Question

1052. 爱生气的书店老板

English Version

题目描述

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 * 104
• 0 <= customers[i] <= 1000
• grumpy[i] == 0 or 1

解法

方法一：滑动窗口

定义 $s$ 表示所有不满意的顾客总数，$cs$ 表示顾客总数。

class Solution:
  def maxSatisfied(
    self, customers: List[int], grumpy: List[int], minutes: int
  ) -> int:
    s = sum(a * b for a, b in zip(customers, grumpy))
    cs = sum(customers)
    t = ans = 0
    for i, (a, b) in enumerate(zip(customers, grumpy), 1):
      t += a * b
      if (j := i - minutes) >= 0:
        ans = max(ans, cs - (s - t))
        t -= customers[j] * grumpy[j]
    return ans

class Solution {
  public int maxSatisfied(int[] customers, int[] grumpy, int minutes) {
    int s = 0, cs = 0;
    int n = customers.length;
    for (int i = 0; i < n; ++i) {
      s += customers[i] * grumpy[i];
      cs += customers[i];
    }
    int t = 0, ans = 0;
    for (int i = 0; i < n; ++i) {
      t += customers[i] * grumpy[i];
      int j = i - minutes + 1;
      if (j >= 0) {
        ans = Math.max(ans, cs - (s - t));
        t -= customers[j] * grumpy[j];
      }
    }
    return ans;
  }
}

class Solution {
public:
  int maxSatisfied(vector<int>& customers, vector<int>& grumpy, int minutes) {
    int s = 0, cs = 0;
    int n = customers.size();
    for (int i = 0; i < n; ++i) {
      s += customers[i] * grumpy[i];
      cs += customers[i];
    }
    int t = 0, ans = 0;
    for (int i = 0; i < n; ++i) {
      t += customers[i] * grumpy[i];
      int j = i - minutes + 1;
      if (j >= 0) {
        ans = max(ans, cs - (s - t));
        t -= customers[j] * grumpy[j];
      }
    }
    return ans;
  }
};

func maxSatisfied(customers []int, grumpy []int, minutes int) int {
  s, cs := 0, 0
  for i, c := range customers {
    s += c * grumpy[i]
    cs += c
  }
  t, ans := 0, 0
  for i, c := range customers {
    t += c * grumpy[i]
    j := i - minutes + 1
    if j >= 0 {
      ans = max(ans, cs-(s-t))
      t -= customers[j] * grumpy[j]
    }
  }
  return ans
}

impl Solution {
  pub fn max_satisfied(customers: Vec<i32>, grumpy: Vec<i32>, minutes: i32) -> i32 {
    let k = minutes as usize;
    let n = customers.len();

    let mut sum = 0;
    for i in 0..k {
      if grumpy[i] == 1 {
        sum += customers[i];
      }
    }
    let mut max = sum;
    for i in k..n {
      if grumpy[i - k] == 1 {
        sum -= customers[i - k];
      }
      if grumpy[i] == 1 {
        sum += customers[i];
      }
      max = max.max(sum);
    }

    sum = 0;
    for i in 0..n {
      if grumpy[i] == 0 {
        sum += customers[i];
      }
    }
    sum + max
  }
}