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

The optimal solution has a time complexity of O(n).

What data structures are used to solve Maximum Number of Robots Within Budget?

This problem uses Array, Binary Search, Queue, Sliding Window, Heap (Priority Queue), Prefix Sum, Monotonic Queue. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Maximum Number of Robots Within Budget?

Insight 1: The problem requires finding the maximum length of a *consecutive* sub-array that satisfies a given condition (total cost <= budget). This strongly suggests using a sliding window approach. Insight 2: Efficiently calculating the maximum charge time within the sliding window is crucial for performance. A monotonic queue (deque) can be used to maintain the maximum element in O(1) time per update.

Which companies ask LeetCode 2398. Maximum Number of Robots Within Budget in interviews?

This problem is frequently asked at InMobi.

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Maximum Number of Robots Within Budget - LeetCode 2398 Solution

Q: How to solve Maximum Number of Robots Within Budget?

The solution uses a sliding window approach combined with a monotonic queue (deque). The sliding window expands from left to right, and the deque maintains the indices of the potential maximum charge times within the current window in descending order. The head of the deque always contains the index of the maximum charge time in the current window. If the cost exceeds the budget, the window shrinks from the left until the cost is within the budget.

Q: Which companies ask LeetCode 2398. Maximum Number of Robots Within Budget in interviews?

This problem is frequently asked at InMobi.

Maximum Number of Robots Within Budget - Complete Solution Guide

Maximum Number of Robots Within Budget is LeetCode problem 2398, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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 n

Detailed Explanation

The problem asks us to find the maximum number of *consecutive* robots we can run within a given budget. Each robot has a charge time and a running cost. The total cost for *k* consecutive robots is calculated as the maximum charge time among those *k* robots, plus *k* times the sum of their running costs. We need to find the largest possible *k* such that the total cost does not exceed the given budget.

Solution Approach

The solution uses a sliding window approach combined with a monotonic queue (deque). The sliding window expands from left to right, and the deque maintains the indices of the potential maximum charge times within the current window in descending order. The head of the deque always contains the index of the maximum charge time in the current window. If the cost exceeds the budget, the window shrinks from the left until the cost is within the budget.

Step-by-Step Algorithm

Step 1: Initialize `left` pointer to 0 (start of the window), `current_sum_costs` to 0, `max_len` to 0, and a deque `dq`.
Step 2: Iterate through the `chargeTimes` and `runningCosts` arrays using the `right` pointer (end of the window).
Step 3: Add the running cost of the robot at index `right` to `current_sum_costs`.
Step 4: Maintain the monotonic queue: Remove elements from the back of the deque if their corresponding charge times are less than or equal to the charge time of the current robot (at index `right`). Then, add `right` to the back of the deque.
Step 5: Check if the cost (maximum charge time from the deque + (window size) * `current_sum_costs`) exceeds the budget.
Step 6: If the cost exceeds the budget, shrink the window from the left. Remove the leftmost element from the deque if it's the leftmost index of the current window, subtract `runningCosts[left]` from `current_sum_costs`, and increment the `left` pointer.
Step 7: Update `max_len` with the maximum window size seen so far (right - left + 1).
Step 8: Return `max_len` after iterating through all the robots.

Key Insights

Insight 1: The problem requires finding the maximum length of a *consecutive* sub-array that satisfies a given condition (total cost <= budget). This strongly suggests using a sliding window approach.
Insight 2: Efficiently calculating the maximum charge time within the sliding window is crucial for performance. A monotonic queue (deque) can be used to maintain the maximum element in O(1) time per update.
Insight 3: The budget is large (up to 10^15), so we need to use `long` or `long long` data types to avoid integer overflow when calculating the total cost.
Insight 4: Realize the Monotonic Queue can maintain a sorted order of elements within the current window, enabling O(1) access to the maximum element. This is key to avoiding O(k) re-computation within each window.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Binary Search, Queue, Sliding Window, Heap (Priority Queue), Prefix Sum, Monotonic Queue.

Companies

Asked at: InMobi.

Maximum Number of Robots Within Budget | LeetCode 2398

Hard

Array Binary Search Queue Sliding Window Heap (Priority Queue)Prefix Sum Monotonic Queue

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

Solution Code

Available in:

import java.util.Deque;
import java.util.LinkedList;

class Solution {
    public int maximumRobots(int[] chargeTimes, int[] runningCosts, long budget) {
        int left = 0;
        long current_sum_costs = 0;
        int max_len = 0;
        Deque<Integer> dq = new LinkedList<>();

        for (int right = 0; right < chargeTimes.length; right++) {
            current_sum_costs += runningCosts[right];
            
            while (!dq.isEmpty() && chargeTimes[dq.peekLast()] <= chargeTimes[right]) {
                dq.removeLast();
            }
            dq.addLast(right);
            
            // Shrink the window from the left if the budget is exceeded
            while (!dq.isEmpty() && (long)chargeTimes[dq.peekFirst()] + (long)(right - left + 1) * current_sum_costs > budget) {
                if (dq.peekFirst() == left) {
                    dq.removeFirst();
                }
                current_sum_costs -= runningCosts[left];
                left += 1;
            }
            
            max_len = Math.max(max_len, right - left + 1);
            
        }
        return max_len;
    }
}

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Problem Info

Description

Problem 2398: Maximum Number of Robots Within Budget - Detailed explanation and solution approaches.

