#2117

Hard

auto_awesomeMath-driven solution strategy

LeetCode Problem Workspace

Abbreviating the Product of a Range

Calculate the abbreviated product of integers in a range efficiently using math-driven analysis of trailing zeros and significant digits.

math

Problem Statement

Given two positive integers left and right where left is less than or equal to right, calculate the product of all integers in the inclusive range [left, right]. The result may be extremely large, so a standard integer calculation may not be feasible.

Return a string representing the abbreviated product. Abbreviation involves removing trailing zeros, capturing the last few digits, determining the first few digits, and appending the count of removed zeros as an exponent, ensuring the representation is compact and precise.

Examples

Example 1

Input: left = 1, right = 4

Output: "24e0"

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"

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"

The product is 7219856259000.

Constraints

1 <= left <= right <= 104

Solution Approach

Count trailing zeros

Iterate through the range and count factors of 5 and 2 to determine the number of trailing zeros. This separates the exponent portion from the significant digits for abbreviation.

Compute first and last digits separately

Use logarithms to calculate the leading digits and modular arithmetic for the last digits without computing the full product. This prevents overflow and aligns with the Math-driven solution strategy.

Construct abbreviated string

Combine the first few digits, last few digits, and the trailing zero count into a string formatted as first_digits + last_digits + e + trailing_zero_count. This produces the final abbreviated product.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time and space depend on the chosen approach: iterative counting and modular arithmetic reduce complexity compared to full multiplication. Logarithmic computation for first digits ensures efficiency for large ranges.

What Interviewers Usually Probe

Expect questions about handling large integer products without overflow.
Look for understanding of separating trailing zeros from significant digits.
Check if the candidate uses math properties like logarithms and modular arithmetic for efficiency.

Common Pitfalls or Variants

Common pitfalls

Attempting to multiply all numbers directly, causing overflow.
Ignoring the correct count of trailing zeros leading to wrong exponent.
Incorrectly computing first or last digits without modular or logarithmic methods.

Follow-up variants

Return only the number of trailing zeros for a range product.
Compute the abbreviated product for non-continuous integer ranges.
Abbreviate products with different digit limits for the first and last significant digits.

FAQ

How do I handle very large products for Abbreviating the Product of a Range?

Use logarithms to get the first digits and modular arithmetic for last digits, separating trailing zeros instead of multiplying all numbers directly.

Why do we count factors of 5 and 2?

Factors of 5 and 2 determine trailing zeros in the product, which are needed for correct abbreviation.

Can the approach work for left and right up to 10^4?

Yes, iterative counting and logarithmic calculations scale efficiently without computing the full product.

What is the format of the abbreviated product?

It is first_digits + last_digits + e + trailing_zero_count, combining the leading digits, last digits, and zero count.

Does the Math-driven solution pattern always apply here?

Yes, separating zeros and computing first/last digits independently follows the Math-driven solution strategy needed for this problem.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def abbreviateProduct(self, left: int, right: int) -> str:
        cnt2 = cnt5 = 0
        for x in range(left, right + 1):
            while x % 2 == 0:
                cnt2 += 1
                x //= 2
            while x % 5 == 0:
                cnt5 += 1
                x //= 5
        c = cnt2 = cnt5 = min(cnt2, cnt5)
        pre = suf = 1
        gt = False
        for x in range(left, right + 1):
            suf *= x
            while cnt2 and suf % 2 == 0:
                suf //= 2
                cnt2 -= 1
            while cnt5 and suf % 5 == 0:
                suf //= 5
                cnt5 -= 1
            if suf >= 1e10:
                gt = True
                suf %= int(1e10)
            pre *= x
            while pre > 1e5:
                pre /= 10
        if gt:
            return str(int(pre)) + "..." + str(suf % int(1e5)).zfill(5) + "e" + str(c)
        return str(suf) + "e" + str(c)

Continue Practicing

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