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] <= 1struct Solution;
use std::i32;
impl Solution {
fn max_satisfied(customers: Vec<i32>, grumpy: Vec<i32>, x: i32) -> i32 {
let mut sum = 0;
let mut max = 0;
let n = customers.len();
for i in 0..n {
if grumpy[i] == 0 {
sum += customers[i];
}
}
let x = x as usize;
for i in 0..x {
if grumpy[i] == 1 {
sum += customers[i];
}
}
max = i32::max(sum, max);
for i in x..n {
if grumpy[i] == 1 {
sum += customers[i];
}
if grumpy[i - x] == 1 {
sum -= customers[i - x];
}
max = i32::max(sum, max);
}
max
}
}
#[test]
fn test() {
let customers = vec![1, 0, 1, 2, 1, 1, 7, 5];
let grumpy = vec![0, 1, 0, 1, 0, 1, 0, 1];
let x = 3;
let res = 16;
assert_eq!(Solution::max_satisfied(customers, grumpy, x), res);
}