Introduction

In this blog post, we will explore how to solve the LeetCode problem "2398" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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. 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

Explanation

Here's the breakdown of the approach, complexities, and the Python code:

• High-Level Approach:

  • Use a sliding window technique to iterate through all possible consecutive subarrays of robots.
  • For each window, efficiently calculate max(chargeTimes) and sum(runningCosts). Use a deque (double-ended queue) to maintain the maximum charge time within the current window in O(1) time. Calculate the sum of running costs in O(1) by adding and subtracting running costs as the window slides.
  • Check if the total cost for each window is within the budget. If yes, update the maximum length.

• Complexity:

  • Runtime Complexity: O(n), where n is the number of robots.
  • Storage Complexity: O(n) due to the deque.

Code

    from collections import deque

def maximumRobots(chargeTimes, runningCosts, budget):
    """
    Calculates the maximum number of consecutive robots that can be run within the given budget.

    Args:
        chargeTimes: A list of charge times for each robot.
        runningCosts: A list of running costs for each robot.
        budget: The total budget.

    Returns:
        The maximum number of consecutive robots that can be run.
    """

    n = len(chargeTimes)
    max_robots = 0
    current_sum = 0
    max_charge_deque = deque()  # Stores indices of chargeTimes in decreasing order
    left = 0

    for right in range(n):
        # Add the current running cost to the sum
        current_sum += runningCosts[right]

        # Maintain the deque for max charge time
        while max_charge_deque and chargeTimes[max_charge_deque[-1]] <= chargeTimes[right]:
            max_charge_deque.pop()
        max_charge_deque.append(right)

        # Check if the current window exceeds the budget
        while max_charge_deque and chargeTimes[max_charge_deque[0]] + (right - left + 1) * current_sum > budget:
            current_sum -= runningCosts[left]
            if max_charge_deque[0] == left:
                max_charge_deque.popleft()
            left += 1

        # Update the maximum number of robots
        max_robots = max(max_robots, right - left + 1)

    return max_robots