Minimum Penalty for a Shop

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ArraysTwo PointersSliding Windows

You are given the customer visit log of a shop represented by a 0-indexed string customers consisting only of characters 'N' and 'Y':

if the i^th character is 'Y', it means that customers come at the i^th hour
whereas 'N' indicates that no customers come at the i^th hour.

If the shop closes at the j^th hour (0 <= j <= n), the penalty is calculated as follows:

For every hour when the shop is open and no customers come, the penalty increases by 1.
For every hour when the shop is closed and customers come, the penalty increases by 1.

Return the earliest hour at which the shop must be closed to incur a minimum penalty.

Note that if a shop closes at the j^th hour, it means the shop is closed at the hour j.

Example 1:

Input: customers = "YYNY"
Output: 2
Explanation: 
- Closing the shop at the 0^th hour incurs in 1+1+0+1 = 3 penalty.
- Closing the shop at the 1^st hour incurs in 0+1+0+1 = 2 penalty.
- Closing the shop at the 2^nd hour incurs in 0+0+0+1 = 1 penalty.
- Closing the shop at the 3^rd hour incurs in 0+0+1+1 = 2 penalty.
- Closing the shop at the 4^th hour incurs in 0+0+1+0 = 1 penalty.
Closing the shop at 2^nd or 4^th hour gives a minimum penalty. Since 2 is earlier, the optimal closing time is 2.

Example 2:

Input: customers = "NNNNN"
Output: 0
Explanation: It is best to close the shop at the 0^th hour as no customers arrive.

Example 3:

Input: customers = "YYYY"
Output: 4
Explanation: It is best to close the shop at the 4^th hour as customers arrive at each hour.

Constraints:

1 <= customers.length <= 10⁵
customers consists only of characters 'Y' and 'N'.

Try coding it in LeetCode

Minimum Penalty for a Shop

Medium

Stripe

8 views

Topics:

ArraysTwo PointersSliding Windows

You are given the customer visit log of a shop represented by a 0-indexed string customers consisting only of characters 'N' and 'Y':

if the i^th character is 'Y', it means that customers come at the i^th hour
whereas 'N' indicates that no customers come at the i^th hour.

If the shop closes at the j^th hour (0 <= j <= n), the penalty is calculated as follows:

For every hour when the shop is open and no customers come, the penalty increases by 1.
For every hour when the shop is closed and customers come, the penalty increases by 1.

Return the earliest hour at which the shop must be closed to incur a minimum penalty.

Note that if a shop closes at the j^th hour, it means the shop is closed at the hour j.

Example 1:

Input: customers = "YYNY"
Output: 2
Explanation: 
- Closing the shop at the 0^th hour incurs in 1+1+0+1 = 3 penalty.
- Closing the shop at the 1^st hour incurs in 0+1+0+1 = 2 penalty.
- Closing the shop at the 2^nd hour incurs in 0+0+0+1 = 1 penalty.
- Closing the shop at the 3^rd hour incurs in 0+0+1+1 = 2 penalty.
- Closing the shop at the 4^th hour incurs in 0+0+1+0 = 1 penalty.
Closing the shop at 2^nd or 4^th hour gives a minimum penalty. Since 2 is earlier, the optimal closing time is 2.

Example 2:

Input: customers = "NNNNN"
Output: 0
Explanation: It is best to close the shop at the 0^th hour as no customers arrive.

Example 3:

Input: customers = "YYYY"
Output: 4
Explanation: It is best to close the shop at the 4^th hour as customers arrive at each hour.

Constraints:

1 <= customers.length <= 10⁵
customers consists only of characters 'Y' and 'N'.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum length of the `customers` string?
Can the `customers` string be empty or null?
Does the `customers` string contain only 'Y' and 'N' characters, or can there be other characters?
If there are multiple hours that result in the same minimum penalty, can I return any one of them, or is there a specific preference (e.g., the earliest hour)?
Can you provide a few more example inputs and their expected outputs to ensure I fully understand the penalty calculation logic?

Brute Force Solution

Approach

The brute force approach to this problem involves checking every possible opening day for the shop. We will try opening the shop on each day and then calculate the total penalty incurred for that choice.

Here's how the algorithm would work step-by-step:

Imagine trying to open the shop on the very first day.
Now, go through each day and see if the shop is open or closed on that day according to the plan we are testing.
Calculate the penalty. If the shop is open and it should be closed, add to the penalty. If the shop is closed and it should be open, add to the penalty.
Record the total penalty for opening the shop on the first day.
Next, imagine opening the shop on the second day instead. Repeat the penalty calculation process.
Continue this process, each time assuming the shop opens on a different day.
Once we have calculated the total penalty for every possible opening day, compare all the penalty scores we've recorded.
Choose the opening day that results in the smallest total penalty. That is our answer.

