LeetCode 1052、爱生气的书店老板

有一个书店老板，他的书店开了 n 分钟。每分钟都有一些顾客进入这家商店。给定一个长度为 n 的整数数组 customers ，其中 customers[i] 是在第 i 分钟开始时进入商店的顾客数量，所有这些顾客在第 i 分钟结束后离开。

在某些时候，书店老板会生气。如果书店老板在第 i 分钟生气，那么 grumpy[i] = 1，否则 grumpy[i] = 0。

当书店老板生气时，那一分钟的顾客就会不满意，若老板不生气则顾客是满意的。

书店老板知道一个秘密技巧，能抑制自己的情绪，可以让自己连续 minutes 分钟不生气，但却只能使用一次。

请你返回 这一天营业下来，最多有多少客户能够感到满意 。

示例 1：

**输入：**customers = [1,0,1,2,1,1,7,5], grumpy = [0,1,0,1,0,1,0,1], minutes = 3 **输出：**16 **解释：**书店老板在最后 3 分钟保持冷静。感到满意的最大客户数量 = 1 + 1 + 1 + 1 + 7 + 5 = 16.

示例 2：

**输入：**customers = [1], grumpy = [0], minutes = 1 **输出：**1

提示：

n == customers.length == grumpy.length
1 <= minutes <= n <= 2 * 10^(4)
0 <= customers[i] <= 1000
grumpy[i] == 0 or 1

二、参考代码

Python

# 题目：LC1052. 爱生气的书店老板
# 难度：中等
# 作者：许老师-闭着眼睛学数理化
# 算法：固定滑窗
# 代码看不懂的地方，请直接在群上提问


class Solution:
    def maxSatisfied(self, customers: List[int], grumpy: List[int], minutes: int) -> int:
        # 假设书店老板无法控制不生气，那么不满意的客人数目是恒定的，
        # 为sum([customers[i]*grumpy[i] for i in range(n)])，n为书店开张分钟
        # 为了使得总的不满意的客人数目尽可能地小，
        # 书店老板在连续的minutes分钟中抑制不生气，使得不满意客人数目的减少量要尽可能地大
        # 因此把问题转化为了在【长度为minutes的滑动窗口中，使得不满意客人减少量尽可能地大】的滑动窗口问题
        
        unsatisfied = [c * g for c, g in zip(customers, grumpy)]
        win_sum = sum(unsatisfied[:minutes])
        max_win_sum = win_sum

        for right, num in enumerate(unsatisfied[minutes:], minutes):
            # A1
            win_sum += num
            # A2
            left = right - minutes
            win_sum -= unsatisfied[left]
            # A3
            max_win_sum = max(max_win_sum, win_sum)

        # 总人数 - 原本总不满意人数 + 克制住没生气挽回的不满意人数 = 最终满意人数
        return sum(customers) - sum(unsatisfied) + max_win_sum

Java

class Solution {
    public int maxSatisfied(int[] customers, int[] grumpy, int minutes) {
        int n = customers.length;
        int[] unsatisfied = new int[n];

        for (int i = 0; i < n; i++) {
            unsatisfied[i] = customers[i] * grumpy[i];
        }

        int winSum = 0;
        for (int i = 0; i < minutes; i++) {
            winSum += unsatisfied[i];
        }
        int maxWinSum = winSum;

        for (int right = minutes; right < n; right++) {
            // A1
            winSum += unsatisfied[right];
            // A2
            int left = right - minutes;
            winSum -= unsatisfied[left];
            // A3
            maxWinSum = Math.max(maxWinSum, winSum);
        }

        int totalCustomers = 0;
        for (int customer : customers) {
            totalCustomers += customer;
        }

        int totalUnsatisfied = 0;
        for (int unsat : unsatisfied) {
            totalUnsatisfied += unsat;
        }

        return totalCustomers - totalUnsatisfied + maxWinSum;
    }
}

C++

class Solution {
public:
    int maxSatisfied(vector<int>& customers, vector<int>& grumpy, int minutes) {
        int n = customers.size();
        vector<int> unsatisfied(n);

        for (int i = 0; i < n; i++) {
            unsatisfied[i] = customers[i] * grumpy[i];
        }

        int winSum = accumulate(unsatisfied.begin(), unsatisfied.begin() + minutes, 0);
        int maxWinSum = winSum;

        for (int right = minutes; right < n; right++) {
            // A1
            winSum += unsatisfied[right];
            // A2
            int left = right - minutes;
            winSum -= unsatisfied[left];
            // A3
            maxWinSum = max(maxWinSum, winSum);
        }

        int totalCustomers = accumulate(customers.begin(), customers.end(), 0);
        int totalUnsatisfied = accumulate(unsatisfied.begin(), unsatisfied.end(), 0);

        return totalCustomers - totalUnsatisfied + maxWinSum;
    }
};