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
class Solution:
def maxSatisfied(self, customers: List[int], grumpy: List[int], X: int) -> int:
s = t = 0
win, n = 0, len(customers)
for i in range(n):
if grumpy[i] == 0:
s += customers[i]
else:
win += customers[i]
if i >= X and grumpy[i - X] == 1:
win -= customers[i - X]
t = max(t, win)
return s + tclass Solution {
public int maxSatisfied(int[] customers, int[] grumpy, int X) {
int s = 0, t = 0;
for (int i = 0, win = 0, n = customers.length; i < n; ++i) {
if (grumpy[i] == 0) {
s += customers[i];
} else {
win += customers[i];
}
if (i >= X && grumpy[i - X] == 1) {
win -= customers[i - X];
}
t = Math.max(t, win);
}
return s + t;
}
}