README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/1000-1099/1052.Grumpy%20Bookstore%20Owner/README_EN.md	1418	Weekly Contest 138 Q2	Array Sliding Window

1052. Grumpy Bookstore Owner

中文文档

Description

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.

During certain minutes, the bookstore owner is grumpy. You are given a binary array grumpy where grumpy[i] is 1 if the bookstore owner is grumpy during the ith minute, and is 0 otherwise.

When the bookstore owner is grumpy, the customers entering during that minute are not satisfied. Otherwise, they are satisfied.

The bookstore owner knows a secret technique to remain not grumpy for minutes consecutive minutes, but this technique can only be used 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], minutes = 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.

Example 2:

Input: customers = [1], grumpy = [0], minutes = 1

Output: 1

 

Constraints:

• n == customers.length == grumpy.length
• 1 <= minutes <= n <= 2 * 104
• 0 <= customers[i] <= 1000
• grumpy[i] is either 0 or 1.

Solutions

Solution 1: Sliding Window

According to the problem description, we only need to count the number of customers when the boss is not angry $tot$, and add the maximum number of customers when the boss is angry within a sliding window of size minutes $mx$.

We define a variable $cnt$ to record the number of customers when the boss is angry within the sliding window, the initial value is the number of customers when the boss is angry in the first minutes. Then we traverse the array, each time we move the sliding window, we update the value of $cnt$, and at the same time update the value of $mx$.

Finally, return $tot + mx$.

The time complexity is $O(n)$, where $n$ is the length of the array customers. The space complexity is $O(1)$.

Python3

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

Java

class Solution {
    public int maxSatisfied(int[] customers, int[] grumpy, int minutes) {
        int cnt = 0;
        int tot = 0;
        for (int i = 0; i < minutes; ++i) {
            cnt += customers[i] * grumpy[i];
            tot += customers[i] * (grumpy[i] ^ 1);
        }
        int mx = cnt;
        int n = customers.length;
        for (int i = minutes; i < n; ++i) {
            cnt += customers[i] * grumpy[i];
            cnt -= customers[i - minutes] * grumpy[i - minutes];
            mx = Math.max(mx, cnt);
            tot += customers[i] * (grumpy[i] ^ 1);
        }
        return tot + mx;
    }
}

C++

class Solution {
public:
    int maxSatisfied(vector<int>& customers, vector<int>& grumpy, int minutes) {
        int cnt = 0;
        int tot = 0;
        for (int i = 0; i < minutes; ++i) {
            cnt += customers[i] * grumpy[i];
            tot += customers[i] * (grumpy[i] ^ 1);
        }
        int mx = cnt;
        int n = customers.size();
        for (int i = minutes; i < n; ++i) {
            cnt += customers[i] * grumpy[i];
            cnt -= customers[i - minutes] * grumpy[i - minutes];
            mx = max(mx, cnt);
            tot += customers[i] * (grumpy[i] ^ 1);
        }
        return tot + mx;
    }
};

Go

func maxSatisfied(customers []int, grumpy []int, minutes int) int {
	var cnt, tot int
	for i, c := range customers[:minutes] {
		cnt += c * grumpy[i]
		tot += c * (grumpy[i] ^ 1)
	}
	mx := cnt
	for i := minutes; i < len(customers); i++ {
		cnt += customers[i] * grumpy[i]
		cnt -= customers[i-minutes] * grumpy[i-minutes]
		mx = max(mx, cnt)
		tot += customers[i] * (grumpy[i] ^ 1)
	}
	return tot + mx
}

TypeScript

function maxSatisfied(customers: number[], grumpy: number[], minutes: number): number {
    let [cnt, tot] = [0, 0];
    for (let i = 0; i < minutes; ++i) {
        cnt += customers[i] * grumpy[i];
        tot += customers[i] * (grumpy[i] ^ 1);
    }
    let mx = cnt;
    for (let i = minutes; i < customers.length; ++i) {
        cnt += customers[i] * grumpy[i];
        cnt -= customers[i - minutes] * grumpy[i - minutes];
        mx = Math.max(mx, cnt);
        tot += customers[i] * (grumpy[i] ^ 1);
    }
    return tot + mx;
}

Rust

impl Solution {
    pub fn max_satisfied(customers: Vec<i32>, grumpy: Vec<i32>, minutes: i32) -> i32 {
        let mut cnt = 0;
        let mut tot = 0;
        let minutes = minutes as usize;
        for i in 0..minutes {
            cnt += customers[i] * grumpy[i];
            tot += customers[i] * (1 - grumpy[i]);
        }
        let mut mx = cnt;
        let n = customers.len();
        for i in minutes..n {
            cnt += customers[i] * grumpy[i];
            cnt -= customers[i - minutes] * grumpy[i - minutes];
            mx = mx.max(cnt);
            tot += customers[i] * (1 - grumpy[i]);
        }
        tot + mx
    }
}