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Formatted question description: https://leetcode.ca/all/1052.html

1052. Grumpy Bookstore Owner (Medium)

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 <= 20000
  • 0 <= customers[i] <= 1000
  • 0 <= grumpy[i] <= 1

Related Topics:
Array, Sliding Window

Solution 1. Sliding Window

// OJ: https://leetcode.com/problems/grumpy-bookstore-owner/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int maxSatisfied(vector<int>& customers, vector<int>& grumpy, int X) {
        int ans = 0, sum = 0;
        for (int i = 0; i < customers.size(); ++i) sum += grumpy[i] == 0 ? customers[i] : 0;
        for (int i = 0; i < customers.size(); ++i) {
            if (grumpy[i] == 1) sum += customers[i];
            if (i >= X && grumpy[i - X] == 1) sum -= customers[i - X];
            if (i >= X - 1) ans = max(ans, sum);
        }
        return ans;
    }
};

  • class Solution {
    public int maxSatisfied(int[] customers, int[] grumpy, int X) {
        int satisfied = 0;
        int length = customers.length;
        for (int i = 0; i < length; i++) {
            if (grumpy[i] == 0)
                satisfied += customers[i];
        }
        int unsatisfiedWindow = 0;
        for (int i = 0; i < X; i++) {
            if (grumpy[i] == 1)
                unsatisfiedWindow += customers[i];
        }
        int maxUnsatisfiedWindow = unsatisfiedWindow;
        for (int i = X; i < length; i++) {
            int prevIndex = i - X;
            if (grumpy[prevIndex] == 1)
                unsatisfiedWindow -= customers[prevIndex];
            if (grumpy[i] == 1)
                unsatisfiedWindow += customers[i];
            maxUnsatisfiedWindow = Math.max(maxUnsatisfiedWindow, unsatisfiedWindow);
        }
        return satisfied + maxUnsatisfiedWindow;
    }
}

############

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;
    }
}

  • // OJ: https://leetcode.com/problems/grumpy-bookstore-owner/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int maxSatisfied(vector<int>& customers, vector<int>& grumpy, int X) {
        int ans = 0, sum = 0;
        for (int i = 0; i < customers.size(); ++i) sum += grumpy[i] == 0 ? customers[i] : 0;
        for (int i = 0; i < customers.size(); ++i) {
            if (grumpy[i] == 1) sum += customers[i];
            if (i >= X && grumpy[i - X] == 1) sum -= customers[i - X];
            if (i >= X - 1) ans = max(ans, sum);
        }
        return ans;
    }
};

  • 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

############

# 1052. Grumpy Bookstore Owner
# https://leetcode.com/problems/grumpy-bookstore-owner/

class Solution:
    def maxSatisfied(self, customers: List[int], grumpy: List[int], minutes: int) -> int:
        n = len(customers)
        A = [i for i in range(n) if grumpy[i] == 1]

        res = customers[A[0]] if A else 0
        if A:
            curr, start = customers[A[0]], 0
            for index in A[1:]:
                if index - A[start] < minutes:
                    curr += customers[index]
                else:
                    while index - A[start] >= minutes:
                        curr -= customers[A[start]]
                        start += 1
                    curr += customers[index]

                res = max(res, curr)
        
        return res + sum(customers[i] for i in range(n) if grumpy[i] == 0)


  • 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
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

  • 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
    }
}

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