Understandable Python Solution with linear time complexity
Intuition\n Describe your first thoughts on how to solve this problem. \nThe intuition behind the above code is to efficiently calculate and represent the produ
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind the above code is to efficiently calculate and represent the product of all integers in the given range in a human-readable format. The product can be very large and contain many trailing zeros, which can make it difficult to read and understand. So this code aims to simplify the representation of the product by removing trailing zeros and truncating the number to the first 5 and last 5 digits, if it has more than 10 digits.\n\n# Approach\n1. Initialize a variable \'product\' to 1 and a variable \'trailing_zeros\' to 0.\n2. Use a for loop to iterate over all integers in the range [left, right]. For each integer in the range, multiply it with the current value of \'product\'\n3. While the last digit of product is zero, count number of trailing zeros and divide product by 10\n4. Convert product to a string, store the number of digits of product in a variable \'digits\'\n5. If the number of digits of the product is greater than 10, return the first 5 digits followed by \'...\' and the last 5 digits, followed by \'e\' and the number of trailing zeros.\n6. If the number of digits of the product is less than or equal to 10, return the original number followed by \'e\' and the number of trailing zeros.\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n    def abbreviateProduct(self, left: int, right: int) -> str:\n        product = 1\n        for i in range(left, right + 1):\n            product *= i\n        trailing_zeros = 0\n        while product % 10 == 0:\n            trailing_zeros += 1\n            product //= 10\n        product_str = str(product)\n        digits = len(product_str)\n        if digits > 10:\n            pre = product_str[:5]\n            suf = product_str[-5:]\n            return f"{pre}...{suf}e{trailing_zeros}"\n        else:\n            return f"{product_str}e{trailing_zeros}"\n```
['Python3']