Simple (5-line) Python solution beats 100% || 5 line solution || beats 100%
\n Describe your first thoughts on how to solve this problem. \n\n# Approach: Simple and step by step approach as mentiones in problem description \nstep 1: cal
\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach: Simple and step by step approach as mentiones in problem description \nstep 1: calculate the product of the given range (left, right)\nrange(left, right+1) will give us a range object or the range which is converted into list by list() function and then to calculate the product of whole list we use prod() function \nso till here the 1st line is explained \n**prod(list(range(left, right+1)))** and converting the result into string with the help of str() function as we have to print our solution in string format and for later opertions\n\nstep 2: removing the trailing zeros form the end of the calculated product with the help of rstrip() method which removes the white space in case of no string is given to it (i.e. "hii   ".rstrip() will become "hii" if we give it any string/parameter it will remove all occurence till any other charater apper for example- "hii00000000000000".rstrip("0") will become "hii")\n\nstep 3: count of traling zeros of the string \nthe experssion goes like this **lenght of string with zeros - length of string without zeros**\n\nstep 4: as there is 2 way to output the result if the lenght of string (without zeros) is less then or equal to 10 the we will output it as it is string (without zeros)+"e"+ count of zeros\n\nstep 5: if the lenght of string after removing zeros is greater then > 10 then we have to print the it in 2 parts the starting 5 digits and the last 5 digits \nexample 123456789987654 consider this as our non zero string \nso we have to print the starting 5 digits which is 12345 and the last 5 digits if it which is 87654 in the given format \nstarting 5 digits + "..." + last 5 digits + "e" + count of zeros\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(1) \n- as we have not used any loops or recursion call so i think the TC will be O(1) **correct me if i\'m wrong** \n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1) \n- as for space complexity, we are not doing any recursion and using only 3 veriables to store our calculation so i think this solution uses constant space \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        summ = str(prod(list(range(left, right+1))))\n        su = summ.rstrip("0")\n        zeros = len(summ)-len(su)\n        if len(su)<=10: return (su+"e"+str(zeros))\n        else: return (su[:5]+"..."+su[-5:]+"e"+str(zeros))\n```
['Python3']