What is the brute force approach and complexity for Gas Station?

Try each station as the start and simulate the whole loop with time complexity O(N²) and space complexity O(1).

What is the optimal approach and complexity for Gas Station?

Track the total and a running tank; when the tank drops below 0, the next station becomes the start with optimal time complexity O(N) and space complexity O(1).

How do you identify the pattern for Gas Station?

Look for key signals in the prompt: "circular route" indicates We drive around once, so the totals must balance, and the start can be any station.; "unlimited gas tank" indicates The fuel never caps out, so only the running balance matters.; "return the starting index, or -1" indicates We need one valid start, and the total tells us early whether any start can exist..

Gas Station

Medium·LeetCode #134 Pro OnlySlight Variation

LeetCode

Problem Statement

There are n gas stations on a circular road. Station i gives gas[i] litres of fuel. Driving from station i to the next station takes cost[i] litres. Your tank has no size limit, and you start with an empty tank at one of the stations.

Return the index of a station where you can start and drive once all the way around the circle, going the same direction at every step. If no start works, return -1. The input is set up so that at most one answer exists.

Input

Two arrays gas and cost of the same length n.

Output

The index of the starting station, or -1 if no start works.

Example 1

Input: gas = [1, 2, 3, 4, 5], cost = [3, 4, 5, 1, 2]

Output: 3

Start at station 3 with 4 litres. Each leg uses its cost and the next station refills the tank. You reach station 3 again with exactly 0 litres left, so the trip works.

Example 2

Input: gas = [2, 3, 4], cost = [3, 4, 3]

Output: -1

The total fuel is 9, but the total cost is 10. No start can cover the whole circle.

Constraints

• 1 ≤ n ≤ 10^5
• `gas` and `cost` both have length `n`.
• 0 ≤ gas[i], cost[i] ≤ 10^4
• The input is set up so that the answer is unique, if there is one.

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“circular route”

→ We drive around once, so the totals must balance, and the start can be any station.

“unlimited gas tank”

→ The fuel never caps out, so only the running balance matters.

“return the starting index, or -1”

→ We need one valid start, and the total tells us early whether any start can exist.

Approach Complexity Ladder

Brute Force — drive from every startO(N²)O(1)

Try each station as the start and simulate the whole loop

Optimal — one pass, move the start after each failureO(N)O(1)

Track the total and a running tank; when the tank drops below 0, the next station becomes the start

Brute Force Intuition

For each start s, drive around the circle, filling up at each station and then paying the cost to reach the next one. If the tank ever goes below 0, this start fails. If you finish the loop with the tank at 0 or more, return s. If no start passes, return -1.

Recognition rule for this sub-pattern:

Keep a running balance; when it fails, the start must move past this point. Two passes handle left and right rules.

Remember this · Greedy

Trigger

Maximise count / minimise cost with a natural "best next step", sorting makes the choice obvious, reachability.

Think

Is there a choice I can always make safely without looking ahead?

Key question

Why can't the greedy choice be worse than the optimal one? (Try to break it with a small example.)

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