Topics

Array Binary Search Queue Sliding Window Heap (Priority Queue)Prefix Sum Monotonic Queue

Companies

InMobi

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2398

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize `left` pointer to 0 (start of the window), `current_sum_costs` to 0, `max_len` to 0, and a deque `dq`.
2
Step 2: Iterate through the `chargeTimes` and `runningCosts` arrays using the `right` pointer (end of the window).
3
Step 3: Add the running cost of the robot at index `right` to `current_sum_costs`.
4
Step 4: Maintain the monotonic queue: Remove elements from the back of the deque if their corresponding charge times are less than or equal to the charge time of the current robot (at index `right`). Then, add `right` to the back of the deque.
5
Step 5: Check if the cost (maximum charge time from the deque + (window size) * `current_sum_costs`) exceeds the budget.
6
Step 6: If the cost exceeds the budget, shrink the window from the left. Remove the leftmost element from the deque if it's the leftmost index of the current window, subtract `runningCosts[left]` from `current_sum_costs`, and increment the `left` pointer.
7
Step 7: Update `max_len` with the maximum window size seen so far (right - left + 1).
8
Step 8: Return `max_len` after iterating through all the robots.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The outer loop iterates `n` times. The inner `while` loops, while nested, perform amortized O(1) operations per element because each element is added and removed from the deque at most once. The space complexity is determined by the deque, which can hold at most `n` indices.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Binary Search

For a given window size `k`, check if it is possible to run `k` robots within the budget. This check involves finding the maximum charge time and the sum of running costs for all possible sub-arrays of size `k`. Use binary search on the window size `k` to find the maximum possible value.

Time: O(n log n), where n is the number of robots. The binary search takes O(log n) time, and checking if a window size `k` is possible takes O(n) time.Space: O(1) if the sum can be calculated in-place, O(n) if prefix sums are used.

Trade-offs: The binary search approach is generally slower than the sliding window approach due to the logarithmic factor. It's useful when the problem requires checking for a specific window size, or when the constraints prevent using a deque. Also, binary search could be beneficial if the cost calculation was more complex making the amortized O(1) operations of the sliding window less effective.

Priority Queue (Heap)

Instead of a monotonic queue, use a priority queue (max heap) to find the maximum charge time in the current window. Re-compute the heap for each window sliding. This will lead to a higher time complexity.

Time: O(n log k), where n is the number of robots and k is the maximum window size.Space: O(k), where k is the maximum window size.

Trade-offs: Heap-based approach is generally less efficient than the deque-based solution because maintaining the heap for each window slide takes O(log k) time, making it slower than the O(1) amortized time of the monotonic queue.

✨

Key Insights

1Insight 1: The problem requires finding the maximum length of a *consecutive* sub-array that satisfies a given condition (total cost <= budget). This strongly suggests using a sliding window approach.
2Insight 2: Efficiently calculating the maximum charge time within the sliding window is crucial for performance. A monotonic queue (deque) can be used to maintain the maximum element in O(1) time per update.
3Insight 3: The budget is large (up to 10^15), so we need to use `long` or `long long` data types to avoid integer overflow when calculating the total cost.
4Insight 4: Realize the Monotonic Queue can maintain a sorted order of elements within the current window, enabling O(1) access to the maximum element. This is key to avoiding O(k) re-computation within each window.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Queue

FIFO structure - First In, First Out. Like a line at a store!

Sliding Window

A moving "window" over data to process subsets efficiently.

Edge Cases

Edge case 1: `budget` is smaller than the smallest `chargeTime`. The algorithm will correctly return 0 because no single robot can be run within the budget. The initial `max_len` is 0, and it's only updated if a valid window is found.
Edge case 2: `chargeTimes` and `runningCosts` are empty. The for loop won't execute, and the initialized `max_len` of 0 is returned, which is correct.
Edge case 3: All running costs are zero. The algorithm correctly identifies the largest sub-array where max(chargeTimes) is within budget.
Edge case 4: Very large budget. The monotonic queue will shrink less often but still function correctly. The `long` type prevents potential overflow.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Using `int` instead of `long` for `current_sum_costs` and the cost calculation, leading to integer overflow and incorrect results for large budgets.

✗Common mistake 2: Not maintaining the monotonic property of the deque, resulting in an incorrect maximum charge time within the window.

✗Common mistake 3: Not shrinking the window when the cost exceeds the budget, leading to an infinite loop or incorrect results.

✗Common mistake 4: Recalculating max charge time within the current window on each slide, which degrades the time complexity.

✗Common mistake 5: Not handling the edge case when deque is empty.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation with a budget constraint. For example, optimizing the number of servers to run concurrently given a power budget, where `chargeTimes` represents the power needed to start a server and `runningCosts` represent the power consumption per unit of time.

🏦Real-world scenario 2: Finding the longest period of uninterrupted service within a financial budget. `chargeTimes` could represent initial setup costs and `runningCosts` could represent ongoing maintenance costs. The algorithm would determine the maximum duration of service achievable without exceeding the budget.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 29, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Maximum Number of Robots Within Budget"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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