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Problem Solution Two strings are anagrams if the frequencies of appearing alphabets are the same. Make a dictionary, where the keys are tuples of alphabet(a-z) frequencies. The values are the strings in the given list. When I initiated an dictionary and used append function with a new key, it gave a KeyError. To avoid this, use defaultdict function and pass list as parameter. Then a defaultdict ..

Problem Solution Start from the center of the string, and expand to left and right until the characters at those positions are not the same. class Solution(object): def longestPalindrome(self, s): """ :type s: str :rtype: str """ if s is None or len(s) < 1: return "" start = 0 end = 0 for i in range(len(s)): len1 = self.expandCenter(s, i, i) len2 = self.expandCenter(s, i, i+1) length = max(len1,..

Problem Solution Make a dictionary of all appearing alphabets and their frequencies. Make a set of unique frequencies. Go through the dictionary and check if the frequency for this character already exists in the set. If it is, decrement the frequency until it is unique or reaches zero. Increment the counter at the same time. Add the frequency to the set. My code exceeded the time limit at first..