Introduction

In this blog post, we will explore how to solve the LeetCode problem "134" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

There are n gas stations along a circular route, where the amount of gas at the ith station is gas[i]. You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from the ith 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 solution, it is guaranteed to be unique. Example 1: Input: gas = [1,2,3,4,5], cost = [3,4,5,1,2] Output: 3 Explanation: Start at station 3 (index 3) and fill up with 4 unit of gas. Your tank = 0 + 4 = 4 Travel to station 4. Your tank = 4 - 1 + 5 = 8 Travel to station 0. Your tank = 8 - 2 + 1 = 7 Travel to station 1. Your tank = 7 - 3 + 2 = 6 Travel to station 2. Your tank = 6 - 4 + 3 = 5 Travel to station 3. The cost is 5. Your gas is just enough to travel back to station 3. Therefore, return 3 as the starting index. Example 2: Input: gas = [2,3,4], cost = [3,4,3] Output: -1 Explanation: You can't start at station 0 or 1, as there is not enough gas to travel to the next station. Let's start at station 2 and fill up with 4 unit of gas. Your tank = 0 + 4 = 4 Travel to station 0. Your tank = 4 - 3 + 2 = 3 Travel to station 1. Your tank = 3 - 3 + 3 = 3 You cannot travel back to station 2, as it requires 4 unit of gas but you only have 3. Therefore, you can't travel around the circuit once no matter where you start. Constraints: n == gas.length == cost.length 1 <= n <= 105 0 <= gas[i], cost[i] <= 104

Explanation

Here's the solution to the gas station problem, focusing on efficiency and clarity:

• Greedy Approach: Iterate through the gas stations. If starting from a station allows a complete circuit, that station is the answer. If a starting station fails, it means no station between the current start and the point of failure can be a valid starting station.
• Track Cumulative Gas: Maintain a running tank representing the gas level. If the tank drops below zero, it indicates a failure from the current starting station.

• Total Gas Check: Before iterating, ensure the total gas available is greater than or equal to the total cost. If not, no solution exists.

• Runtime Complexity: O(n), Storage Complexity: O(1)

Code

    def canCompleteCircuit(gas: list[int], cost: list[int]) -> int:
    """
    Finds the starting gas station's index if you can travel around the circuit once,
    otherwise returns -1.
    """
    n = len(gas)

    # Check if a solution is even possible
    if sum(gas) < sum(cost):
        return -1

    start = 0
    tank = 0
    total = 0  # Added total to track overall gas availability which also can solve the issue if total(gas) < total(cost)

    for i in range(n):
        tank += gas[i] - cost[i]
        total += gas[i] - cost[i] # Accumulate overall gas balance
        if tank < 0:
            start = i + 1
            tank = 0

    return start