What is the time complexity of Minimum Number of Refueling Stops?

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

What data structures are used to solve Minimum Number of Refueling Stops?

This problem uses Array, Dynamic Programming, Greedy, Heap (Priority Queue). Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Minimum Number of Refueling Stops?

Insight 1: We should use a greedy approach, prioritizing stations with the most fuel within our reachable distance. Insight 2: A max heap (priority queue) is the ideal data structure to efficiently track the stations with the most fuel within reachable range.

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Minimum Number of Refueling Stops - LeetCode 871 Solution

Q: Which companies ask LeetCode 871. Minimum Number of Refueling Stops in interviews?

This problem is frequently asked at Barclays, DE Shaw, Flipkart, Morgan Stanley, Rubrik and 1 more companies.

Minimum Number of Refueling Stops - Complete Solution Guide

Minimum Number of Refueling Stops is LeetCode problem 871, 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

A car travels from a starting position to a destination which is target miles east of the starting position. There are gas stations along the way. The gas stations are represented as an array stations where stations[i] = [position i , fuel i ] indicates that the i th gas station is position i miles east of the starting position and has fuel i liters of gas. The car starts with an infinite tank of gas, which initially has startFuel liters of fuel in it. It uses one liter of gas per one mile that

Detailed Explanation

The problem asks us to find the minimum number of refueling stops required for a car to reach a target distance, starting with a given amount of fuel. The car consumes one unit of fuel per unit of distance traveled. We are provided with a list of gas stations, each having a specific location (distance from the start) and fuel capacity. We can stop at any station and refuel completely. The goal is to determine the fewest stops necessary to reach the target. If it's impossible to reach the target, we should return -1.

Solution Approach

The solution utilizes a greedy approach combined with a max heap. We iterate through the gas stations. At each step, we add all the stations within our current reachable distance to the max heap. The max heap stores the fuel capacity of these stations. Then, if our current reachable distance is less than the target, we refuel at the station with the most fuel (the top element of the max heap), incrementing our stops counter and updating our reachable distance. If the heap is empty and we haven't reached the target, it implies there are no reachable stations, so we return -1. We repeat this process until we reach the target distance.

Step-by-Step Algorithm

Step 1: Initialize `stops` to 0, `reachable` to `startFuel`, and an empty max heap `max_heap`.
Step 2: Iterate through the `stations` array using index `i`.
Step 3: While `i` is within the bounds of the `stations` array AND the station's position `stations[i][0]` is less than or equal to the `reachable` distance, add the station's fuel `stations[i][1]` to the `max_heap` and increment `i`.
Step 4: If the `max_heap` is empty AND `reachable` is still less than the `target`, return -1 (unreachable).
Step 5: If the `reachable` distance is less than the `target`, refuel at the station with the most fuel: increment `stops` by 1 and add the top element from `max_heap` to `reachable` (and remove the top element from the heap).
Step 6: Repeat steps 3-5 until the `reachable` distance is greater than or equal to the `target`.
Step 7: Return the final value of `stops`.

Key Insights

Insight 1: We should use a greedy approach, prioritizing stations with the most fuel within our reachable distance.
Insight 2: A max heap (priority queue) is the ideal data structure to efficiently track the stations with the most fuel within reachable range.
Insight 3: It's crucial to consider scenarios where we run out of fuel before reaching any further stations, implying the destination is unreachable.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Dynamic Programming, Greedy, Heap (Priority Queue).

Companies

Asked at: Barclays, DE Shaw, Flipkart, Morgan Stanley, Rubrik, Snap.

