Table of Contents

## Maximum Number of Robots Within Budget solution leetcode

**Maximum Number of Robots Within Budget solution leetcode** – You have `n`

robots. You are given two **0-indexed** integer arrays, `chargeTimes`

and `runningCosts`

, both of length `n`

. The `i`

robot costs ^{th}`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`

.

## Maximum Number of Robots Within Budget solution leetcode

Input:chargeTimes = [3,6,1,3,4], runningCosts = [2,1,3,4,5], budget = 25Output:3Explanation: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.

## Maximum Number of Robots Within Budget solution leetcode

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

## Maximum Number of Robots Within Budget solution leetcode

`chargeTimes.length == runningCosts.length == n`

`1 <= n <= 5 * 10`

^{4}`1 <= chargeTimes[i], runningCosts[i] <= 10`

^{5}`1 <= budget <= 10`

^{15}