Asked in: Unstop
Image of the Question
All Testcases Passed ✔
Solution
def maxBottles(n,x,costs):
sm = sum(costs)
costs.sort()
i,j = 0,n-1
// ... rest of solution available after purchaseExplanation
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To solve this problem effectively, the key is to understand the dynamic nature of beer prices and how buying from shops over multiple days affects your budget constraints. The goal is to maximize the number of beer bottles you can buy from N shops with a fixed daily budget X before all shop prices become unaffordable.
1. Understanding Price Dynamics:
Each shop starts with an initial price given in the input. After every purchase from a shop, its price increases by 1 coin for the next day. You can buy at most one bottle from each shop per day, but you can visit multiple shops in the same day if their prices fit within your daily budget X.
2. Clarifying the Constraints:
- You have a fixed daily budget X.
- You can buy multiple bottles in a day, but at most one from each shop.
- After buying a bottle from a shop, its price increases by 1 coin for the next day.
- The process continues day by day until you cannot afford a bottle from any shop on a given day (i.e., all shop prices exceed X).
- Your task is to compute the total number of bottles bought before prices become unaffordable across all shops.
3. Breaking Down the Problem:
The price at each shop changes over time, which makes this a dynamic pricing problem. On day d (starting from 0), the price of a bottle at shop i will be:
initial_price[i] + d
because every purchase at that shop on previous days increments the price by 1 each day. But remember, you can only buy once from a shop per day, so the price increment depends on how many times you've bought from that particular shop.
4. Key Observations:
- Since prices only increase by 1 each time you buy from a shop, the cost of buying k bottles from a shop with initial price p would be: p + 0, p + 1, p + 2, ..., p + (k-1).
- The total cost for k bottles from a single shop is the sum of an arithmetic progression: k * p + (k * (k - 1)) / 2.
- However, the daily budget X restricts how many bottles you can buy each day across all shops, and crucially, the prices increase day by day, which means prices evolve in a predictable linear fashion.
5. Strategy:
To maximize the total number of bottles:
- Consider that you buy bottles in rounds (days). On each day, you can buy from any shop whose price on that day does not exceed X.
- The price of shop i on day d is: initial_price[i] + d (since you buy only once per day).
- So, for day d, the shops you can afford are those with initial_price[i] + d <= X.
- To maximize the number of bottles per day, buy from all shops that satisfy this condition.
- Continue this day-by-day purchase process until on a day, no shop is affordable (i.e., initial_price[i] + d > X for all shops).
6. Simplifying the Approach:
- Sort the shops by their initial prices.
- On day 0, buy from all shops with initial_price <= X.
- On day 1, buy from all shops with initial_price + 1 <= X.
- And so forth, incrementing day until no shops meet the criteria.
- The total bottles bought is the sum over all days of the count of shops you could afford.
7. How to Efficiently Compute the Result:
- Sorting the initial prices helps quickly identify how many shops are affordable on each day.
- For each day d, find the number of shops where initial_price[i] <= X - d.
- The total bottles bought is the sum of these counts for d = 0 up to the day where no shops are affordable.
- Since prices increase linearly and initial prices are fixed, this process is equivalent to counting how many initial prices are less than or equal to (X - d) for each d.
8. Algorithm Outline:
- Sort the array of initial prices.
- Initialize day = 0, total_bottles = 0.
- For day d = 0, 1, 2, ..., while there exists shops with initial_price[i] <= X - d:
- Use binary search or a pointer to find how many shops satisfy initial_price[i] <= X - d.
- Add this count to total_bottles.
- Increment day.
- Stop when no shops meet the condition.
9. Edge Cases:
- If all shops have initial prices greater than X, you cannot buy any bottles.
- If the daily budget is very high, you might be able to buy bottles for many days, so be mindful of efficiency.
- Input arrays where initial prices are all equal or widely varied.
10. Time Complexity:
- Sorting the initial prices takes O(N log N).
- For each day, binary searching for the number of affordable shops takes O(log N).
- The maximum number of days is limited by X and the minimum initial price, so the total complexity depends on these values.
- To optimize, you can precompute the maximum number of days as (X - min(initial_price)), limiting the loop.
11. Summary:
- Model the price growth over days and use sorting and binary search to efficiently count affordable shops per day.
- Sum these counts over all valid days to find the total number of bottles bought.
- This approach avoids simulating every purchase explicitly and uses the problem's linear price increase property for efficient computation.
By thinking carefully about how prices evolve and using sorted data with binary searches, you can solve the problem efficiently and correctly.
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