Brute Force Approach Python3 (Accepted) (Commented)
\nclass Solution:\n    def abbreviateProduct(self, left: int, right: int) -> str:\n        ans = 1                                                              
```\nclass Solution:\n    def abbreviateProduct(self, left: int, right: int) -> str:\n        ans = 1                                                                  //Initialise the product with 1\n        while left <= right:                                               // Start the multiplying numbers in\n            if left == right:                                                 // Exception case when left is right else the number will be multiplied 2 times\n                ans *= left                                                   //then only multiply either left or right\n            else:\n                ans *= left * right                                        // Else multiply left and right numbers and multiply with ans\n            left += 1                                                         // Increment left by one\n            right -= 1                                                        //Decrement right by 1\n        count = 0                                                           // Initialise count of trailing zeroes\n        ans = str(ans)                                                    // Converting integer to string\n        i = len(ans) - 1                                                  // Start the pointer with end of string\n        while i >= 0 and ans[i] == \'0\':                          // Decrement pointer by one while the value at pointer is 0\n            count += 1                                                   //and increase the count of trailing zeroes\n            i -= 1\n        fans = \'\'                                                            //Empty string which will store the number without the trailing zeroes\n        for j in range(i+1):                                            // will use the i pointer which stored the last location of the trailing zero\n            fans += ans[j]                                              //store each character until the trailing zero isn\'t reached\n        final = \'\'                                                           //Final ans which will give the required result \n        if len(fans) > 10:                                              //If the length of the number without the trailing zeroes has a length greater than 10\n            temp1 = \'\'                                                   //Will store the first 5 character of the number\n            for j in range(5):                                          // Adding the first 5 characters\n                temp1 += fans[j]\n            temp2 = \'\'                                                   //Will store the last 5 characters of the number\n            for j in range(-5,0):                                      // Add the last 5 characters\n                temp2 += fans[j]\n            final = temp1 + \'...\' + temp2 + \'e\' + str(count)           //Final ans with first 5 character, last 5 characters + e + count of trailing zeroes\n        else:                                                                    //If length of the number is less than 10\n            final = fans + \'e\' + str(count)                         // Final ans with number without trailing zeroes + e + count of trailing zeroes\n        return final                                                      //Return the final string\n        \n```
['Python', 'Python3']