Easy to understand Python solution for Abbreviating the Product of a Range ,
Intuition\n Describe your first thoughts on how to solve this problem. \nGood challenging problem \n\n# Approach\n Describe your approach to solving the problem
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nGood challenging problem \n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nnormal method , solution is easy to understand just used simple string slicing methods to solve it \n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->O(right-left+logN) \nN = value of product of all numbers between left and right\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->O(1)\n\n# Code\n```\nclass Solution(object):\n    def abbreviateProduct(self, left, right):\n        """\n        :type left: int\n        :type right: int\n        :rtype: str\n        """\n        prod=1\n        for i in range(left,right+1):\n            prod*=i\n        str_prod = str(prod)\n        count_digit=0\n        count_zeros=0\n        str5=str_prod\n        while (str5[-1] == \'0\'):\n            str5 = str5[:-1]\n            count_zeros = count_zeros+1\n        count_digit = len(str5)\n        res=""\n        if count_digit<=10:\n            str2="e"+str(count_zeros)\n            if(count_zeros==0):\n                res = str_prod+str2\n            else:\n                res = str_prod[:-count_zeros]\n                res = res+str2\n            #print(res)\n        else:\n            str4 = str_prod[:-count_zeros]\n            str3 = "e"+str(count_zeros)\n            pre = str5[:5]\n            suf = str5[len(str5)-5:]\n            res = pre+"..."+suf+str3\n            #print("res = ",res)\n        return res\n```
['Python']