Maximum Number of Robots Within Budget

Hard

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ArraysSliding Windows

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.

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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible ranges for the robot costs and battery power values?
Can the costs or battery power values be negative?
If no robots can be used within the budget, what value should be returned?
Is the input array guaranteed to be non-empty?
What data types are we using? For example, are costs and budget integers or floating point numbers?

Brute Force Solution

Approach

The brute force approach means checking every single possible group of robots to find the largest group that fits within the budget. We look at every possible starting point and length of a robot group, without skipping any.

Here's how the algorithm would work step-by-step:

Start by considering the first robot by itself.
Then consider the first two robots as a group.
Keep adding robots to the group, one at a time, checking the cost of that group each time.
If the cost of the group is more than the budget, stop adding robots to that group.
Now, start with the second robot by itself, and repeat the process of adding robots and checking the cost.
Continue doing this, starting with each robot as the beginning of a new group.
Keep track of the largest group of robots that has a cost less than or equal to the budget.
After considering all possible groups, return the size of the largest group you found.

Code Implementation

def maximum_robots_within_budget_brute_force(
    robot_costs, running_costs, budget
):
    max_robots = 0
    number_of_robots = len(robot_costs)

    for start_index in range(number_of_robots):
        current_cost = 0
        number_of_current_robots = 0
        max_robot_cost = 0

        for end_index in range(start_index, number_of_robots):
            # Determine the number of robots in current group.
            number_of_current_robots = end_index - start_index + 1

            # Find max cost for current window
            max_robot_cost = max(max_robot_cost, robot_costs[end_index])

            current_cost = (
                max_robot_cost
                + number_of_current_robots * running_costs[end_index]
            )

            # Stop if the current cost exceeds the budget.
            if current_cost > budget:
                break

            # Update maximum number of robots if valid
            max_robots = max(max_robots, number_of_current_robots)

    return max_robots

Big(O) Analysis

Time Complexity

O(n²) – The brute force solution iterates through each robot as a potential starting point for a team. For each starting robot, it expands the team by adding subsequent robots one by one until the cost exceeds the budget. In the worst case, for each of the 'n' robots, the algorithm might need to consider up to 'n' robots to its right, leading to roughly n * n/2 operations. Thus, the time complexity is O(n²).

Space Complexity

O(1) – The brute force approach described iterates through subarrays, calculating the cost of each. It doesn't create any auxiliary data structures like lists, arrays, or hash maps to store intermediate results, visited elements, or other information related to the subarrays other than single variables for calculations and maximum size. The space used is limited to a few scalar variables (e.g., to track the current subarray's cost and the maximum robots found so far). Therefore, the space complexity remains constant, regardless of the input size N (number of robots).

Optimal Solution

Approach

The goal is to find the largest possible group of robots we can afford. The efficient solution involves considering increasing sizes of robot groups and efficiently checking if each group can be afforded within the budget using a sliding window approach.

Here's how the algorithm would work step-by-step:

Imagine lining up all the robots in a row.
Start by considering a group of just one robot and see if we can afford it.
Now, consider a group of two robots next to each other in the line. Check if we can afford this group.
Continue increasing the size of the group one robot at a time, always checking if the current group is within budget.
To efficiently calculate the cost of each group, use a 'sliding window'. As you add a robot to the end of the group, remove the robot from the beginning, so you're always looking at consecutive robots.
To determine if we can afford the group, we need to know the maximum 'build cost' and the total 'running cost' within that group.
The trick is to efficiently find the maximum build cost within each sliding window. We can use a clever data structure to keep track of the maximum, so we don't have to recalculate it every time.
Keep track of the largest group you've found so far that is within budget. Whenever you find a bigger affordable group, update your record.
Eventually, you will have checked all possible group sizes. The largest affordable group you recorded is your answer.

Code Implementation

def maximum_robots(robot_build_costs, robot_running_costs, budget):
    max_robots = 0
    window_start = 0
    current_cost = 0
    deque = []

    for window_end in range(len(robot_build_costs)): 
        # Maintain a decreasing deque to track max build cost
        while deque and robot_build_costs[deque[-1]] <= robot_build_costs[window_end]:
            deque.pop()
        deque.append(window_end)

        current_cost += robot_running_costs[window_end]

        #Calculate current group length
        window_length = window_end - window_start + 1

        #Check if the current window is within the budget
        if robot_build_costs[deque[0]] + window_length * current_cost <= budget:
            max_robots = max(max_robots, window_length)
        else:
            #Shrink the window from the left
            while robot_build_costs[deque[0]] + window_length * current_cost > budget:
                current_cost -= robot_running_costs[window_start]
                window_start += 1

                #Remove outdated max element if it is out of the window
                if deque and deque[0] < window_start:
                    deque.pop(0)
                window_length = window_end - window_start + 1

                # We can't afford any more robots, even after shrinking
                if robot_build_costs[deque[0]] + window_length * current_cost > budget and window_start > window_end:
                    return max_robots
                    
                if robot_build_costs[deque[0]] + window_length * current_cost <= budget:
                   max_robots = max(max_robots, window_length)

    return max_robots

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the robots once using a sliding window approach. While the window slides, determining the maximum build cost within the window can be done in O(1) time using a data structure like a deque. Calculating the total running cost within the window also takes O(1) time per window. Because each robot is visited a constant number of times, the time complexity is directly proportional to the number of robots, n. Therefore, the overall time complexity is O(n).

Space Complexity

O(N) – The dominant space complexity comes from the need to efficiently find the maximum build cost within each sliding window as mentioned in step 7. This is typically achieved using a data structure like a deque (double-ended queue) to store indices of build costs. In the worst case, the deque could potentially store indices for all N robots if the build costs are monotonically increasing or decreasing, resulting in space proportional to the number of robots. Additionally, some constant amount of extra space will be created, but this will be dominated by the space required for the deque, which is O(N).

Edge Cases

Case	How to Handle
Empty cost array or empty charge array	Return 0 immediately as no robots can be selected.
Null cost or charge array	Throw an IllegalArgumentException or return 0 based on API contract.
Budget is negative	Return 0 since the budget cannot be negative.
Cost or charge array contains negative numbers	Throw an IllegalArgumentException as cost and charge cannot be negative.
Very large cost or charge values leading to integer overflow during sum calculation	Use long data type for intermediate sums to prevent overflow or use a modular arithmetic if appropriate.
All charge values are zero	The problem turns into checking minimum cost <= budget, so handle that boundary case appropriately.
Input arrays are of different lengths	Throw IllegalArgumentException as arrays must be of the same size.
Large input arrays nearing memory limits	Optimize the algorithm and data structures to minimize memory usage; consider using sliding window technique.

Empty cost array or empty charge array

How to Handle:

Return 0 immediately as no robots can be selected.

Null cost or charge array

How to Handle:

Throw an IllegalArgumentException or return 0 based on API contract.

Budget is negative

How to Handle:

Return 0 since the budget cannot be negative.

Cost or charge array contains negative numbers

How to Handle:

Throw an IllegalArgumentException as cost and charge cannot be negative.

Very large cost or charge values leading to integer overflow during sum calculation

How to Handle:

Use long data type for intermediate sums to prevent overflow or use a modular arithmetic if appropriate.

All charge values are zero

How to Handle:

The problem turns into checking minimum cost <= budget, so handle that boundary case appropriately.

Input arrays are of different lengths

How to Handle:

Throw IllegalArgumentException as arrays must be of the same size.

Large input arrays nearing memory limits

How to Handle:

Optimize the algorithm and data structures to minimize memory usage; consider using sliding window technique.