Algorithm
Problem Name: 1052. Grumpy Bookstore Owner
There is a bookstore owner that has a store open for n minutes. Every minute, some number of customers enter the store. You are given an integer array customers of length n where customers[i] is the number of the customer that enters the store at the start of the ith minute and all those customers leave after the end of that minute.
On some 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 of that minute are not satisfied, otherwise, they are satisfied.
The bookstore owner knows a secret technique to keep themselves not grumpy for minutes consecutive minutes, 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], 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.length1 <= minutes <= n <= 2 * 1040 <= customers[i] <= 1000grumpy[i]is either0or1.
Code Examples
#1 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int maxSatisfied(int[] customers, int[] grumpy, int minutes) {
int totalSum = 0;
int maxDiff = 0;
int grumpySum = 0;
for (int i = 0; i < customers.length; i++) {
totalSum += grumpy[i] == 1 ? 0 : customers[i];
grumpySum += grumpy[i] == 1 ? customers[i] : 0;
if (i >= minutes) {
grumpySum -= grumpy[i - minutes] == 1 ? customers[i - minutes] : 0;
}
maxDiff = Math.max(maxDiff, grumpySum);
}
return totalSum + maxDiff;
}
}
Input
Output
#2 Code Example with Javascript Programming
Code -
Javascript Programming
const maxSatisfied = function(customers, grumpy, X) {
if (customers.length === 1) return customers[0]
const totalSatisfiedCustomers = customers.reduce(
(ac, el, idx) => ac + (grumpy[idx] === 0 ? el : 0),
0
)
const arr = customers.map((el, idx) => (grumpy[idx] === 1 ? el : 0))
const acArr = []
let ac = 0
for (let i = 0, len = arr.length; i < len; i++) {
acArr[i] = ac = ac + arr[i]
}
let max = 0
for (let i = X - 1, len = grumpy.length; i < len; i++) {
let tmp = i - X < 0 ? 0 : acArr[i - X]
if (acArr[i] - tmp > max) max = acArr[i] - tmp
}
return totalSatisfiedCustomers + max
}
Input
Output
#3 Code Example with Python Programming
Code -
Python Programming
class Solution:
def maxSatisfied(self, customers: List[int], grumpy: List[int], x: int) -> int:
dif = mx = sum(c * g for c, g in zip(customers[:x], grumpy[:x]))
for j in range(x, len(grumpy)):
dif += (grumpy[j] * customers[j]) - (grumpy[j - x] * customers[j - x])
mx = max(mx, dif)
return mx + sum(c * (1- g) for c, g in zip(customers, grumpy))
Input
Output