Maximum Number of Robots Within Budget

arraybinary searchqueue

Problem Statement

You have n robots. You are given two 0-indexed integer arrays, chargeTimes and runningCosts , both of length n . The i ^th 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 .

Examples

Example 1

Input: chargeTimes = [3,6,1,3,4], runningCosts = [2,1,3,4,5], budget = 25
Output: 3
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
No robot can be run that does not exceed the budget, so we return 0.

Constraints

chargeTimes.length == runningCosts.length == n

1 <= n <= 5 * 10 ^4

1 <= chargeTimes[i], runningCosts[i] <= 10 ^5

1 <= budget <= 10 ^15

😤

Hard

Difficulty

Topic Info

ModuleDSA

CategoryArray

Sub-topicBinary Search