Problem Description

You have a certain amount of money (total) and want to buy pens and pencils. Each pen costs cost1 and each pencil costs cost2.

Your task is to find how many different ways you can spend your money on these items. You can:

• Buy any number of pens (including zero)
• Buy any number of pencils (including zero)
• Spend part of your money or all of it
• Leave some money unspent

Each unique combination of pens and pencils counts as one distinct way.

For example, if you have total = 20, cost1 = 10 (pen price), and cost2 = 5 (pencil price), you could:

• Buy 0 pens and 0 pencils
• Buy 0 pens and 1 pencil
• Buy 0 pens and 2 pencils
• Buy 0 pens and 3 pencils
• Buy 0 pens and 4 pencils
• Buy 1 pen and 0 pencils
• Buy 1 pen and 1 pencil
• Buy 1 pen and 2 pencils
• Buy 2 pens and 0 pencils

This gives us 9 distinct ways to make purchases.

The solution works by iterating through all possible numbers of pens you can afford (x ranges from 0 to total // cost1). For each number of pens, it calculates how many pencils you can buy with the remaining money: (total - x * cost1) // cost2. Since you can also choose to buy 0 pencils up to this maximum number, you add 1 to include the zero option. The sum of all these possibilities gives the total number of distinct ways.

Intuition

The key insight is that we can break this problem down by fixing one variable and calculating the possibilities for the other.

Think about it this way: if we decide to buy exactly x pens, we've already spent x * cost1 dollars. This leaves us with total - x * cost1 dollars for pencils. With this remaining money, we can buy anywhere from 0 to (total - x * cost1) // cost2 pencils.

Why does this enumeration strategy work? Because once we fix the number of pens, the problem becomes one-dimensional - we just need to count how many valid pencil quantities exist. For each fixed number of pens, the number of valid ways equals the maximum number of pencils we can afford plus one (to account for buying zero pencils).

The beauty of this approach is that we avoid checking every possible combination of pens and pencils. Instead of nested loops that would check if x * cost1 + y * cost2 <= total for all possible (x, y) pairs, we iterate through pen quantities only once. For each pen quantity, we directly calculate the valid range of pencil quantities using integer division.

This transforms what could be a two-dimensional search space into a series of one-dimensional calculations. We're essentially asking: "If I buy 0 pens, how many pencil options do I have? If I buy 1 pen, how many pencil options remain?" and so on. The sum of all these options gives us our answer.

The formula (total - x * cost1) // cost2 + 1 captures this perfectly - it tells us the count of valid pencil quantities (including zero) for a given number of pens x.

Learn more about Math patterns.

Solution Approach

The implementation follows a straightforward enumeration strategy where we iterate through all possible quantities of pens and calculate the corresponding number of ways to buy pencils.

We start by initializing a counter ans = 0 to track the total number of distinct ways.

The main loop iterates through all possible quantities of pens: for x in range(total // cost1 + 1). Here, x represents the number of pens we're buying. The upper bound total // cost1 + 1 ensures we consider all affordable quantities of pens, from 0 up to the maximum number we can buy with our total money.

For each fixed number of pens x, we calculate:

1. The money spent on pens: x * cost1
2. The remaining money for pencils: total - (x * cost1)
3. The maximum number of pencils we can buy: (total - (x * cost1)) // cost2

The number of ways to buy pencils with the remaining money is y = (total - (x * cost1)) // cost2 + 1. We add 1 because we can choose to buy anywhere from 0 to the maximum number of pencils. Each of these choices represents a distinct way.

We accumulate these counts by adding y to our answer for each iteration: ans += y.

The algorithm runs in O(total / cost1) time complexity since we iterate through at most total // cost1 + 1 values. The space complexity is O(1) as we only use a constant amount of extra space for variables.

This approach efficiently counts all valid combinations without explicitly generating them, making it both time and space efficient for the given constraints.

Example Walkthrough

Let's walk through a small example with total = 15, cost1 = 4 (pen price), and cost2 = 3 (pencil price).

We'll iterate through each possible number of pens and count how many ways we can buy pencils with the remaining money.

