What is the time complexity of Number of Ways to Buy Pens and Pencils?

The optimal solution has a time complexity of O(total/cost1).

What data structures are used to solve Number of Ways to Buy Pens and Pencils?

This problem uses Math, Enumeration. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Number of Ways to Buy Pens and Pencils?

Insight 1: We can iterate through the possible number of pens we can buy, and for each number of pens, calculate how many pencils we can buy with the remaining money. Insight 2: The number of pencils we can buy with the remaining money is simply the remaining money divided by the cost of a pencil, plus one (to include the case where we buy zero pencils).

Which companies ask LeetCode 2240. Number of Ways to Buy Pens and Pencils in interviews?

This problem is frequently asked at Reddit.

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Number of Ways to Buy Pens and Pencils - LeetCode 2240 Solution

Number of Ways to Buy Pens and Pencils - Complete Solution Guide

Number of Ways to Buy Pens and Pencils is LeetCode problem 2240, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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. - I

Detailed Explanation

The problem asks us to find the number of ways to buy pens and pencils, given a total amount of money and the cost of each pen and pencil. We can buy any number of pens and pencils, including zero, as long as the total cost does not exceed the given total amount. The goal is to count all the possible combinations of pens and pencils we can buy.

Solution Approach

The solution iterates through the possible number of pens we can buy. For each possible number of pens, it calculates the remaining money. It then calculates the number of pencils we can buy with the remaining money. The total number of ways is the sum of the number of possible pencil combinations for each pen count. The code first ensures `cost1` represents the larger cost to reduce the number of iterations in the main loop.

Step-by-Step Algorithm

Step 1: Ensure `cost1` represents the more expensive item. If `cost1 < cost2`, swap them.
Step 2: Initialize a variable `ways` to 0. This variable will store the total number of ways to buy pens and pencils.
Step 3: Initialize a variable `num_item1` to 0. This variable represents the number of pens being purchased in the current iteration.
Step 4: Iterate while `num_item1 * cost1 <= total`. This loop goes through the possible number of pens we can buy.
Step 5: Inside the loop, calculate the `remaining_money` after buying `num_item1` pens: `remaining_money = total - (num_item1 * cost1)`.
Step 6: Calculate the number of pencils we can buy with the remaining money: `(remaining_money // cost2) + 1`. We add 1 to include the possibility of buying zero pencils.
Step 7: Add the number of pencil combinations to the `ways` variable.
Step 8: Increment `num_item1` to consider the next possible number of pens.
Step 9: After the loop finishes, return the `ways` variable, which contains the total number of ways to buy pens and pencils.

Key Insights

Insight 1: We can iterate through the possible number of pens we can buy, and for each number of pens, calculate how many pencils we can buy with the remaining money.
Insight 2: The number of pencils we can buy with the remaining money is simply the remaining money divided by the cost of a pencil, plus one (to include the case where we buy zero pencils).
Insight 3: Swapping the cost of pens and pencils can improve the efficiency slightly by iterating over the more expensive item. This can slightly reduce iterations.

Complexity Analysis

Time Complexity: O(total/cost1)

Space Complexity: O(1)

Topics

This problem involves: Math, Enumeration.

Companies

Asked at: Reddit.

Number of Ways to Buy Pens and Pencils | LeetCode 2240

Medium

Math Enumeration

Time: O(total/cost1)Space: O(1)

Solution Code

Available in:

class Solution {
    public long waysToBuyPensPencils(int total, int cost1, int cost2) {
        if (cost1 < cost2) {
            int temp = cost1;
            cost1 = cost2;
            cost2 = temp;
        }
        
        long ways = 0;
        long num_item1 = 0;
        
        while (num_item1 * cost1 <= total) {
            long remaining_money = total - (num_item1 * cost1);
            ways += (remaining_money / cost2) + 1;
            num_item1 += 1;
        }
        
        return ways;
    }
}

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Problem Info

Description

Problem 2240: Number of Ways to Buy Pens and Pencils - Detailed explanation and solution approaches.

