Two working solutions, easy python solution.
Intuitionjust obtain the product of range value and convert it into string, check number of '0' in the string , and return the output as per condition.Complexit
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
just obtain the product of range value and convert it into string, check number of '0' in the string , and return the output as per condition.

# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
O(Right−Left)
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
O(1)
# Code
```python3 []
import sys
sys.set_int_max_str_digits(100000)
class Solution:
    def abbreviateProduct(self, left: int, right: int) -> str:
        #this logic also works but it takes a time a lot --1530ms
        # x,k = 1,0
        # for i in range(left, right+1):
        #     x*=i
        # x = str(x)
        # n =  len(x)
        # for j in x:
        #     if j == '0':
        #         k+=1
        #     else:
        #         k = 0
        # x = x[:n-k]
        # if len(x)<=10: return x+'e'+str(k)
        # else: return (x[:5]+'...'+x[-5:]+'e'+str(k))
        

        #**************************************************************************************
        k = str(prod(list(range(left, right+1)))) #1524ms
        i = k.rstrip('0')
        zeros = len(k) - len(i)
        if len(i)<=10: return (i+'e'+str(zeros))
        else: return (i[:5]+'...'+i[-5:]+'e'+str(zeros))

```
['Python3']