What is the time complexity of Gas Station?

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

What data structures are used to solve Gas Station?

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

What are the key insights for solving Gas Station?

Insight 1: If the total gas available is less than the total cost, a solution is impossible. This can be determined by summing `gas[i] - cost[i]` for all i. If the sum is negative, there's no solution. Insight 2: If a starting point 's' allows you to reach station 'i' but not station 'i+1', then any starting point between 's' and 'i' cannot be a valid starting point. This is because you'd have even less gas when starting at stations between s and i. This insight allows for the greedy approach.

Which companies ask LeetCode 134. Gas Station in interviews?

This problem is frequently asked at BNY Mellon, BitGo, Cisco, CureFit, Dream11 and 11 more companies.

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Gas Station - LeetCode 134 Solution

Gas Station - Complete Solution Guide

Gas Station is LeetCode problem 134, 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

There are n gas stations along a circular route, where the amount of gas at the i th station is gas[i] . You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from the i th station to its next (i + 1) th station. You begin the journey with an empty tank at one of the gas stations. Given two integer arrays gas and cost , return the starting gas station's index if you can travel around the circuit once in the clockwise direction, otherwise return -1 . If there exists a so

Detailed Explanation

The Gas Station problem presents a scenario with 'n' gas stations arranged in a circle. Each station 'i' has a certain amount of gas, `gas[i]`. You have a car with an unlimited gas tank and need to travel around the circuit once, starting from one of the stations. The cost to travel from station 'i' to the next station '(i + 1)' is `cost[i]`. The goal is to find the starting station index that allows you to complete the circular route. If no such starting station exists, return -1. The problem guarantees a unique solution if one exists.

Solution Approach

The solution employs a greedy approach to efficiently find the starting station. It maintains two variables: `total_tank` to track the overall gas surplus (or deficit) and `current_tank` to track the gas in the tank while traversing the circuit. The algorithm iterates through the stations, calculating the net gain (gas - cost) at each station. If `current_tank` becomes negative, it indicates that the current starting station is not feasible. The algorithm resets `current_tank` to 0 and updates `start_station` to the next station (i+1). If `total_tank` remains negative after iterating through all stations, it means no solution exists. Otherwise, the final `start_station` is the answer.

Step-by-Step Algorithm

Step 1: Initialize `total_tank` and `current_tank` to 0, and `start_station` to 0.
Step 2: Iterate through the gas and cost arrays from i = 0 to n-1.
Step 3: Calculate `net_gain = gas[i] - cost[i]` for the current station.
Step 4: Update `total_tank` by adding `net_gain` to it: `total_tank += net_gain`.
Step 5: Update `current_tank` by adding `net_gain` to it: `current_tank += net_gain`.
Step 6: If `current_tank` becomes negative, it means we cannot reach the next station starting from the current `start_station`. Therefore, update `start_station = i + 1` and reset `current_tank = 0`.
Step 7: After the loop finishes, check if `total_tank` is negative. If it is, return -1 because there's no solution.
Step 8: If `total_tank` is non-negative, return `start_station` as the starting station.

Key Insights

Insight 1: If the total gas available is less than the total cost, a solution is impossible. This can be determined by summing `gas[i] - cost[i]` for all i. If the sum is negative, there's no solution.
Insight 2: If a starting point 's' allows you to reach station 'i' but not station 'i+1', then any starting point between 's' and 'i' cannot be a valid starting point. This is because you'd have even less gas when starting at stations between s and i. This insight allows for the greedy approach.
Insight 3: The problem guarantees a unique solution, which greatly simplifies the search. Without this guarantee, we would need to test all possible starting locations.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Greedy.

Companies

Asked at: BNY Mellon, BitGo, Cisco, CureFit, Dream11, Flipkart, Freecharge, Goldman Sachs, Infosys, Lucid, Mastercard, Oracle, PhonePe, ServiceNow, Yahoo, Zepto.

Gas Station | LeetCode 134

Medium

Array Greedy

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int canCompleteCircuit(int[] gas, int[] cost) {
        int n = gas.length;
        
        int total_tank = 0;
        int current_tank = 0;
        int start_station = 0;
        
        for (int i = 0; i < n; i++) {
            int net_gain = gas[i] - cost[i];
            
            total_tank += net_gain;
            current_tank += net_gain;
            
            if (current_tank < 0) {
                start_station = i + 1;
                current_tank = 0;
            }
        }
        
        if (total_tank < 0) {
            return -1;
        } else {
            return start_station;
        }
    }
}

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Problem Info

Description

Problem 134: Gas Station - Detailed explanation and solution approaches.

