What is the time complexity of Maximum Running Time of N Computers?

The optimal solution has a time complexity of O(m log(S/n)).

What data structures are used to solve Maximum Running Time of N Computers?

This problem uses Array, Binary Search, Greedy, Sorting. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Maximum Running Time of N Computers?

Insight 1: Binary search is appropriate because we are looking for the maximum possible running time within a certain range. The problem's monotonic property allows for efficient searching: If a time `T` is possible, then all times less than `T` are also possible; if a time `T` is impossible, then all times greater than `T` are also impossible. Insight 2: The greedy approach of using the batteries optimally involves distributing their power to maximize the shared runtime among all computers. The `check` function is crucial. It determines if a given runtime `T` is achievable by summing the usable time from each battery (up to `T` minutes) and comparing it against the total required time (`n * T`).

Which companies ask LeetCode 2141. Maximum Running Time of N Computers in interviews?

This problem is frequently asked at Deutsche Bank, Flipkart.

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Maximum Running Time of N Computers - LeetCode 2141 Solution

Maximum Running Time of N Computers - Complete Solution Guide

Maximum Running Time of N Computers is LeetCode problem 2141, 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 computers. You are given the integer n and a 0-indexed integer array batteries where the i th battery can run a computer for batteries[i] minutes. You are interested in running all n computers simultaneously using the given batteries. Initially, you can insert at most one battery into each computer. After that and at any integer time moment, you can remove a battery from a computer and insert another battery any number of times . The inserted battery can be a totally new battery or a

Detailed Explanation

The problem asks us to find the maximum time we can run `n` computers simultaneously, given an array `batteries` where each element represents the running time of a battery. We can insert at most one battery into each computer initially, and we can swap batteries between computers at any time. The goal is to maximize the running time of all computers, assuming they all run simultaneously until the batteries are exhausted.

Solution Approach

The solution employs a binary search algorithm to find the maximum possible running time. The search space is from 1 to `sum(batteries) // n`. For each `mid` value in the binary search, a `check` function determines if it is possible to run all `n` computers for `mid` minutes. The `check` function iterates through the `batteries` array and accumulates the total usable time (either the battery's full capacity or the target time `mid`, whichever is smaller). If the total usable time is greater than or equal to the total required time (`n * mid`), it means `mid` is a feasible running time, and we can potentially find a larger one by searching in the upper half of the search space. Otherwise, `mid` is too large, and we need to search in the lower half.

Step-by-Step Algorithm

Step 1: Initialize `low` to 1 and `high` to `sum(batteries) // n`. `ans` is initialized to 0. `low` represents the minimum possible running time and `high` the maximum.
Step 2: Perform binary search while `low <= high`.
Step 3: Calculate the middle value `mid = (low + high) // 2`.
Step 4: Call the `check` function with `mid`, `n`, and `batteries` to determine if running all computers for `mid` minutes is possible.
Step 5: If `check(mid)` returns `True`, update `ans` to `mid` and move `low` to `mid + 1` to search for a larger possible running time.
Step 6: If `check(mid)` returns `False`, move `high` to `mid - 1` to search for a smaller possible running time.
Step 7: After the binary search completes, return the value of `ans`.

Key Insights

Insight 1: Binary search is appropriate because we are looking for the maximum possible running time within a certain range. The problem's monotonic property allows for efficient searching: If a time `T` is possible, then all times less than `T` are also possible; if a time `T` is impossible, then all times greater than `T` are also impossible.
Insight 2: The greedy approach of using the batteries optimally involves distributing their power to maximize the shared runtime among all computers. The `check` function is crucial. It determines if a given runtime `T` is achievable by summing the usable time from each battery (up to `T` minutes) and comparing it against the total required time (`n * T`).
Insight 3: The upper bound for the binary search range is `sum(batteries) // n`. If the total battery power is less than `n`, the computer can only run the shortest battery length time. It would be meaningless to search beyond this value.

Complexity Analysis

Time Complexity: O(m log(S/n))

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search, Greedy, Sorting.

Companies

Asked at: Deutsche Bank, Flipkart.

Maximum Running Time of N Computers | LeetCode 2141

Hard

Array Binary Search Greedy Sorting

Time: O(m log(S/n))Space: O(1)

Solution Code

Available in:

class Solution {
    public long maxRunTime(int n, int[] batteries) {
        long low = 1;
        long sum = 0;
        for (int b : batteries) {
            sum += b;
        }
        long high = sum / n;
        long ans = 0;

        while (low <= high) {
            long mid = low + (high - low) / 2;
            if (check(mid, n, batteries)) {
                ans = mid;
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }
        return ans;
    }

    private boolean check(long T, int n, int[] batteries) {
        if (T == 0) {
            return true;
        }

        long target = (long)n * T;
        long current_sum = 0;
        for (int b : batteries) {
            current_sum += Math.min((long)b, T);
        }

        return current_sum >= target;
    }
}

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Problem Info

Description

Problem 2141: Maximum Running Time of N Computers - Detailed explanation and solution approaches.

