How to Solve Maximum Number of Robots Within Budget Leetcode Problem - Interview Coder Guide
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How to Solve Maximum Number of Robots Within Budget Problem

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Hard#2449
LeetCode Problem

Maximum Number of Robots Within Budget

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...

ArrayBinary SearchQueue+4 more

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Problem Breakdown

Understanding the Maximum Number of Robots Within Budget Problem

Let's break down this LeetCode problem and understand what makes it challenging in interview settings.

Problem Statement

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.

HardProblem #2449
LeetCode

Maximum Number of Robots Within Budget

Related Topics
ArrayBinary SearchQueueSliding WindowHeap (Priority Queue)Prefix SumMonotonic Queue
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Examples

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

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