LeetCode #2240 — MEDIUM

Number of Ways to Buy Pens and Pencils

Move from brute-force thinking to an efficient approach using math strategy.

The Problem

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. 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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math

Example 1

20
10
5

Example 2

5
10
10

Core Insight

What unlocks the optimal approach

Fix the number of pencils purchased and calculate the number of ways to buy pens.
Sum up the number of ways to buy pens for each amount of pencils purchased to get the answer.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2240: Number of Ways to Buy Pens and Pencils
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;
    }
}

// Accepted solution for LeetCode #2240: Number of Ways to Buy Pens and Pencils
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
}

# Accepted solution for LeetCode #2240: Number of Ways to Buy Pens and Pencils
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

// Accepted solution for LeetCode #2240: Number of Ways to Buy Pens and Pencils
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
    }
}

// Accepted solution for LeetCode #2240: Number of Ways to Buy Pens and Pencils
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(\frac\textittotal\textitcost1)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.