Simple python3 solution | Hash Table with different by prefix mark
Complexity\n- Time complexity: O(n \cdot m)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(n)\n Add your space complexity here, e.g. O(n) 
# Complexity\n- Time complexity: $$O(n \\cdot m)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n``` python3 []\nclass Solution:\n    def minDeletionSize(self, strs: List[str]) -> int:\n        n = len(strs)\n        m = len(strs[0])\n\n        different_by_prefix = set()\n        result = 0\n        for j in range(m):\n            new_different_by_index = set()\n            for i in range(n - 1):\n                if i in different_by_prefix:\n                    continue\n                a = strs[i][j]\n                b = strs[i + 1][j]\n                if a > b:\n                    result += 1\n                    break\n                elif a < b:\n                    new_different_by_index.add(i)\n            else:\n                different_by_prefix.update(new_different_by_index)\n        return result\n```
['Hash Table', 'Greedy', 'Python3']