- Difficulty: Hard
- Tags: LeetCode, Hard, Queue, Array, Binary Search, Prefix Sum, Sliding Window, Heap (Priority Queue), leetcode-2398, O(n), Mono Deque, Two Pointers
Problem
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