Gas Station

Dec 18, 2024 • 5 mins read
Hero Image

Problem Overview

The LeetCode problem Gas Station involves finding a starting gas station from which we can complete a circular route around all stations. The challenge is rooted in optimizing the utilization of gas in accordance with its availability and consumption requirements at multiple stations.

Here's a brief summary of the problem:

  • You are at a series of gas stations arranged in a circle and you want to know if you can make a full circuit once by starting at any one of the stations.
  • Each station provides a certain amount of gas.
  • Traveling from one station to the next consumes a specific amount of gas.
  • The task is to determine a starting station index where you can begin the circuit and return to the same station, using only the given gas at each station.

Understanding the Problem

The problem can be summarized as having an array of gas stations where:

  • gas[i] is the gas amount provided by the ith station.
  • cost[i] is the gas required to travel from the ith station to the (i+1)th station.

To solve this, you need to find out if there exists a station i such that, starting with 0 gas in your tank, you can complete the journey. If multiple answers exist, any valid one will suffice. If it’s impossible to complete the circuit, return -1.

Key DSA Concepts and Theory

1. Greedy Algorithm

A Greedy Algorithm builds up a solution piece by piece, always choosing the next piece that offers the most immediate benefit. The core idea is that choosing locally optimal solutions can result in a globally optimal solution to some problems.

2. Circular Arrays

This problem embodies a circular nature where the end of the array connects to the beginning, posing challenges in linear traversal. By envisioning it as a circular queue, we can effectively traverse without prematurely ceasing due to the linear end condition.

Solution Approach

The solution to this problem is best approached using a Greedy Strategy:

Explanation

  • If the total gas is less than the total cost, it's clearly impossible to complete the circuit, so return -1.
  • If the total gas is more than or equal to the total cost, there is a guaranteed solution. The strategy is to traverse and check each station to see if we can start from there with a net positive or zero gas till the end. If not, reset and move the starting point forward.

Step-by-Step Approach

  1. Initialization:

    • Track the total gas and total cost across all stations.
    • Track the current surplus of gas as you simulate the journey.
    • Keep a record of the starting station.

  2. Simulation:

    • Traverse each station. For each station:
      • Add the available gas and subtract the needed cost.
      • If ever this sum is negative, reset the current surplus and set the next station as the new starting point.
    • This reset implies all stations before the current one cannot be a valid start because they fail to provide enough gas to reach the current station.

  3. Result Determination:

    • If the total gas is greater than or equal to the total cost after one full traversal, return the starting index. Otherwise, return -1.

C++ Code

#include <vector>

int canCompleteCircuit(std::vector<int>& gas, std::vector<int>& cost) {
    int totalGas = 0, totalCost = 0, currentSurplus = 0, start = 0;

    for (int i = 0; i < gas.size(); ++i) {
        totalGas += gas[i];
        totalCost += cost[i];
        currentSurplus += gas[i] - cost[i];

        if (currentSurplus < 0) {
            start = i + 1;
            currentSurplus = 0;
        }
    }

    return (totalGas >= totalCost) ? start : -1;
}

Java Code

public int canCompleteCircuit(int[] gas, int[] cost) {
    int totalGas = 0, totalCost = 0, currentSurplus = 0, start = 0;

    for (int i = 0; i < gas.length; ++i) {
        totalGas += gas[i];
        totalCost += cost[i];
        currentSurplus += gas[i] - cost[i];

        if (currentSurplus < 0) {
            start = i + 1;
            currentSurplus = 0;
        }
    }

    return (totalGas >= totalCost) ? start : -1;
}

Time and Space Complexity Analysis

  • Time Complexity: O(N), where N is the number of stations. We traverse each station once.
  • Space Complexity: O(1), meaning no additional space proportional to input size is used beyond fixed variables.

Common Mistakes to Avoid

  1. Assuming Linear Traversal: The problem is circular, so it’s crucial to reset the journey's start if a negative surplus is encountered.
  2. Ignoring Total Gas vs Total Cost Check: Always check if total gas available is less than total gas needed to prevent unnecessary computations.
  3. Incorrect Index Management: Be careful when determining the starting index post reset in the algorithm.

Similar Problems on LeetCode

  1. LeetCode 134 - Gas Station
  2. LeetCode 330 - Patching Array
  3. LeetCode 135 - Candy

Additional Resources and References