Problem Statement
In this scenario, we are tasked with managing a shop's operating hours based on a record of customer visits for each hour. The record, represented as a string customers, uses 'Y' to signify visits during an hour and 'N' for hours without visits.
Our goal is to determine the earliest hour to close the shop to minimize the penalties incurred from being open during hours with no visitors and being closed during hours with visitors. Each unoptimized hour—whether it's an open hour without visitors or a closed hour with visitors—adds a penalty of one point.
For instance, if we close the shop at hour "j", we accumulate penalties for all 'N' characters (no customers) from the start up to hour "j-1" (because the shop remains open during these hours) and for all 'Y' characters (customers present) from hour "j" to the end of the string (as the shop would be closed then). The challenge is thus to analyze different closing times from 0 to n and choose the earliest one that results in the least penalty.
Examples
Example 1
Input:
customers = "YYNY"
Output:
2
Explanation:
- Closing the shop at the 0th hour incurs in 1+1+0+1 = 3 penalty. - Closing the shop at the 1st hour incurs in 0+1+0+1 = 2 penalty. - Closing the shop at the 2nd hour incurs in 0+0+0+1 = 1 penalty. - Closing the shop at the 3rd hour incurs in 0+0+1+1 = 2 penalty. - Closing the shop at the 4th hour incurs in 0+0+1+0 = 1 penalty. Closing the shop at 2nd or 4th 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 0th hour as no customers arrive.
Example 3
Input:
customers = "YYYY"
Output:
4
Explanation:
It is best to close the shop at the 4th hour as customers arrive at each hour.
Constraints
1 <= customers.length <= 105customersconsists only of characters'Y'and'N'.
Approach and Intuition
To solve this problem efficiently, follow these steps:
- Calculate the total number of 'Y' characters (this represents the total customers visiting all day).
- Keep a running count of penalties incurred if the shop was open from the start until just before each hour i.e., compute penalties for all possible closing times.
- Start with a penalty of all customers (all 'Y's) being missed as if the shop was closed from hour 0 onward.
- Iterate over each character in the
customersstring:- Decrease the penalty for hours after the current if it's a 'Y'—this is because we are considering the shop to close later, hence one less hour will have a penalty of a missed customer.
- Increase the penalty for the current hour if it's 'N'—this represents an extra penalty for being open an additional hour without customers.
- Track the minimum penalty observed and the earliest hour at which this minimum occurs.
- The approach leverages cumulative additions and subtractions as the loop progresses to adjust the open and closed hours' penalties dynamically, rather than recalculating from scratch for each possible closing time.
- By keeping the operations linear with respect to the number of hours/customer entries (length of the
customersstring), the solution remains efficient and scalable.
This method ensures a time complexity of O(n), where n is the length of the customers string, allowing it to handle large inputs effectively within the given constraints.
Solutions
- C++
- Java
- Python
class Solution {
public:
int calculateOptimalClosingTime(string customerStatus) {
int bestPenalty = 0, currentPenalty = 0;
int optimalHour = 0;
for (int hour = 0; hour < customerStatus.length(); hour++) {
if (customerStatus[hour] == 'Y') {
currentPenalty--;
} else {
currentPenalty++;
}
if (currentPenalty < bestPenalty) {
optimalHour = hour + 1;
bestPenalty = currentPenalty;
}
}
return optimalHour;
}
};
The provided solution in C++ addresses the issue of determining the minimum penalty time for a shop based on its customers' presence. This is achieved by calculating the optimal closing time such that the penalty is minimized.
- Define the
calculateOptimalClosingTimefunction which receives a stringcustomerStatusindicating hourly customer presence ('Y' for present, 'N' for absent). - Initialize
bestPenaltyto 0 to track the lowest penalty encountered andcurrentPenaltyto manage the running penalty calculation. - Employ a loop to iterate through each hour, represented by each character in
customerStatus. - Update
currentPenaltywith a decrement for 'Y' and an increment for 'N'. - Within the loop, check if the
currentPenaltyis less thanbestPenalty. If so, updateoptimalHourto the current hour plus one and setbestPenaltyto this newcurrentPenalty. - After iterating through the hours, the function concludes by returning
optimalHour, the hour with the minimum penalty.
Key points to consider:
- The penalty decreases when there are customers (denoted 'Y') and increases when absent ('N').
- Tracking the smallest penalty during iteration helps determine the earliest time the shop should close to minimize its penalty.
- The optimal closing time is not the end of the string length but is rather the hour after the minimum penalty hour since hours typically start counting from 0 in programming.
class Solution {
public int calculateOptimalTime(String visitorData) {
int lowestCumulativePenalty = 0, currentPenalty = 0;
int optimalHour = 0;
for (int index = 0; index < visitorData.length(); index++) {
char visitorStatus = visitorData.charAt(index);
if (visitorStatus == 'Y') {
currentPenalty--;
} else {
currentPenalty++;
}
if (currentPenalty < lowestCumulativePenalty) {
optimalHour = index + 1;
lowestCumulativePenalty = currentPenalty;
}
}
return optimalHour;
}
}
The provided Java solution is designed to find the minimum penalty time for a shop based on the given visitor data. The calculateOptimalTime method processes a string where each character ('Y' or 'N') represents whether a visitor's experience was positive or negative, respectively.
Here’s how the algorithm works:
- Initialize
lowestCumulativePenaltyandcurrentPenaltyat zero andoptimalHouralso as zero. - Iterate through each character of the
visitorDatastring. - If the character is 'Y', decrement the
currentPenaltyby one. If it's 'N', increment thecurrentPenalty. - Continuously update the
lowestCumulativePenaltyandoptimalHourif a new lowest penalty is found during each iteration.
This approach effectively tracks the point in time (or after the optimal hour) with the least cumulative penalty. The function returns the optimalHour which indicates the best time to minimize negative impact due to the previous visitor data. This could be used to determine strategic operational adjustments or improvements in customer service.
class Solution:
def earliestClosing(self, schedule: str) -> int:
optimal_time = 0
least_penalty = 0
current_penalty = 0
for index, status in enumerate(schedule):
if status == "Y":
current_penalty -= 1
else:
current_penalty += 1
if current_penalty < least_penalty:
least_penalty = current_penalty
optimal_time = index + 1
return optimal_time
Find the optimal closing time for a shop to minimize penalty based on customer arrival and departure times. The method earliestClosing accepts a string schedule, where each character represents a minute and is either 'Y' (indicating customer presence) or 'N' (indicating no customers). The goal is to determine the minimum index where closing the shop afterwards incurs the least customer dissatisfaction.
- Initialize
optimal_timeto zero, which tracks the best minute to close the shop. - Start with
least_penaltyandcurrent_penaltyas zero. - Iterate through each minute in the
schedulestring:- Decrease
current_penaltyby one for 'Y' (a customer leaves unhappily if closed). - Increase
current_penaltyby one for 'N' (benefit as no customer is disturbed). - If
current_penaltydrops belowleast_penalty, updateleast_penaltyand setoptimal_timeto the current minute index plus one.
- Decrease
- Return
optimal_timereflecting the minute to minimize negative impact on customers.
No comments yet.