Skip to content

2398. Maximum Number of Robots Within Budget

Difficulty Topics

Description

You have n robots. You are given two 0-indexed integer arrays, chargeTimes and runningCosts, both of length n. The ith robot costs chargeTimes[i] units to charge and costs runningCosts[i] units to run. You are also given an integer budget.

The total cost of running k chosen robots is equal to max(chargeTimes) + k * sum(runningCosts), where max(chargeTimes) is the largest charge cost among the k robots and sum(runningCosts) is the sum of running costs among the k robots.

Return the maximum number of consecutive robots you can run such that the total cost does not exceed budget.

 

Example 1:

Input: chargeTimes = [3,6,1,3,4], runningCosts = [2,1,3,4,5], budget = 25
Output: 3
Explanation: 
It is possible to run all individual and consecutive pairs of robots within budget.
To obtain answer 3, consider the first 3 robots. The total cost will be max(3,6,1) + 3 * sum(2,1,3) = 6 + 3 * 6 = 24 which is less than 25.
It can be shown that it is not possible to run more than 3 consecutive robots within budget, so we return 3.

Example 2:

Input: chargeTimes = [11,12,19], runningCosts = [10,8,7], budget = 19
Output: 0
Explanation: No robot can be run that does not exceed the budget, so we return 0.

 

Constraints:

  • chargeTimes.length == runningCosts.length == n
  • 1 <= n <= 5 * 104
  • 1 <= chargeTimes[i], runningCosts[i] <= 105
  • 1 <= budget <= 1015

Solution

maximum-number-of-robots-within-budget.py
from sortedcontainers import SortedList

class Solution:
    def maximumRobots(self, chargeTimes: List[int], runningCosts: List[int], budget: int) -> int:
        n = len(chargeTimes)
        largest = SortedList()
        res = i = costs = 0

        for j in range(n):
            largest.add(chargeTimes[j])
            costs += runningCosts[j]

            total = largest[-1] + (j - i + 1) * costs

            while total > budget:
                largest.remove(chargeTimes[i])
                costs -= runningCosts[i]
                i += 1
                total = largest[-1] if largest else 0 + (j - i + 1) * costs

            res = max(res, j - i + 1)

        return res