Topics

Array Binary Search Greedy Sorting

Companies

Deutsche BankFlipkart

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:2141

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

7 steps

1
Step 1: Initialize `low` to 1 and `high` to `sum(batteries) // n`. `ans` is initialized to 0. `low` represents the minimum possible running time and `high` the maximum.
2
Step 2: Perform binary search while `low <= high`.
3
Step 3: Calculate the middle value `mid = (low + high) // 2`.
4
Step 4: Call the `check` function with `mid`, `n`, and `batteries` to determine if running all computers for `mid` minutes is possible.
5
Step 5: If `check(mid)` returns `True`, update `ans` to `mid` and move `low` to `mid + 1` to search for a larger possible running time.
6
Step 6: If `check(mid)` returns `False`, move `high` to `mid - 1` to search for a smaller possible running time.
7
Step 7: After the binary search completes, return the value of `ans`.

💡

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(m log(S/n))

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The binary search's logarithmic component stems from halving the search space in each iteration. The linear component is due to the `check` function's iteration over the battery array in each step of the binary search.

📖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

Sorting and Greedy

Sort the batteries in ascending order. Iterate through the batteries, and for each battery, calculate the remaining power needed to keep all computers running. If a battery's power is insufficient, discard it and reduce the number of computers that need power. This is difficult to implement correctly and doesn't necessarily provide improved complexity.

Time: O(m log m), where m is the number of batteries due to sortingSpace: O(1) or O(m), depending on the sorting algorithm used

Trade-offs: Sorting adds overhead and the greedy approach to assigning to computers is complex to implement correctly. Not typically preferred over binary search.

✨

Key Insights

1Insight 1: Binary search is appropriate because we are looking for the maximum possible running time within a certain range. The problem's monotonic property allows for efficient searching: If a time `T` is possible, then all times less than `T` are also possible; if a time `T` is impossible, then all times greater than `T` are also impossible.
2Insight 2: The greedy approach of using the batteries optimally involves distributing their power to maximize the shared runtime among all computers. The `check` function is crucial. It determines if a given runtime `T` is achievable by summing the usable time from each battery (up to `T` minutes) and comparing it against the total required time (`n * T`).
3Insight 3: The upper bound for the binary search range is `sum(batteries) // n`. If the total battery power is less than `n`, the computer can only run the shortest battery length time. It would be meaningless to search beyond this value.

📚

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!

Greedy Algorithm

Making the best choice at each step without looking ahead.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: `n == batteries.length`. In this case, the problem simplifies to finding the minimum value in the `batteries` array. The binary search correctly handles this as `sum(batteries)//n` is usually higher than the lowest value, eventually converging to the correct answer.
Edge case 2: `batteries` contains very large values. The code uses `long` in Java/C++ and Python handles large numbers well, preventing integer overflow when calculating the sum and the target running time in `check` function.
Edge case 3: `n` is very large, close to or greater than `batteries.length`. In such cases, the initial `high` value of `sum(batteries) // n` might be small. The algorithm correctly handles these cases because binary search will converge to the largest possible value.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer overflow. When calculating the sum of batteries or the required running time (`n * mid`), using `int` can lead to overflow. Use `long` to prevent this. (Addressed in Java/C++ solutions).

✗Common mistake 2: Incorrect initial bounds for the binary search. Starting the `low` at 0 will lead to divide by zero errors and other logical problems, so starting from 1 is crucial. Not considering `sum(batteries) // n` as a maximum for `high` will create unnecessary iterations.

✅

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 in data centers. Optimizing the runtime of servers with varying power sources (batteries) is applicable to scheduling tasks efficiently in a data center.

🏦Real-world scenario 2: Emergency power management. Allocating emergency power from different generators to critical systems to maximize uptime, considering that batteries degrade over time. This can be applied to hospitals, disaster relief efforts, etc.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ 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: November 9, 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

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

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

@CodeNinjaJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Can binary search be applied to non-sorted arrays?

Binary search requires a monotonic property, not necessarily sorted data. It can be applied to any search space where you can determine if the answer is in the left or right half. Examples include searching rotated arrays or finding minimum in optimization problems.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Maximum Running Time of N Computers"?

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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