Solution Explanation:

The problem asks to calculate the product of a range of numbers, abbreviate it by removing trailing zeros and shortening long numbers, and finally represent it as a string in a specific format.

The solution involves several steps:

1. Counting Trailing Zeros: The number of trailing zeros in the product is determined by the number of times 10 is a factor. Since 10 = 2 * 5, we count the number of factors of 2 and 5 in the product. The minimum of these counts represents the number of trailing zeros (c).

2. Calculating the Product: The product is calculated while simultaneously removing trailing zeros. We iterate through the numbers in the range, and for each number, we divide out factors of 2 and 5 based on the count of trailing zeros c calculated in step 1. This ensures the final product doesn't contain these trailing zeros.

3. Handling Large Numbers: If the product is too large (more than 10 digits after removing trailing zeros), we abbreviate it to show the first 5 digits and the last 5 digits.

4. Constructing the Result String: The result string is built in the format <pre>...<suf>eC, where <pre> represents the first 5 digits, <suf> represents the last 5 digits (after removing trailing zeros), and C is the count of trailing zeros.

Time Complexity Analysis:

The solution iterates through the range [left, right] once to count trailing zeros and once to compute the abbreviated product. The loop has a linear time complexity of O(right - left). The other operations (removing trailing zeros, abbreviating if needed, constructing the output string) take constant time relative to the input size. Therefore, the overall time complexity is O(n), where n = (right - left + 1).

Space Complexity Analysis:

The space used depends on the length of the final string representation of the product. In the worst case, this is a constant amount of space (the space needed to store the first 5 digits, the last 5 digits, the '...', and the exponent 'eC'). Therefore, the space complexity is O(1).

Code Implementation (Python):

The Python code below implements the solution efficiently:

import numpy
 
 
class Solution:
    def abbreviateProduct(self, left: int, right: int) -> str:
        cnt2 = cnt5 = 0
        z = numpy.float128(0) # Use numpy.float128 for higher precision
        for x in range(left, right + 1):
            z += numpy.log10(x)
            while x % 2 == 0:
                x //= 2
                cnt2 += 1
            while x % 5 == 0:
                x //= 5
                cnt5 += 1
        c = cnt2 = cnt5 = min(cnt2, cnt5)
        suf = y = 1
        gt = False
        for x in range(left, right + 1):
            while cnt2 and x % 2 == 0:
                x //= 2
                cnt2 -= 1
            while cnt5 and x % 5 == 0:
                x //= 5
                cnt5 -= 1
            suf = suf * x % 100000
            if not gt:
                y *= x
                gt = y >= 1e10
        if not gt:
            return str(y) + "e" + str(c)
        pre = int(pow(10, z - int(z) + 4)) # Calculate first 5 digits using logarithm
        return str(pre) + "..." + str(suf).zfill(5) + "e" + str(c)
 

Other languages like Java, C++, and Go would follow similar logic with appropriate data type adjustments (e.g., using long long in C++ or float64 in Go for large numbers). Note that the use of numpy.float128 in Python and similar high-precision floating-point types in other languages improves the accuracy of calculating the first 5 digits for very large products.