Leetcode #2483: Minimum Penalty for a Shop
In this guide, we solve Leetcode #2483 Minimum Penalty for a Shop in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: String, Prefix Sum
Intuition
Range queries become simple once we precompute cumulative sums.
We can transform subarray conditions into prefix comparisons.
Approach
Compute prefix sums and use a map to find matching prefixes.
This avoids nested loops while keeping the logic clear.
Steps:
- Compute prefix sums.
- Use a map to find valid ranges.
- Update the answer.
Example
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.
Python Solution
class Solution:
def bestClosingTime(self, customers: str) -> int:
ans = 0
mn = cost = customers.count("Y")
for j, c in enumerate(customers, 1):
cost += 1 if c == "N" else -1
if cost < mn:
ans, mn = j, cost
return ans
Complexity
The time complexity is , where is the length of the string . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.