Code Implementation

def minimum_penalty_shop_brute_force(customers):
    minimum_total_penalty = float('inf')
    best_opening_hour = -1

    # Iterate through each possible opening hour
    for opening_hour_index in range(len(customers) + 1):
        current_total_penalty = 0

        # Iterate through each hour to calculate penalty
        for current_hour_index in range(len(customers)):
            # Calculate penalty before opening
            if current_hour_index < opening_hour_index:

                # Add penalty if customer arrives but shop is closed
                if customers[current_hour_index] == 'Y':
                    current_total_penalty += 1

            # Calculate penalty after opening
            else:
                # Add penalty if customer does not arrive but shop is open
                if customers[current_hour_index] == 'N':
                    current_total_penalty += 1

        # Update minimum penalty and best opening hour if needed
        if current_total_penalty < minimum_total_penalty:
            minimum_total_penalty = current_total_penalty
            best_opening_hour = opening_hour_index

    return best_opening_hour

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through each possible opening day of the shop. For each potential opening day, it iterates through all days to calculate the penalty. Therefore, if 'n' is the number of days, the outer loop runs 'n' times, and the inner loop also runs 'n' times. This results in approximately n * n operations. Thus, the time complexity is O(n²).

Space Complexity

O(1) – The brute force approach iterates through each possible opening day and calculates the penalty. While iterating, it only needs to store a few variables such as the current opening day being tested and the calculated penalty for that opening day. The amount of extra memory needed to store these variables does not scale with the input size N (where N is the number of days). Therefore, the space complexity is constant.

Optimal Solution

Approach

The goal is to find the minimum penalty a shop incurs based on whether it's open or closed each day. The optimal approach involves calculating cumulative penalties for being open and closed and using those to efficiently determine the best day to close the shop for good.

Here's how the algorithm would work step-by-step:

Calculate how much penalty you'd accumulate if the shop were always open up to each day.
Calculate how much penalty you'd accumulate if the shop were always closed starting from each day (going backwards).
For each day, imagine closing the shop from that day onward. Calculate the total penalty: the penalty from always being open before that day PLUS the penalty from always being closed from that day onward.
Keep track of the minimum total penalty you find across all days.
The day that results in the absolute minimum total penalty is when the shop should be closed for good to minimize losses.

Code Implementation

def minimum_penalty(customers): 
    number_of_days = len(customers)
    open_penalty = [0] * (number_of_days + 1)
    closed_penalty = [0] * (number_of_days + 1)

    for i in range(number_of_days): 
        open_penalty[i + 1] = open_penalty[i] + (1 if customers[i] == 'N' else 0)

    # Accumulate penalties going backwards, closing the shop
    for i in range(number_of_days - 1, -1, -1):
        closed_penalty[i] = closed_penalty[i + 1] + (1 if customers[i] == 'Y' else 0)

    minimum_total_penalty = float('inf')
    best_closing_day = -1

    # Iterate to determine best closing day
    for i in range(number_of_days + 1):
        total_penalty = open_penalty[i] + closed_penalty[i]

        # Find smallest total penalty by closing on 'i'th day
        if total_penalty < minimum_total_penalty:
            minimum_total_penalty = total_penalty
            best_closing_day = i

    return best_closing_day

Big(O) Analysis

Time Complexity

O(n) – The algorithm calculates prefix sums for the 'always open' penalty in O(n) time. It similarly computes suffix sums for the 'always closed' penalty, also in O(n) time. Finally, it iterates through each day to determine the minimum total penalty, which takes O(n) time. Therefore, the dominant operation is a single pass through the input of size n, resulting in a linear time complexity.

Space Complexity

O(N) – The algorithm creates two auxiliary arrays, one to store the cumulative penalty for being always open up to each day and another to store the cumulative penalty for being always closed starting from each day. Both these arrays have a size proportional to N, where N is the number of days. Therefore, the dominant space usage is driven by these two arrays, resulting in O(N) space complexity.

Edge Cases

Case	How to Handle
Null or empty input string	Return 0 as there are no customers and no penalty can be incurred.
String containing characters other than 'Y' and 'N'	Throw an IllegalArgumentException as the input is invalid.
String with a single character 'Y'	Opening at hour 0 results in a penalty of 0 since we can open the shop to avoid it.
String with a single character 'N'	Opening at hour 0 results in a penalty of 1 since we will open the store with no customer.
String with all 'Y' characters	Opening at hour n (closing immediately) results in minimum (0) penalty.
String with all 'N' characters	Opening at hour 0 results in minimum (n) penalty which will be same for all open hours.
Maximum string length (scalability)	The prefix sum approach provides an O(n) solution, ensuring it scales well for large input sizes.
Integer overflow when calculating penalties with very long strings	Use long data type to store the penalty values to avoid potential overflow.