This is a leetcode interview problem that requires a hash table with strings. Please refer to this link for more detail of the problem description.
Let me first show you the solution code.
class Solution: def checkAlmostEquivalent(self, word1: str, word2: str) -> bool: def default_val(): return list((0, 0)) count = defaultdict(default_val) for letter in word1: count[letter] += 1 for letter in word2: count[letter] += 1 for letter, freq in count.items(): if abs(freq - freq) > 3: return False return True
A hash table is necessary in order to keep track of the frequencies of the letter in the word. Note that I used a list of which size is 2 – the first index is frequencies for word1 and the second index is frequencies for word2.
Once you collected the frequencies, you just need to check if the frequency difference is greater than 3.