[Python3, Java, C++] Counting Mirror Words O(n)
\nExplanation:\n\n2 letter words can be of 2 types:\n1. Where both letters are same\n2. Where both letters are different\n\nBased on the above information:\n If
\n***Explanation:***\n\n2 letter words can be of 2 types:\n1. Where both letters are same\n2. Where both letters are different\n\nBased on the above information:\n* If we are able to find the mirror of a word, `ans += 4`\n* The variable `unpaired` is used to store the number of unpaired words with both letters same.\n* Unpaired here means a word that has not found its mirror word.\n* At the end if unpaired same letter words are > 0, we can use one of them as the center of the palindromic string. \n\n**Solution 1: With HashMap**\n<iframe src="https://leetcode.com/playground/2BoNSYTg/shared" frameBorder="0" width="500" height="570"></iframe>\n\n**Solution 2: Without HashMap**\n<iframe src="https://leetcode.com/playground/LH7fNiWY/shared" frameBorder="0" width="500" height="400"></iframe>
['C', 'Java', 'Python3']