Topics

Math Enumeration

Companies

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2240

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Ensure `cost1` represents the more expensive item. If `cost1 < cost2`, swap them.
2
Step 2: Initialize a variable `ways` to 0. This variable will store the total number of ways to buy pens and pencils.
3
Step 3: Initialize a variable `num_item1` to 0. This variable represents the number of pens being purchased in the current iteration.
4
Step 4: Iterate while `num_item1 * cost1 <= total`. This loop goes through the possible number of pens we can buy.
5
Step 5: Inside the loop, calculate the `remaining_money` after buying `num_item1` pens: `remaining_money = total - (num_item1 * cost1)`.
6
Step 6: Calculate the number of pencils we can buy with the remaining money: `(remaining_money // cost2) + 1`. We add 1 to include the possibility of buying zero pencils.
7
Step 7: Add the number of pencil combinations to the `ways` variable.
8
Step 8: Increment `num_item1` to consider the next possible number of pens.
9
Step 9: After the loop finishes, return the `ways` variable, which contains the total number of ways to buy pens and pencils.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(total/cost1)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the number of iterations in the `while` loop, which depends on how many of the more expensive item can be bought within the total budget. The space complexity is constant because the number of variables used does not scale with the input size.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Mathematical Formula

Attempting to derive a direct mathematical formula might be possible but would likely involve complex case analysis and might not be simpler than the iterative approach. The complexity of finding such a formula and ensuring its correctness could outweigh the benefit.

Time: Potentially O(1) if a formula could be derived.Space: O(1)

Trade-offs: Requires significant mathematical derivation, prone to errors, and may not be easier to understand or implement. The iterative solution is simpler and more maintainable.

✨

Key Insights

1Insight 1: We can iterate through the possible number of pens we can buy, and for each number of pens, calculate how many pencils we can buy with the remaining money.
2Insight 2: The number of pencils we can buy with the remaining money is simply the remaining money divided by the cost of a pencil, plus one (to include the case where we buy zero pencils).
3Insight 3: Swapping the cost of pens and pencils can improve the efficiency slightly by iterating over the more expensive item. This can slightly reduce iterations.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: `cost1` or `cost2` is greater than `total`. In this case, the loop will execute at most once (when `num_item1` is 0), and the remaining money will be `total`. The number of pencils we can buy will be `(total // cost2) + 1` if `cost2 <= total`, or 1 if `cost2 > total` which corresponds to buying 0 pens and 0 pencils.
Edge case 2: `cost1` and `cost2` are both greater than `total`. In this scenario, the `while` loop will execute only once where `num_item1` is zero. The remaining money will be equal to total, and we add `(total // cost2) + 1` to the `ways`. If `cost2` is also greater than `total`, then `total // cost2` will be zero, so we will add `0 + 1 = 1` to `ways`, so `ways` will be one, representing the single way where we buy 0 pens and pencils. The code handles this implicitly without any special checks.
Edge case 3: total is 0. The loop condition `num_item1 * cost1 <= total` will only run once with `num_item1 = 0`, and remaining money is 0. We add (0//cost2) + 1 which is 1. Therefore, 1 way to buy 0 pens and 0 pencils

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where zero pens or pencils are bought. The `+ 1` in the calculation of the number of pencils handles this case.

✗Common mistake 2: Integer overflow when calculating `num_item1 * cost1`. The provided solution correctly uses `long` data type (Java and C++) or no explicit types (Python), which prevents integer overflows since the total, cost1 and cost2 can be up to 10^6. It is handled because `total` is also declared as a 32 bit integer.

✗Common mistake 3: Forgetting to account for the remaining money when the pen cost is close to or equal to the total. The integer division `remaining_money // cost2` takes care of the fractional part appropriately.

✗Common mistake 4: Swapping the costs might seem unnecessary, but it does provide a very minor optimization by reducing the number of iterations in cases where cost2 > cost1, the while loop will execute fewer times.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Budget allocation - determining how many of two different resources can be acquired with a limited budget.

🏦Real-world scenario 2: Inventory management - deciding how many units of two different products to stock, given a limited storage space and the sizes of each product.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 20, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

❓

Frequently Asked Questions

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Number of Ways to Buy Pens and Pencils"?

For this medium-level Math problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether enumeration techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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