Leetcode #2117: Abbreviating the Product of a Range
In this guide, we solve Leetcode #2117 Abbreviating the Product of a Range in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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].
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Math
Intuition
There is a mathematical invariant or formula that directly leads to the result.
Using math avoids unnecessary loops and reduces complexity.
Approach
Derive the formula or update rule, then compute the answer directly.
Handle edge cases like overflow or zero carefully.
Steps:
- Identify the math relationship.
- Compute the result with a loop or formula.
- Handle edge cases.
Example
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".
Python Solution
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)
Complexity
The time complexity is O(n) or O(1). The space complexity is O(1).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.