Leetcode #2240: Number of Ways to Buy Pens and Pencils
In this guide, we solve Leetcode #2240 Number of Ways to Buy Pens and Pencils 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 an integer total indicating the amount of money you have. You are also given two integers cost1 and cost2 indicating the price of a pen and pencil respectively.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Math, Enumeration
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: total = 20, cost1 = 10, cost2 = 5
Output: 9
Explanation: The price of a pen is 10 and the price of a pencil is 5.
- If you buy 0 pens, you can buy 0, 1, 2, 3, or 4 pencils.
- If you buy 1 pen, you can buy 0, 1, or 2 pencils.
- If you buy 2 pens, you cannot buy any pencils.
The total number of ways to buy pens and pencils is 5 + 3 + 1 = 9.
Python Solution
class Solution:
def waysToBuyPensPencils(self, total: int, cost1: int, cost2: int) -> int:
ans = 0
for x in range(total // cost1 + 1):
y = (total - (x * cost1)) // cost2 + 1
ans += y
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
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.