What is the time complexity of Minimum Time to Repair Cars?

The optimal solution has a time complexity of O(nlog(cars^2 * min_rank)).

What data structures are used to solve Minimum Time to Repair Cars?

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

What are the key insights for solving Minimum Time to Repair Cars?

Insight 1: The problem can be solved using binary search on the time taken to repair all the cars. The minimum time can be 1 and the maximum time can be min(ranks) * cars * cars. Insight 2: For a given time 't', we can calculate the number of cars each mechanic can repair by taking the square root of (t / r), where 'r' is the mechanic's rank. Summing up the number of cars each mechanic can repair gives the total number of cars repaired within time 't'.

Which companies ask LeetCode 2594. Minimum Time to Repair Cars in interviews?

This problem is frequently asked at Deloitte, HashedIn.

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Minimum Time to Repair Cars - LeetCode 2594 Solution

Minimum Time to Repair Cars - Complete Solution Guide

Minimum Time to Repair Cars is LeetCode problem 2594, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given an integer array ranks representing the ranks of some mechanics. ranks i is the rank of the i th mechanic . A mechanic with a rank r can repair n cars in r * n 2 minutes. You are also given an integer cars representing the total number of cars waiting in the garage to be repaired. Return the minimum time taken to repair all the cars. Note: All the mechanics can repair the cars simultaneously. Example 1: Input: ranks = [4,2,3,1], cars = 10 Output: 16 Explanation: - The first mechani

Detailed Explanation

The problem asks us to find the minimum time required to repair a given number of cars by a group of mechanics with different ranks. Each mechanic with rank 'r' can repair 'n' cars in 'r * n^2' minutes. We need to find the smallest time such that the mechanics can collectively repair all the cars.

Solution Approach

The solution uses binary search to find the minimum time. We initialize the lower bound 'low' to 1 and the upper bound 'high' to the maximum possible time (min_rank * cars * cars). In each iteration, we calculate the middle value 'mid' and determine how many cars can be repaired within that time. If the number of repaired cars is greater than or equal to the total number of cars, we update the upper bound to 'mid - 1' and store the current 'mid' as a potential answer. Otherwise, we update the lower bound to 'mid + 1'. The loop continues until the lower bound is greater than the upper bound. The final 'ans' holds the minimum time taken to repair all the cars.

Step-by-Step Algorithm

Step 1: Initialize 'low' to 1 and 'high' to min(ranks) * cars * cars.
Step 2: Initialize 'ans' to 'high'.
Step 3: While 'low' is less than or equal to 'high':
Step 4: Calculate 'mid' as 'low + (high - low) // 2'.
Step 5: Initialize 'repaired_count' to 0.
Step 6: Iterate through each mechanic's rank 'r' in 'ranks':
Step 7: Calculate the number of cars the mechanic can repair within time 'mid' as 'int(sqrt(mid / r))' and add it to 'repaired_count'.
Step 8: If 'repaired_count' is greater than or equal to 'cars':
Step 9: Update 'ans' to 'mid' and update 'high' to 'mid - 1'.
Step 10: Else:
Step 11: Update 'low' to 'mid + 1'.
Step 12: Return 'ans'.

Key Insights

Insight 1: The problem can be solved using binary search on the time taken to repair all the cars. The minimum time can be 1 and the maximum time can be min(ranks) * cars * cars.
Insight 2: For a given time 't', we can calculate the number of cars each mechanic can repair by taking the square root of (t / r), where 'r' is the mechanic's rank. Summing up the number of cars each mechanic can repair gives the total number of cars repaired within time 't'.
Insight 3: Since the time required is always an integer, we can perform an integer binary search and check for each 'mid' value (time) if all 'cars' can be repaired or not.

Complexity Analysis

Time Complexity: O(nlog(cars^2 * min_rank))

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search.

Companies

Asked at: Deloitte, HashedIn.