Topics

Array Greedy

Companies

BNY MellonBitGoCiscoCureFitDream11FlipkartFreechargeGoldman Sachs

Stats

Difficulty:Medium

Languages:4

Companies:16

Problem #:134

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 `total_tank` and `current_tank` to 0, and `start_station` to 0.
2
Step 2: Iterate through the gas and cost arrays from i = 0 to n-1.
3
Step 3: Calculate `net_gain = gas[i] - cost[i]` for the current station.
4
Step 4: Update `total_tank` by adding `net_gain` to it: `total_tank += net_gain`.
5
Step 5: Update `current_tank` by adding `net_gain` to it: `current_tank += net_gain`.
6
Step 6: If `current_tank` becomes negative, it means we cannot reach the next station starting from the current `start_station`. Therefore, update `start_station = i + 1` and reset `current_tank = 0`.
7
Step 7: After the loop finishes, check if `total_tank` is negative. If it is, return -1 because there's no solution.
8
Step 8: If `total_tank` is non-negative, return `start_station` as the starting station.

💡

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(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is linear because the main loop iterates through the input arrays once. The space complexity is constant because the amount of memory used does not grow 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

Brute Force

Try each gas station as a starting point and simulate the journey to see if the circuit can be completed. For each starting point, iterate through the stations, keeping track of the current gas level. If at any point the gas level drops below zero, that starting point is invalid.

Time: O(n^2), where n is the number of gas stations. For each starting point, we might iterate through all n stations.Space: O(1). Uses constant extra space.

Trade-offs: This is a straightforward but inefficient approach. It's simple to implement but results in a quadratic time complexity, making it unsuitable for large datasets. The provided solution is much more efficient.

✨

Key Insights

1Insight 1: If the total gas available is less than the total cost, a solution is impossible. This can be determined by summing `gas[i] - cost[i]` for all i. If the sum is negative, there's no solution.
2Insight 2: If a starting point 's' allows you to reach station 'i' but not station 'i+1', then any starting point between 's' and 'i' cannot be a valid starting point. This is because you'd have even less gas when starting at stations between s and i. This insight allows for the greedy approach.
3Insight 3: The problem guarantees a unique solution, which greatly simplifies the search. Without this guarantee, we would need to test all possible starting locations.

📚

Key Terms Explained

For Students

Array

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

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: If the total gas available is less than the total cost (i.e., `total_tank < 0`), there's no solution. The code explicitly checks this condition and returns -1.
Edge case 2: A circular route with only one gas station. The algorithm works correctly in this case because the loop will execute once, and if `gas[0] >= cost[0]`, the `start_station` will remain 0 and `total_tank` will be positive. Otherwise, `total_tank` will be negative, and -1 will be returned.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not realizing that the `start_station` should be updated to `i + 1` when `current_tank` becomes negative, rather than just reseting `current_tank` to 0.

✗Common mistake 2: Incorrectly calculating the `net_gain` or not accounting for the circular nature of the route when using the brute-force approach.

✅

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: Optimizing delivery routes. Imagine a fleet of delivery trucks that need to visit multiple locations. The `gas` array could represent the fuel available at each location, and the `cost` array could represent the fuel needed to travel between locations. The algorithm can help determine the optimal starting location to minimize fuel consumption.

🏦Real-world scenario 2: Resource allocation in a distributed system. Consider a network of computers where each computer has a certain amount of available resources (like CPU or memory). The `gas` array could represent the resource surplus at each computer, and the `cost` array could represent the resource requirements of a task that needs to be executed across the network. The algorithm can help identify a suitable starting point for the task so that it can be completed successfully.

💡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: October 13, 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 14

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

@DevExpert92Jul 7

❓

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 verify if a greedy approach is correct?

Prove the greedy choice property: the locally optimal choice leads to globally optimal solution. Use exchange argument: show that swapping any element with the greedy choice doesn't improve the solution. If you can't prove it, test extensively or consider DP.

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 "Gas Station"?

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

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