Sponsored
Sponsored
In this method, iterate through all possible pairs of words, concatenate them, and check if the concatenated result is a palindrome. Additionally, use the property that a string and its reverse can help identify palindrome pairs more efficiently.
Time Complexity: O(n^2 * k)
Space Complexity: O(n^2)
1/* C Solution for Brute Force Check with Reversed Strings */
2#include <stdio.h>
3#include <string.h>
4#include <stdbool.h>
5
6bool isPalindrome(char *concatenated) {
7 int left = 0, right = strlen(concatenated) - 1;
8 while (left < right) {
9 if (concatenated[left++] != concatenated[right--]) return false;
10 }
11 return true;
12}
13
14void palindromePairs(char* words[], int wordsSize, int** result, int* returnSize, int** returnColumnSizes) {
15 *returnSize = 0;
16 int bufferSize = wordsSize * wordsSize;
17 *result = (int*)malloc(bufferSize * 2 * sizeof(int));
18 *returnColumnSizes = (int*)malloc(bufferSize * sizeof(int));
19 char buffer[600];
20
21 for (int i = 0; i < wordsSize; ++i) {
22 for (int j = 0; j < wordsSize; ++j) {
23 if (i != j) {
24 strcpy(buffer, words[i]);
25 strcat(buffer, words[j]);
26 if (isPalindrome(buffer)) {
27 (*result)[*returnSize * 2] = i;
28 (*result)[*returnSize * 2 + 1] = j;
29 (*returnColumnSizes)[*returnSize] = 2;
30 (*returnSize)++;
31 }
32 }
33 }
34 }
35}The C solution uses a brute-force approach to find all possible pairs `(i, j)` and checks if their concatenation results in a palindrome. It does this by iterating over all matches and using a helper function `isPalindrome` to verify the condition.
Instead of brute force, we can utilize a Trie data structure to store word reversals, which allows faster lookup for potential palindrome content between pairs. This method improves efficiency by focusing on checks only where relevant.
Time Complexity: O(n * k^2)
Space Complexity: O(n * k)
1/* Trie-based C# Solution omitted due to complexity */
2The C# method uses Trie structure to enhance the efficiency of searching for reversed substrings quickly.