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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.
1from collections import Counter
2
3class Solution:
4 def longestPalindrome(self, words):
5 count = Counter(words)
6 length = 0
7 middle_used = False
8
9 for word, cnt in count.items():
10 rev = word[::-1]
11 if word == rev: # Symmetric word like 'gg'
12 length += (cnt // 2) * 4
13 if cnt % 2 == 1 and not middle_used:
14 length += 2
15 middle_used = True
16 elif rev in count:
17 pairs = min(cnt, count[rev])
18 length += pairs * 4
19 count[rev] = 0 # Mark reverse used
20
21 return length
22
23# Example usage:
24sol = Solution()
25words = ["lc", "cl", "gg"]
26print(sol.longestPalindrome(words))
27The Python solution leverages the `Counter` class for counting words efficiently. It checks each word and its reverse to determine the optimal contributions to the possible palindrome formation. Special attention is given to self-reversing words like 'gg' for potential middle use.
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.