Leetcode #1052: Grumpy Bookstore Owner
In this guide, we solve Leetcode #1052 Grumpy Bookstore Owner 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
There is a bookstore owner that has a store open for n minutes. You are given an integer array customers of length n where customers[i] is the number of the customers that enter the store at the start of the ith minute and all those customers leave after the end of that minute.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, Sliding Window
Intuition
We are looking for a contiguous region that satisfies a constraint, which is a classic sliding-window signal.
Expanding and shrinking the window lets us maintain validity without restarting the scan.
Approach
Grow the window with a right pointer, and shrink from the left only when the constraint is violated.
Track the best window as you go to keep the solution linear.
Steps:
- Expand the right end of the window.
- While invalid, move the left end to restore constraints.
- Update the best window found.
Python Solution
class Solution:
def maxSatisfied(
self, customers: List[int], grumpy: List[int], minutes: int
) -> int:
mx = cnt = sum(c * g for c, g in zip(customers[:minutes], grumpy))
for i in range(minutes, len(customers)):
cnt += customers[i] * grumpy[i]
cnt -= customers[i - minutes] * grumpy[i - minutes]
mx = max(mx, cnt)
return sum(c * (g ^ 1) for c, g in zip(customers, grumpy)) + mx
Complexity
The time complexity is , where is the length of the array customers. 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.