Introduction

In this blog post, we will explore how to solve the LeetCode problem "2117" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given two positive integers left and right with left <= right. Calculate the product of all integers in the inclusive range [left, right]. Since the product may be very large, you will abbreviate it following these steps: Count all trailing zeros in the product and remove them. Let us denote this count as C. For example, there are 3 trailing zeros in 1000, and there are 0 trailing zeros in 546. Denote the remaining number of digits in the product as d. If d > 10, then express the product as

... where 
 denotes the first 5 digits of the product, and  denotes the last 5 digits of the product after removing all trailing zeros. If d <= 10, we keep it unchanged.      For example, we express 1234567654321 as 12345...54321, but 1234567 is represented as 1234567.   Finally, represent the product as a string "
...eC".      For example, 12345678987600000 will be represented as "12345...89876e5".    Return a string denoting the abbreviated product of all integers in the inclusive range [left, right].   Example 1:  Input: left = 1, right = 4 Output: "24e0" Explanation: The product is 1 × 2 × 3 × 4 = 24. There are no trailing zeros, so 24 remains the same. The abbreviation will end with "e0". Since the number of digits is 2, which is less than 10, we do not have to abbreviate it further. Thus, the final representation is "24e0".  Example 2:  Input: left = 2, right = 11 Output: "399168e2" Explanation: The product is 39916800. There are 2 trailing zeros, which we remove to get 399168. The abbreviation will end with "e2". The number of digits after removing the trailing zeros is 6, so we do not abbreviate it further. Hence, the abbreviated product is "399168e2".  Example 3:  Input: left = 371, right = 375 Output: "7219856259e3" Explanation: The product is 7219856259000.    Constraints:  1 <= left <= right <= 104  

Explanation

• Calculate the product of numbers in the range [left, right]. Instead of directly calculating the full product (which can lead to integer overflow), we can compute the product modulo a large number (10^10) during calculation to keep the intermediate results manageable and find prefix and suffix. The number of trailing zeros can be calculated by finding the minimum number of factors 2 and 5 in the product.
  • Remove the trailing zeros from the product and determine whether further abbreviation (using prefix and suffix) is needed based on the length of remaining digits.
  • Format the result string with prefix, suffix, and the count of trailing zeros.

• Runtime Complexity: O(right - left) which simplifies to O(n), where n is the size of the input range. Storage Complexity: O(1) since we only store a few integer variables like product, number of 2's and 5's.

Code

    def abbreviate_product(left: int, right: int) -> str:
    """
    Calculates the abbreviated product of all integers in the inclusive range [left, right].
    """

    product = 1
    trailing_zeros = 0
    num_twos = 0
    num_fives = 0
    prefix = 1
    suffix = 1

    for i in range(left, right + 1):
        num = i
        while num % 2 == 0:
            num_twos += 1
            num //= 2
        while num % 5 == 0:
            num_fives += 1
            num //= 5

        prefix = (prefix * i) % 100000
        suffix = (suffix * i) % 10000000000

    trailing_zeros = min(num_twos, num_fives)
    suffix //= (10**trailing_zeros)

    suffix_str = str(suffix)
    d = len(suffix_str)
    if d > 10:
        prefix_str = str(prefix).zfill(5)
        suffix_str = str(suffix).zfill(10)[-5:]
        result = prefix_str + "..."+ suffix_str + "e" + str(trailing_zeros)
    else:
        result = str(suffix) + "e" + str(trailing_zeros)

    return result