main(string[] args) {
Solution sol = new Solution();
string[] words = { "lc", "cl", "gg" };
Console.WriteLine(sol.LongestPalindrome(words));
}
}
C#'s solution embraces array manipulation for fast pair matching and reverse existence checking, facilitating efficient palindrome construction via combined indexed element-wise operations on constants size arrays.