How do you solve Abbreviating the Product of a Range on LeetCode?

Abbreviating the Product of a Range (LeetCode #2117) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding trailing zeros is key to simplifying large products.

What is the brute force solution for Abbreviating the Product of a Range?

The brute force approach for Abbreviating the Product of a Range is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach calculates the product of all integers in the range directly. While simple, this method can lead to extremely large numbers that are difficult to manage. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Abbreviating the Product of a Range?

Abbreviating the Product of a Range uses the following concepts: Math, Number Theory. The recommended patterns to study are: Mathematical Properties, Logarithmic Functions.

What are the interview tips for Abbreviating the Product of a Range?

Explain your thought process clearly while coding. Discuss edge cases, such as very small or very large ranges. Practice breaking down the problem into smaller parts.

What are common mistakes when solving Abbreviating the Product of a Range?

Forgetting to count trailing zeros correctly. Not handling large numbers properly, leading to overflow.

#2117

Abbreviating the Product of a Range

Hard

MathNumber TheoryMathematical PropertiesLogarithmic Functions

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

This approach leverages mathematical properties to avoid computing the full product directly. By using logarithms and modular arithmetic, we can efficiently find the required digits and trailing zeros.

⚙️

Algorithm

5 steps

1Step 1: Count the number of trailing zeros by finding the minimum of the counts of factors 2 and 5 in the range.
2Step 2: Use logarithms to calculate the sum of logarithms of numbers in the range to find the total number of digits.
3Step 3: Calculate the last 5 digits using modular arithmetic.
4Step 4: Calculate the first 5 digits using exponentiation and logarithms.
5Step 5: Format the result based on the number of digits and trailing zeros.

solution.py41 lines

1# Full working Python code
2import math
3
4def abbreviate_product(left, right):
5    trailing_zeros = 0
6    count_2 = count_5 = 0
7    for i in range(left, right + 1):
8        while i % 2 == 0:
9            count_2 += 1
10            i //= 2
11        while i % 5 == 0:
12            count_5 += 1
13            i //= 5
14    trailing_zeros = min(count_2, count_5)
15
16    # Calculate the last 5 digits
17    last_digits = 1
18    for i in range(left, right + 1):
19        last_digits = (last_digits * (i % 100000)) % 100000
20
21    # Calculate the total number of digits
22    total_log = sum(math.log10(i) for i in range(left, right + 1))
23    total_digits = int(total_log) + 1
24
25    # Calculate first 5 digits
26    first_digits = 1
27    for i in range(left, right + 1):
28        first_digits *= i
29        while first_digits >= 100000:
30            first_digits //= 10
31
32    first_digits_str = str(first_digits)[:5]
33    last_digits_str = str(last_digits).rstrip('0')[-5:]
34
35    if total_digits > 10:
36        result = f'{first_digits_str}...{last_digits_str}e{trailing_zeros}'
37    else:
38        result = f'{first_digits_str}e{trailing_zeros}'
39    return result
40
41print(abbreviate_product(1, 4))

ℹ

Complexity note: The time complexity is O(n) because we iterate through the range only a few times to calculate the necessary values without storing large products.

1Understanding trailing zeros is key to simplifying large products.
2Logarithmic properties help manage large numbers effectively.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.