Find the anagram substring

anagram_substring.py

from collections import Counter, defaultdict


class Solution(object):

    def solve(self, s1, s2):

        s1 = list(s1)
        s2 = list(s2)

        map1 = Counter(s1)
        n = len(s1)
        map2 = defaultdict(int)

        for i in range(0, n):
            map2[s2[i]] = map2[s2[i]] + 1

        are_anagrams = self.check_anagrams(map1, map2)

        if are_anagrams:
            return "".join(s2)

        for i, ch in enumerate(s2):

            if i < n:
                continue

            are_anagrams = self.check_anagrams(map1, map2)

            if are_anagrams:
                return "".join(s2[i-n:i])

            # remove what's out of the window
            index_to_remove = i - n
            character_to_remove = s2[index_to_remove]
            map2[character_to_remove] = map2[character_to_remove] - 1
            if map2[character_to_remove] == 0:
                map2.pop(character_to_remove)

            # insert the new character
            map2[ch] = map2[ch] + 1

        return None

    def check_anagrams(self, map1, map2):

        n = len(map1)
        m = len(map2)

        if n != m:
            return False

        for k, v in map1.items():
            if v != map2[k]:
                return False

        return True


if __name__ == "__main__":
    s1 = "hyzx"
    s2 = "abcdegfhxzypop"

    ans = Solution().solve(s1, s2)

    print(ans)