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Abbreviating the Product of a Range

Difficulty: Hard

Problem Description

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:

  1. Count all trailing zeros in the product and remove them. Let us denote this count as C.
  2. Denote the remaining number of digits in the product as d. If d > 10, express the product as <pre>...<suf> where <pre> denotes the first 5 digits of the product, and <suf> denotes the last 5 digits of the product after removing all trailing zeros. If d <= 10, keep it unchanged.
  3. Represent the product as a string "<pre>...<suf>eC".

Return a string denoting the abbreviated product of all integers in the inclusive range [left, right].

Key Insights

  • The product of a range of integers can grow very large, making it impractical to calculate directly.
  • Trailing zeros are created by pairs of factors of 2 and 5; hence counting these factors allows us to determine the number of trailing zeros.
  • To abbreviate, we can focus on the first and last few digits of the product without calculating the entire product.
  • The manipulation of digits can be efficiently handled with string operations once the significant digits are identified.

Space and Time Complexity

Time Complexity: O(n), where n is the number of integers in the range [left, right]. This accounts for iterating through the range to compute the product and count factors. Space Complexity: O(1), as we only use a fixed amount of extra space for counting and string manipulation.

Solution

To solve the problem, we can use the following approach:

  1. Initialize variables to hold the product's significant digits and count the number of trailing zeros.
  2. Iterate from left to right, updating the product and counting the factors of 2 and 5 to determine the number of trailing zeros.
  3. After calculating the product, convert it to a string to extract the first 5 and last 5 digits.
  4. Format the final output based on the number of remaining digits.

Code Solutions

def abbreviate_product(left: int, right: int) -> str:
    product = 1
    count_zeros = 0
    factors_of_2 = 0
    factors_of_5 = 0
    
    for num in range(left, right + 1):
        product *= num
        
        # Count factors of 2
        while num % 2 == 0:
            factors_of_2 += 1
            num //= 2
            
        # Count factors of 5
        while num % 5 == 0:
            factors_of_5 += 1
            num //= 5
            
    count_zeros = min(factors_of_2, factors_of_5)

    # Remove trailing zeros and get significant digits
    product //= 10 ** count_zeros
    product_str = str(product)

    if len(product_str) > 10:
        pre = product_str[:5]
        suf = product_str[-5:]
        return f"{pre}...{suf}e{count_zeros}"
    else:
        return f"{product_str}e{count_zeros}"
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