Step 1: Buy 0 pens

• Money spent on pens: 0 * 4 = 0
• Money remaining: 15 - 0 = 15
• Maximum pencils affordable: 15 // 3 = 5
• Ways to buy pencils: 5 + 1 = 6 (can buy 0, 1, 2, 3, 4, or 5 pencils)
• Running total: ans = 6

Step 2: Buy 1 pen

• Money spent on pens: 1 * 4 = 4
• Money remaining: 15 - 4 = 11
• Maximum pencils affordable: 11 // 3 = 3
• Ways to buy pencils: 3 + 1 = 4 (can buy 0, 1, 2, or 3 pencils)
• Running total: ans = 6 + 4 = 10

Step 3: Buy 2 pens

• Money spent on pens: 2 * 4 = 8
• Money remaining: 15 - 8 = 7
• Maximum pencils affordable: 7 // 3 = 2
• Ways to buy pencils: 2 + 1 = 3 (can buy 0, 1, or 2 pencils)
• Running total: ans = 10 + 3 = 13

Step 4: Buy 3 pens

• Money spent on pens: 3 * 4 = 12
• Money remaining: 15 - 12 = 3
• Maximum pencils affordable: 3 // 3 = 1
• Ways to buy pencils: 1 + 1 = 2 (can buy 0 or 1 pencil)
• Running total: ans = 13 + 2 = 15

We stop here because buying 4 pens would cost 4 * 4 = 16, which exceeds our total of 15.

Final Answer: 15 distinct ways

This gives us combinations like:

• (0 pens, 0 pencils), (0 pens, 1 pencil), ..., (0 pens, 5 pencils)
• (1 pen, 0 pencils), (1 pen, 1 pencil), ..., (1 pen, 3 pencils)
• (2 pens, 0 pencils), (2 pens, 1 pencil), (2 pens, 2 pencils)
• (3 pens, 0 pencils), (3 pens, 1 pencil)

Solution Implementation

Time and Space Complexity

The time complexity is O(total / cost1). The algorithm iterates through all possible quantities of the first item (pens) that can be purchased, from 0 up to total // cost1. This gives us at most total / cost1 + 1 iterations. For each iteration, we perform constant time operations: calculating the remaining budget after buying x pens and determining how many pencils can be bought with that remainder. Since the loop dominates the runtime, the overall time complexity is O(total / cost1).

The space complexity is O(1). The algorithm only uses a fixed amount of extra space regardless of the input size. We maintain only a few integer variables: ans for counting valid combinations, x as the loop counter, and y for calculating the number of pencils. No additional data structures that grow with the input are used, resulting in constant space complexity.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Integer Overflow in Other Languages

While Python handles large integers automatically, in languages like C++ or Java, calculating num_pens * cost1 could cause integer overflow when dealing with large values.

Solution: Use appropriate data types (like long long in C++ or long in Java) or check for overflow conditions before multiplication.

2. Off-by-One Error in Range

A common mistake is forgetting to include either the case where we buy 0 pens or the maximum number of pens we can afford. Writing range(total // cost1) instead of range(total // cost1 + 1) would miss the case where we spend all our money on pens.

Solution: Always include both boundaries - start from 0 pens and go up to and including total // cost1.

3. Forgetting the Zero Purchase Option for Pencils

When calculating the number of ways to buy pencils, developers might return just remaining_money // cost2 without adding 1, forgetting that buying 0 pencils is also a valid option.

Solution: Always add 1 to account for the option of buying 0 pencils: (remaining_money // cost2) + 1.

4. Attempting to Optimize by Swapping cost1 and cost2

Some might try to optimize by ensuring cost1 >= cost2 to minimize iterations. However, this adds unnecessary complexity and could introduce bugs if not handled carefully with the variable swapping logic.

Solution: Keep the solution simple and straightforward. The performance difference is negligible for reasonable input sizes, and clarity is more important than minor optimizations.

5. Using Nested Loops Unnecessarily

Beginners might use two nested loops to iterate through both pens and pencils explicitly, which would result in O(n²) time complexity:

# Inefficient approach - avoid this
for pens in range(total // cost1 + 1):
    for pencils in range(total // cost2 + 1):
        if pens * cost1 + pencils * cost2 <= total:
            total_ways += 1

Solution: Use the mathematical approach shown in the correct solution where we calculate the number of valid pencil options directly for each pen count, achieving O(n) complexity.