Today, the bookstore owner has a store open for customers.length minutes. Every minute, some number of customers (customers[i]) enter the store, and all those customers leave after the end of that minute.
On some minutes, the bookstore owner is grumpy. If the bookstore owner is grumpy on the i-th minute, grumpy[i] = 1, otherwise grumpy[i] = 0. When the bookstore owner is grumpy, the customers of that minute are not satisfied, otherwise they are satisfied.
The bookstore owner knows a secret technique to keep themselves not grumpy for X minutes straight, but can only use it once.
Return the maximum number of customers that can be satisfied throughout the day.
Example 1:
Input: customers = [1,0,1,2,1,1,7,5], grumpy = [0,1,0,1,0,1,0,1], X = 3 Output: 16 Explanation: The bookstore owner keeps themselves not grumpy for the last 3 minutes. The maximum number of customers that can be satisfied = 1 + 1 + 1 + 1 + 7 + 5 = 16.
Note:
1 <= X <= customers.length == grumpy.length <= 200000 <= customers[i] <= 10000 <= grumpy[i] <= 1
Solution: Sliding Window
Sum the costumers of non grumpy minutes, recording the max sum of the sliding window of size X.
Time complexity: O(n)
Space complexity: o(n)
C++
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// Author: Huahua class Solution { public: int maxSatisfied(vector<int>& customers, vector<int>& grumpy, int X) { int base = 0; int window = 0; int best_window = 0; for (int i = 0; i < grumpy.size(); ++i) { (grumpy[i] ? window : base) += customers[i]; if (i >= X && grumpy[i - X]) window -= customers[i - X]; best_window = max(best_window, window); } return base + best_window; } }; |
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