Algorithms in Depth
Intuition\nTo create the longest palindrome, we need to find pairs of words that can be concatenated to form palindromes, as well as any single words that have 
# Intuition\nTo create the longest palindrome, we need to find pairs of words that can be concatenated to form palindromes, as well as any single words that have the same letters. If we can find a single word with an odd number of occurences, we can use it as the center of the palindrome. Otherwise, we will have to use pairs of words to form the palindrome.\n\n# Approach\n1. Use a counter to count the number of occurences of each word in the input list words.\n2. Initialize variables `res` and `odd` to 0.\n3. Iterate through the words and their counts in the counter. If a word has the same first and last letters, we can use it to form palindromes. In this case, we add half of the count of the word (rounded down) to res and set odd to 1 if the count is odd.\n4. If a word does not have the same first and last letters, check if the reverse of the word is in the counter. If it is, we can use the word and its reverse to form palindromes. In this case, we add the minimum of the count of the word and the count of its reverse to res.\n5. Return `(res + odd) * 2`, which is the maximum number of pairs of words that can be concatenated to form a palindrome, plus 1 odd pair if it exists (to account for the single pair at the center of the palindrome).\n\n# Complexity\nLet $n$ be the length of the input array `words`.\n\n- Time complexity: $$O(n)$$\nWe traverse the list of words words once and use a linear-time operation (Counter) to count the occurrences of each word.\n\n- Space complexity: $$O(n)$$\nWe use a dictionary (Counter) to store the counts of the words in words, so the space complexity is linear in the size of the input.\n\n# Code\n\n```python []\nclass Solution:\n    def longestPalindrome(self, words: List[str]) -> int:\n        pairs, res, odd = Counter(words), 0, 0\n        for key, freq in pairs.items():\n            if key[0] == key[-1]:\n                res += freq & ~1\n                odd = freq % 2 or odd\n            elif key[::-1] in pairs:\n                res += min(freq, pairs[key[::-1]])\n        return (res + odd) * 2\n```
['Hash Table', 'Counting', 'Python3']