Minimum Time to Repair Cars | LeetCode 2594

Medium

Array Binary Search

Time: O(nlog(cars^2 * min_rank))Space: O(1)

Solution Code

Available in:

class Solution {
    public long repairCars(int[] ranks, int cars) {
        long low = 1;
        long high = (long) ranks[0] * cars * cars;
        for (int r : ranks) {
            high = Math.min(high, (long) r * cars * cars);
        }

        long ans = high;
        while (low <= high) {
            long mid = low + (high - low) / 2;
            
            long repaired_count = 0;
            for (int r : ranks) {
                repaired_count += (long) Math.sqrt((double) mid / r);
            }
            
            if (repaired_count >= cars) {
                ans = mid;
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }
        
        return ans;
    }
}

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Problem Info

Description

Problem 2594: Minimum Time to Repair Cars - Detailed explanation and solution approaches.

Topics

Array Binary Search

Companies

DeloitteHashedIn

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:2594

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

12 steps

1
Step 1: Initialize 'low' to 1 and 'high' to min(ranks) * cars * cars.
2
Step 2: Initialize 'ans' to 'high'.
3
Step 3: While 'low' is less than or equal to 'high':
4
Step 4: Calculate 'mid' as 'low + (high - low) // 2'.
5
Step 5: Initialize 'repaired_count' to 0.
6
Step 6: Iterate through each mechanic's rank 'r' in 'ranks':
7
Step 7: Calculate the number of cars the mechanic can repair within time 'mid' as 'int(sqrt(mid / r))' and add it to 'repaired_count'.
8
Step 8: If 'repaired_count' is greater than or equal to 'cars':
9
Step 9: Update 'ans' to 'mid' and update 'high' to 'mid - 1'.
10
Step 10: Else:
11
Step 11: Update 'low' to 'mid + 1'.
12
Step 12: Return '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(nlog(cars^2 * min_rank))

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the binary search, which reduces the search space by half in each iteration. The loop condition is 'low <= high', so the number of iterations is logarithmic with base 2 of the search space. The space complexity is constant because we are not using any auxiliary data structures that scale with the input size.

📖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

Dynamic Programming

A dynamic programming approach could be considered where dp[i][j] stores the minimum time to repair 'j' cars using the first 'i' mechanics. However, this approach is likely to be less efficient due to the large constraints on the number of cars.

Time: O(n * cars^2), where n is the number of mechanicsSpace: O(n * cars)

Trade-offs: Dynamic programming would be significantly slower and require more memory compared to the binary search approach, especially with the large constraints. Therefore, binary search is superior in this scenario.

✨

Key Insights

1Insight 1: The problem can be solved using binary search on the time taken to repair all the cars. The minimum time can be 1 and the maximum time can be min(ranks) * cars * cars.
2Insight 2: For a given time 't', we can calculate the number of cars each mechanic can repair by taking the square root of (t / r), where 'r' is the mechanic's rank. Summing up the number of cars each mechanic can repair gives the total number of cars repaired within time 't'.
3Insight 3: Since the time required is always an integer, we can perform an integer binary search and check for each 'mid' value (time) if all 'cars' can be repaired or not.

📚

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!

Edge Cases

Edge case 1: When ranks array is empty or cars is zero. The problem constraints ensure this won't happen, as ranks.length >= 1 and cars >= 1.
Edge case 2: When a mechanic has a very low rank. The binary search and the square root calculation in the loop ensure the correctness of result even in such cases, as the solution considers the minimum time that satisfies the condition.
Edge case 3: All mechanics have the same rank. The binary search approach correctly calculates the optimal time, distributing the work evenly across all mechanics.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer overflow in the calculation of 'high'. Using long long (Java, C++) or appropriate large integer type is crucial.

✗Common mistake 2: Not correctly calculating the number of cars repaired by each mechanic. The square root calculation and integer casting must be done accurately.

✗Common mistake 3: Incorrectly setting the initial values for 'low' and 'high'. 'low' should start at 1, and 'high' should be the maximum possible time (min_rank * cars * cars).

✗Common mistake 4: Off-by-one errors in the binary search update steps for 'low' and 'high'.

✅

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 problems where different resources (mechanics) have different efficiencies (ranks) and we need to minimize the time to complete a task (repair cars).

🏦Real-world scenario 2: Task scheduling in a distributed computing environment where different machines (mechanics) have different processing powers (ranks) and we need to minimize the overall execution time for a set of jobs (cars).

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ 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: August 14, 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

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

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 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.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "Minimum Time to Repair Cars"?

For this medium-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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