🔥🔥EASY BEGINNER FRIENDLY SOLUTION USING FREQUENCY HASHMAP !! 🔥🔥
Complexity\n- Time complexity:\nO(N)\n\n- Space complexity:\nO(N)\n\n# Code\n\nfrom collections import Counter\nclass Solution:\n    def digitCount(self, num: s
# Complexity\n- Time complexity:\nO(N)\n\n- Space complexity:\nO(N)\n\n# Code\n```\nfrom collections import Counter\nclass Solution:\n    def digitCount(self, num: str) -> bool:\n        dic=Counter(num)\n        c=0\n        for i in range(len(num)):\n            if dic[str(i)]==int(num[i]):\n                c+=1\n                \n        if c==len(num):\n            return 1\n        return 0\n\n```
['Hash Table', 'Python3']