Beats 93.71%||43ms||Easy to understand||Simple code||HashMap||Counter
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n1. Counting each charac
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Counting each character using hashmap.\n2. Checking  if each count is divisible by the number of words in the given list hence can be distributed equally among the words.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n$$O(N)$$ N is the total number of characters in all words combined.\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n$$O(U)$$ where U is the number of unique characters in the string s. \n$$O(N)$$ : worst case, when all characters are unique, U is equal to N.\n\n# Code\n```\nclass Solution:\n    def makeEqual(self, words: List[str]) -> bool:\n        s="".join(words) #joining the words\n        d1=Counter(s) #counting char\n        l=len(words)\n        for i in d1:\n            if d1[i]%l!=0:#checking divisibility\n                return False\n        return True\n```
['Python3']