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
Related Topics:
Array, Sliding Window
// OJ: https://leetcode.com/problems/grumpy-bookstore-owner/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
int maxSatisfied(vector<int>& customers, vector<int>& grumpy, int X) {
int ans = 0, sum = 0;
for (int i = 0; i < customers.size(); ++i) sum += grumpy[i] == 0 ? customers[i] : 0;
for (int i = 0; i < customers.size(); ++i) {
if (grumpy[i] == 1) sum += customers[i];
if (i >= X && grumpy[i - X] == 1) sum -= customers[i - X];
if (i >= X - 1) ans = max(ans, sum);
}
return ans;
}
};