Grumpy Bookstore Owner

Description

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

Solution(javascript)

/*
 * @lc app=leetcode id=1052 lang=javascript
 *
 * [1052] Grumpy Bookstore Owner
 */

// @lc code=start
/**
 * @param {number[]} customers
 * @param {number[]} grumpy
 * @param {number} X
 * @return {number}
 */
const maxSatisfied = function (customers, grumpy, X) {
  let sum = customers.reduce((acc, num, index) => {
    if (grumpy[index] === 0) {
      return acc + num
    }
    return acc
  }, 0)
  let max = sum
  for (let i = 0; i < X; i++) {
    if (grumpy[i] === 1) {
      sum += customers[i]
    }
  }
  max = Math.max(max, sum)
  for (let i = X; i < customers.length; i++) {
    if (grumpy[i] === 1) {
      sum += customers[i]
    }
    if (grumpy[i - X] === 1) {
      sum -= customers[i - X]
    }
    max = Math.max(max, sum)
  }
  return max
}
// @lc code=end