Maximum Number of Robots Within Budget

Problem Description

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.

Key Insights

• The total cost is determined by the maximum charge time of the selected robots and the sum of their running costs.
• We need to find a sliding window of consecutive robots such that the total cost remains within budget.
• Efficiently managing the maximum charge time and the sum of running costs is crucial for optimizing performance.

Space and Time Complexity

Time Complexity: O(n)
Space Complexity: O(n)

Solution

To solve the problem, we can use a sliding window approach with two pointers to maintain the range of consecutive robots. We also utilize a max-heap (or a suitable data structure) to keep track of the maximum charge time efficiently. The algorithm works as follows:

1. Initialize two pointers, left and right, both set to the start of the robot arrays.
2. Maintain a variable to store the current sum of running costs and a structure to keep track of the maximum charge time in the current window.
3. Expand the right pointer to include more robots and update the current sum and maximum charge time.
4. If the total cost exceeds the budget, move the left pointer to shrink the window until the total cost is within budget.
5. Throughout the process, keep track of the maximum number of robots that can be run within the budget.

Code Solutions