Number of Ways to Buy Pens and Pencils - Practice Coding Problems

Number of Ways to Buy Pens and Pencils - Problem

Math Enumeration Medium

You're planning to stock up on writing supplies for the upcoming school year! 📝

Given your budget of total dollars, you can purchase pens at cost1 dollars each and pencils at cost2 dollars each. You want to find out how many different ways you can spend your money.

Here's the twist: you can buy any quantity of each item (including zero), and you don't have to spend all your money. Each different combination of pens and pencils counts as a distinct way.

Example: If you have $20, pens cost $1, and pencils cost $2, you could buy:

0 pens, 0 pencils (spend $0)
1 pen, 0 pencils (spend $1)
0 pens, 1 pencil (spend $2)
And many more combinations...

Goal: Return the total number of distinct ways to buy some combination of pens and pencils within your budget.

Input & Output

example_1.py — Basic case

$ Input: total = 20, cost1 = 1, cost2 = 1

› Output: 231

💡 Note: With $20 and both items costing $1 each, we can buy any combination from (0,0) to (20,0), (19,1), (18,2), ..., (0,20). This forms a triangular counting pattern: 21 + 20 + 19 + ... + 1 = 231 ways.

example_2.py — Different costs

$ Input: total = 5, cost1 = 10, cost2 = 10

› Output: 1

💡 Note: Since both pens and pencils cost $10 each but we only have $5, we can't afford any items. The only way is to buy 0 pens and 0 pencils, giving us 1 way total.

example_3.py — Asymmetric costs

$ Input: total = 20, cost1 = 1, cost2 = 2

› Output: 121

💡 Note: We can buy 0-20 pens ($1 each). For each pen count: 0 pens allows 0-10 pencils (11 ways), 1 pen allows 0-9 pencils (10 ways), etc. Total: 11+10+9+8+7+6+5+4+3+2+1 = 66... Wait, let me recalculate: 11+10+10+9+9+8+8+7+7+6+6+5+5+4+4+3+3+2+2+1+1 = 121 ways.

Visualization

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Understanding the Visualization

Set up the problem space

Create a grid where x-axis = pens, y-axis = pencils. Each point (x,y) costs x×cost1 + y×cost2

Identify valid region

All points where total cost ≤ budget form a triangular/curved region

Count efficiently

Instead of checking each point, count points in each column (fixed pen count) mathematically

Key Takeaway

🎯 Key Insight: Instead of checking every (pen, pencil) combination, fix the pen count and mathematically calculate how many pencil options fit the remaining budget. This transforms a 2D enumeration problem into efficient 1D calculation!

Time & Space Complexity

Time Complexity

⏱️

O(total/cost1)

We only iterate through possible pen counts once, calculating pencil combinations mathematically

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables for calculations

✓ Linear Space

Constraints

0 ≤ total ≤ 10⁶
1 ≤ cost1, cost2 ≤ 10⁶
The answer will fit in a 64-bit signed integer

Asked in

a Amazon 15 G Google 12 ⊞ Microsoft 8 f Meta 6

The optimal solution uses a mathematical approach: iterate through possible pen counts and calculate (don't enumerate) the number of valid pencil combinations for each. This reduces complexity from O(n²) to O(n) by recognizing that for each pen count, the pencil combinations form a simple arithmetic sequence that can be counted directly.

Common Approaches

Approach	Time	Space	Notes
✓ Optimized Math (Single Loop)	O(total/cost1)	O(1)	Calculate combinations mathematically without nested loops
Brute Force (Nested Loops)	O((total/cost1) × (total/cost2))	O(1)	Try every possible combination of pens and pencils within budget

Optimized Math (Single Loop) — Algorithm Steps

Calculate maximum pens we can afford: total // cost1
For each pen count from 0 to max_pens:
Calculate remaining money after buying pens
Calculate max pencils possible with remaining money
Add (max_pencils + 1) to total ways (includes 0 to max_pencils)

Visualization

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Step-by-Step Walkthrough

Fix pen count

For each possible number of pens

Calculate remaining budget

Subtract pen costs from total budget

Count pencil options

Remaining budget ÷ pencil cost + 1 gives all valid pencil counts

Code -

solution.c — C

#include <stdio.h>

long long waysToBuyPensAndPencils(int total, int cost1, int cost2) {
    long long ways = 0;
    int maxPens = total / cost1;
    
    for (int pens = 0; pens <= maxPens; pens++) {
        int remaining = total - pens * cost1;
        int maxPencils = remaining / cost2;
        // Can buy 0, 1, 2, ..., maxPencils pencils
        ways += maxPencils + 1;
    }
    
    return ways;
}

int main() {
    int total, cost1, cost2;
    scanf("%d %d %d", &total, &cost1, &cost2);
    printf("%lld\n", waysToBuyPensAndPencils(total, cost1, cost2));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(total/cost1)

We only iterate through possible pen counts once, calculating pencil combinations mathematically

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables for calculations

✓ Linear Space

Constraints

0 ≤ total ≤ 10⁶
1 ≤ cost1, cost2 ≤ 10⁶
The answer will fit in a 64-bit signed integer

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Optimized Math (Single Loop) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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