LeetCode //134. Gas Station

134. Gas Station

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 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
• $1<=n<=10^5$
• $0<=gas[i],cost[i]<=10^4$

From: LeetCode
Link: 134. Gas Station

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Solution:

Ideas：

The underlying idea behind this algorithm is to find out if the total amount of gas is sufficient to cover the total cost. If it is not, that means we can’t complete the circuit, regardless of where we start.

If the total amount of gas is sufficient to cover the total cost, then there must be a starting point from which we can complete the circuit. The algorithm finds this starting point by maintaining a “tank” variable that simulates the amount of gas in the car’s tank as we move from station to station.

At each station, we add the gas available at that station to the tank and subtract the cost to move to the next station. If at any point the tank goes below zero, that means we can’t start at the current starting point (as stored in the “start” variable), so we update the starting point to the next station and reset the tank to zero.

In essence, the algorithm tries to find a point from which we can start the journey such that we never run out of gas as we go around the circuit.

It’s important to note that the algorithm assumes that the input is such that if a solution exists, it is unique. This is a key part of why the algorithm can confidently return the starting point that it finds, without needing to check other potential starting points.

Code:

int canCompleteCircuit(int* gas, int gasSize, int* cost, int costSize){
    int total_gas = 0;
    int total_cost = 0;
    int tank = 0;
    int start = 0;
    
    for(int i = 0; i < gasSize; i++){
        total_gas += gas[i];
        total_cost += cost[i];
        tank += gas[i] - cost[i];
        if(tank < 0){
            start = i+1;
            tank = 0;
        }
    }
    
    if(total_gas < total_cost)
        return -1;
    else
        return start;
}