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apps/docs/content/(05-decision-space)/greedy/134-gas-station.mdx

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+---
+title: '134. Gas Station'
+description: There are n gas stations along a circular route, where the amount of gas at the i^th station is gas[i]
+icon: dot
+topics:
+  - { name: "Array", slug: "array" }
+  - { name: "Greedy", slug: "greedy" }
+sidebar:
+  label: 'Gas Station'
+  badge: 'Medium'
+---
+
+### 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 <= 10^5`
+- `0 <= gas[i], cost[i] <= 10^4`
+- The input is generated such that the answer is unique.
+
+## Solution
+
+```py
+class Solution:
+    def canCompleteCircuit(self, gas: list[int], cost: list[int]) -> int:
+```