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Abbreviating the Product of a Range

Number: 2222

Difficulty: Hard

Paid? No

Companies: Avalara

Problem Description

Given two positive integers left and right (with left ≤ right), compute the product of all integers in the inclusive range [left, right]. Because the product can be huge, we “abbreviate” it as follows:

  1. Count the number C of trailing zero digits (i.e. count the maximum power of 10 that divides the product) and remove those trailing zeros.
  2. Let d be the number of digits in the product after removal. If d > 10 use the abbreviated form: the string starts with the first 5 digits of the product, followed by "..." then the last 5 digits; otherwise, keep the number unchanged.
  3. Finally, append “eC” to the result so that the final string format is "...eC" (or simply the full number if d ≤ 10).

For example, if the product is 12345678987600000 then after removing the 5 trailing zeros we get 123456789876; since it has more than 10 digits, we abbreviate it to "12345...89876e5".

Key Insights

  • Direct multiplication will result in an enormous number so we must avoid computing the whole product explicitly.
  • Count the total factors of 2 and 5 in the range to determine the number of trailing zeros C (since each pair (2,5) gives a factor 10).
  • Work with a “stripped” product by removing factors of 2 and 5 as they are encountered. Later, after knowing C, multiply back only the extra unmatched 2’s or 5’s.
  • Use logarithms to compute the total number of digits and to extract the first 5 digits without computing the entire number.
  • For the last 5 digits (suffix), use modular arithmetic with a modulus of 10^5 carefully since after removal the number is coprime with 10.
  • When d ≤ 10, the full number (after removal of trailing zeros) is small and can be computed exactly.

Space and Time Complexity

Time Complexity: O(n) where n = right - left + 1 (we loop through all numbers in the range) Space Complexity: O(1) (only a constant amount of extra space is needed)

Solution

We solve the problem in two conceptual parts:

  1. Compute the overall logarithm, count of 2’s and 5’s, and an accumulation for the “suffix” (modular product) while “stripping” factors of 2 and 5 from each number. Also add the base-10 logarithm of every integer to later compute the number of digits.
  2. After processing, let C be the minimum of total factors 2 and total factors 5. This is the number of trailing zeros. The “stripped” product becomes (product_over_range_without_removed_2s_and_5s * 2^(extra2)*5^(extra5)) where extra2 and extra5 are the remaining unmatched factors after removing C from each.
  3. Compute d as floor(total_log - C) + 1. If d ≤ 10 then the full product (after dividing by 10^C) is small and we can compute it exactly.
  4. If d > 10, compute the prefix (first 5 digits) by looking at the fractional part of (total_log – C) and using exponentiation (prefix = floor(10^(fraction + 4)) because that gives the first 5 digits). Compute the suffix by performing the product multiplication (with extra factors included) modulo 10^5. Format the suffix to ensure it has exactly 5 digits (adding leading zeros if needed).

We use modular arithmetic and logarithms to avoid overflow while ensuring accuracy for the abbreviated representation.

Code Solutions

Below are solutions in Python, JavaScript, C++, and Java with line-by-line comments.

import math

def abbreviateProduct(left, right):
    # Initialize counters for factors of 2 and 5, logarithm sum and suffix modulo product.
    total_log = 0.0   # Sum of log10 of all numbers in the range.
    count2 = 0
    count5 = 0
    mod = 10**5      # We only care about last 5 digits.
    product_mod = 1  # Will hold the product modulo 'mod' after removal of factors of 2 and 5.

    # For exact product when result is small (d <= 10)
    product_exact = 1

    # Process each number in the range.
    for num in range(left, right + 1):
        total_log += math.log10(num)  # Update total logarithm.
        
        current = num
        # Remove factors of 2.
        while current % 2 == 0:
            current //= 2
            count2 += 1
        # Remove factors of 5.
        while current % 5 == 0:
            current //= 5
            count5 += 1
        
        # Multiply the reduced number into our suffix product modulo mod.
        product_mod = (product_mod * (current % mod)) % mod
        
        # For exact product, we can multiply directly.
        product_exact *= num

    # Number of trailing zeros to remove.
    trailingZeros = min(count2, count5)
    # Calculate extra factors that remain after removing trailing zeros.
    extra2 = count2 - trailingZeros
    extra5 = count5 - trailingZeros

    # For product_mod, multiply back the extra factors.
    product_mod = (product_mod * pow(2, extra2, mod)) % mod
    product_mod = (product_mod * pow(5, extra5, mod)) % mod

    # Calculate total digits of the product after removal.
    digits_after_removal = math.floor(total_log - trailingZeros) + 1

    # If the number of digits is small, compute the exact product after removing trailing zeros.
    if digits_after_removal <= 10:
        # Remove trailing zeros by dividing.
        product_without_zeros = product_exact // (10**trailingZeros)
        return str(product_without_zeros) + "e" + str(trailingZeros)
    else:
        # Compute prefix: extract the fractional part to determine the first 5 digits.
        fractional_part = (total_log - trailingZeros) - math.floor(total_log - trailingZeros)
        prefix_val = int(pow(10, fractional_part + 4))  # 10^(fraction + (5-1))
        
        # Ensure suffix is exactly 5 digits (add leading zeros if needed).
        suffix_str = str(product_mod).zfill(5)
        
        return str(prefix_val) + "..." + suffix_str + "e" + str(trailingZeros)


# Example usage:
print(abbreviateProduct(1, 4))    # Expected "24e0"
print(abbreviateProduct(2, 11))   # Expected "399168e2"
print(abbreviateProduct(371, 375))# Expected "7219856259e3"
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