2483. Minimum Penalty For A Shop
class Solution:
"""LeetCode 2483: Minimum Penalty for a Shop.
PROBLEM DESCRIPTION:
-------------------
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 ith character is 'Y', it means that customers come at the ith hour
- whereas 'N' indicates that no customers come at the ith hour.
If the shop closes at the jth 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 jth hour, it means the shop is closed at the hour j.
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 <= 10^5
- customers consists only of characters 'Y' and 'N'.
SOLUTION APPROACH:
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The key insight is that we can calculate the penalty for each possible closing time
incrementally, rather than recalculating from scratch each time.
1. Initial Setup:
- Start by assuming we close at hour 0 (immediately)
- The penalty for this is the count of all 'Y's (customers we miss)
2. Incremental Updates:
- As we iterate through each hour, we update the penalty for closing at the next hour
- If current hour has 'Y': closing one hour later REDUCES penalty by 1
(we serve this customer instead of missing them)
- If current hour has 'N': closing one hour later INCREASES penalty by 1
(we're open during an hour with no customers)
3. Track Minimum:
- Keep track of the minimum penalty seen and the corresponding closing time
- Return the earliest hour that achieves minimum penalty
Time Complexity: O(n) where n is the length of customers string
Space Complexity: O(1) as we only use constant extra space
Example walkthrough with "YYNY":
- Close at 0: penalty = 4 Y's = 4 (but wait, we need to count properly)
Actually: penalty = count of Y's after position 0 = 3
- After hour 0 (Y): penalty = 3 - 1 = 2 (close at 1)
- After hour 1 (Y): penalty = 2 - 1 = 1 (close at 2) ← minimum!
- After hour 2 (N): penalty = 1 + 1 = 2 (close at 3)
- After hour 3 (Y): penalty = 2 - 1 = 1 (close at 4)
Result: 2 (earliest hour with minimum penalty of 1)
"""
def bestClosingTime(self, customers: str) -> int:
# Start by counting the penalty of closing at 0
penalty = customers.count("Y")
best_penalty = penalty
best_j = 0
for idx, customer in enumerate(customers):
# After each customer, the penalty of closing at next idx
# is reduced, while it is increased the penalty if no one arrives
if customer == "Y":
penalty -= 1
else: # 'N'
penalty += 1
j = idx + 1 # closing after hour i
if penalty < best_penalty:
best_penalty = penalty
best_j = j
return best_j