Minimum Number of Refueling Stops | LeetCode 871

Hard

Array Dynamic Programming Greedy Heap (Priority Queue)

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

Solution Code

Available in:

import java.util.PriorityQueue;

class Solution {
    public int minRefuelStops(int target, int startFuel, int[][] stations) {
        int stops = 0;
        long reachable = startFuel;
        PriorityQueue<Integer> max_heap = new PriorityQueue<>((a, b) -> b - a);
        int i = 0;
        
        while (reachable < target) {
            while (i < stations.length && stations[i][0] <= reachable) {
                max_heap.offer(stations[i][1]);
                i++;
            }
            
            if (max_heap.isEmpty()) {
                return -1;
            }
            
            reachable += max_heap.poll();
            stops++;
        }
        
        return stops;
    }
}

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Problem Info

Description

Problem 871: Minimum Number of Refueling Stops - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Greedy Heap (Priority Queue)

Companies

BarclaysDE ShawFlipkartMorgan StanleyRubrikSnap

Stats

Difficulty:Hard

Languages:4

Companies:6

Problem #:871

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 `stops` to 0, `reachable` to `startFuel`, and an empty max heap `max_heap`.
2
Step 2: Iterate through the `stations` array using index `i`.
3
Step 3: While `i` is within the bounds of the `stations` array AND the station's position `stations[i][0]` is less than or equal to the `reachable` distance, add the station's fuel `stations[i][1]` to the `max_heap` and increment `i`.
4
Step 4: If the `max_heap` is empty AND `reachable` is still less than the `target`, return -1 (unreachable).
5
Step 5: If the `reachable` distance is less than the `target`, refuel at the station with the most fuel: increment `stops` by 1 and add the top element from `max_heap` to `reachable` (and remove the top element from the heap).
6
Step 6: Repeat steps 3-5 until the `reachable` distance is greater than or equal to the `target`.
7
Step 7: Return the final value of `stops`.

💡

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 log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity stems from the fact that each gas station may potentially be pushed and popped from the heap, leading to n heap operations, each taking O(log n) time. The space complexity is determined by the maximum size of the heap, which can be n in the worst case.

📖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

We can use dynamic programming to solve this problem. dp[i][j] represents the farthest distance we can reach with i stops using the first j stations. The time complexity would be O(n^2) and space complexity O(n^2).

Time: O(n^2)Space: O(n^2)

Trade-offs: This approach is generally less efficient than the greedy approach using a max heap due to its higher time and space complexity. However, it might be useful if modifications to the problem require storing information about intermediate states or have dependencies that don't easily lend themselves to a greedy solution.

✨

Key Insights

1Insight 1: We should use a greedy approach, prioritizing stations with the most fuel within our reachable distance.
2Insight 2: A max heap (priority queue) is the ideal data structure to efficiently track the stations with the most fuel within reachable range.
3Insight 3: It's crucial to consider scenarios where we run out of fuel before reaching any further stations, implying the destination is unreachable.

📚

Key Terms Explained

For Students

Array

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: `target` is reachable with the `startFuel` and no refueling is needed (empty `stations` list, or the target is within the `startFuel` distance). The algorithm correctly handles this, returning 0.
Edge case 2: It's impossible to reach even the first gas station. The algorithm detects this because the inner `while` loop doesn't execute, the `max_heap` remains empty, and the condition `if not max_heap:` becomes true, returning -1.
Edge case 3: We can reach some stations but cannot reach the target because we exhaust all reachable stations' fuel and still can't reach a further station. The `if not max_heap:` condition within the main `while` loop correctly detects this scenario and returns -1.
Edge case 4: Start fuel is zero. If startFuel is zero, the algorithm will immediately check if the max_heap is empty, and since no stations could possibly be reached, it will return -1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where the target is unreachable (not checking for an empty heap). This leads to incorrect results or infinite loops.

✗Common mistake 2: Not considering the initial `startFuel`. Forgetting to account for the initial fuel available will cause you to miss the simple cases where you can reach the target without stopping at any station.

✗Common mistake 3: Not using a max heap. If you use a simple list to store the fuel and then find the maximum each time, the time complexity will be much higher.

✅

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: Route planning for delivery vehicles or long-haul trucks where minimizing refueling stops is important for efficiency and cost savings.

🏦Real-world scenario 2: Determining the feasibility of a journey for an electric vehicle given a limited battery capacity and a network of charging stations along the route. The stations are locations to 'refuel', or recharge the batteries.

💡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: October 30, 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.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

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 "Minimum Number of Refueling Stops"?

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 dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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