How do you solve Maximum Number of Robots Within Budget on LeetCode?

Maximum Number of Robots Within Budget (LeetCode #2398) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The maximum charge time among selected robots significantly influences the total cost.

What is the time complexity of Maximum Number of Robots Within Budget?

The optimal solution for Maximum Number of Robots Within Budget runs in O(n) time and uses O(1) space. This complexity is achieved because we only traverse the array once with two pointers, making it efficient.

What is the brute force solution for Maximum Number of Robots Within Budget?

The brute force approach for Maximum Number of Robots Within Budget is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible subarray of robots to see if they can run within the budget. It calculates the total cost for each subarray and keeps track of the maximum length that fits within the budget. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Maximum Number of Robots Within Budget?

Always explain your thought process clearly while coding. Consider edge cases and test your solution with them. Practice optimizing brute force solutions to gain confidence in identifying patterns.

What are common mistakes when solving Maximum Number of Robots Within Budget?

Failing to update the maximum charge time correctly when adjusting the window. Not considering edge cases where no robots can be run within the budget.

#2398

Maximum Number of Robots Within Budget

Hard

ArrayBinary SearchQueueSliding WindowHeap (Priority Queue)Prefix SumMonotonic QueueSliding WindowTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal solution uses a sliding window approach to maintain a dynamic range of robots. It keeps track of the maximum charge time and the sum of running costs efficiently, allowing us to check if the current window is within budget without recalculating everything.

⚙️

Algorithm

4 steps

1Step 1: Initialize pointers for the sliding window and variables for the maximum charge time and total running cost.
2Step 2: Expand the right pointer to include more robots while checking if the total cost exceeds the budget.
3Step 3: If it exceeds, move the left pointer to reduce the window size until the cost is within budget.
4Step 4: Update the maximum length of the window whenever a valid configuration is found.

solution.py14 lines

1def maxRobots(chargeTimes, runningCosts, budget):
2    n = len(chargeTimes)
3    left = 0
4    total_running_cost = 0
5    max_charge = 0
6    max_length = 0
7    for right in range(n):
8        total_running_cost += runningCosts[right]
9        max_charge = max(max_charge, chargeTimes[right])
10        while max_charge + (right - left + 1) * total_running_cost > budget:
11            total_running_cost -= runningCosts[left]
12            left += 1
13        max_length = max(max_length, right - left + 1)
14    return max_length

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Complexity note: This complexity is achieved because we only traverse the array once with two pointers, making it efficient.

1The maximum charge time among selected robots significantly influences the total cost.
2Using a sliding window allows for efficient management of the current set of robots being considered.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.