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2240. Number of Ways to Buy Pens and Pencils

Description

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. You can spend part or all of your money to buy multiple quantities (or none) of each kind of writing utensil.

Return the number of distinct ways you can buy some number of pens and pencils.

 

Example 1:

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.

Example 2:

Input: total = 5, cost1 = 10, cost2 = 10
Output: 1
Explanation: The price of both pens and pencils are 10, which cost more than total, so you cannot buy any writing utensils. Therefore, there is only 1 way: buy 0 pens and 0 pencils.

 

Constraints:

  • 1 <= total, cost1, cost2 <= 106

Solutions

Solution 1: Enumeration

We can enumerate the number of pens to buy, denoted as $x$. For each $x$, the maximum number of pencils we can buy is $\frac{\textit{total} - x \times \textit{cost1}}{\textit{cost2}}$. The number of ways for each $x$ is this value plus 1. We sum up the number of ways for all $x$ to get the answer.

The time complexity is $O(\frac{\textit{total}}{\textit{cost1}})$, and the space complexity is $O(1)$.

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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

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class Solution {
    public long waysToBuyPensPencils(int total, int cost1, int cost2) {
        long ans = 0;
        for (int x = 0; x <= total / cost1; ++x) {
            int y = (total - x * cost1) / cost2 + 1;
            ans += y;
        }
        return ans;
    }
}

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class Solution {
public:
    long long waysToBuyPensPencils(int total, int cost1, int cost2) {
        long long ans = 0;
        for (int x = 0; x <= total / cost1; ++x) {
            int y = (total - x * cost1) / cost2 + 1;
            ans += y;
        }
        return ans;
    }
};

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func waysToBuyPensPencils(total int, cost1 int, cost2 int) (ans int64) {
    for x := 0; x <= total/cost1; x++ {
        y := (total-x*cost1)/cost2 + 1
        ans += int64(y)
    }
    return
}

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function waysToBuyPensPencils(total: number, cost1: number, cost2: number): number {
    let ans = 0;
    for (let x = 0; x <= Math.floor(total / cost1); ++x) {
        const y = Math.floor((total - x * cost1) / cost2) + 1;
        ans += y;
    }
    return ans;
}

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impl Solution {
    pub fn ways_to_buy_pens_pencils(total: i32, cost1: i32, cost2: i32) -> i64 {
        let mut ans: i64 = 0;
        for pen in 0..=total / cost1 {
            ans += (((total - pen * cost1) / cost2) as i64) + 1;
        }
        ans
    }
}

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