134, Gas Station

I Problem

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

Related Topics

• Greedy
• Array

II Solution

Approach 1: Brute Force

pub fn can_complete_circuit(gas: Vec<i32>, cost: Vec<i32>) -> i32 {
    let (len, mut start, mut curr_gas) = (gas.len(), 0, 0);

    'outer: for i in 0..len {
        start = len;
        for j in i..len {
            if gas[j] >= cost[j] {
                start = j;
                break;
            }
        }

        if start == len {
            break;
        }

        curr_gas = gas[start] - cost[start];
        for i in (start + 1..len).chain(0..start) {
            curr_gas += gas[i];
            if curr_gas < cost[i] {
                continue 'outer;
            }
            curr_gas -= cost[i];
        }

        return start as i32;
    }

    return -1;
}

public int canCompleteCircuit(int[] gas, int[] cost) {
    int len = gas.length, start, curr_gas;

    outer: for (int i = 0; i < len; i++) {
        start = len;
        for (int j = i; j < len; j++) {
            if (gas[j] >= cost[j]) {
                start = j;
                break;
            }
        }

        if (start == len) {
            break;
        }

        curr_gas = gas[start] - cost[start];
        for (int j = 1; j < len; j++) {
            int k = (start + j) % len;
            curr_gas += gas[k];
            if (curr_gas < cost[k]) {
                continue outer;
            }
            curr_gas -= cost[k];
        }

        return start;
    }

    return -1;
}

func canCompleteCircuit(gas []int, cost []int) int {
    size, start, currGas := len(gas), 0, 0

outer:
    for i := 0; i < size; i++ {
        start = size
        for j := i; j < size; j++ {
            if gas[j] >= cost[j] {
                start = j
                break
            }
        }
        
        if start == size {
            break
        }
        
        currGas = gas[start] - cost[start]
        for j := 1; j < size; j++ {
            k := (start + j) % size
            currGas += gas[k]
            if currGas < cost[k] {
                continue outer
            }
            currGas -= cost[k]
        }
        
        return start
    }

    return -1
}

Approach 2: Traverse Once

pub fn can_complete_circuit(gas: Vec<i32>, cost: Vec<i32>) -> i32 {
    let (len, mut i) = (gas.len(), 0);

    while i < len {
        let (mut sum_of_gas, mut sum_of_cost, mut cnt) = (0, 0, 0);

        while cnt < len {
            let j = (i + cnt) % len;
            sum_of_gas += gas[j];
            sum_of_cost += cost[j];
            if sum_of_cost > sum_of_gas {
                break;
            }
            cnt += 1;
        }

        if cnt == len {
            return i as i32;
        } else {
            i += cnt + 1;
        }
    }

    return -1;
}

public int canCompleteCircuit(int[] gas, int[] cost) {
    int len = gas.length, i = 0;

    while (i < len) {
        int sumOfGas = 0, sumOfCost = 0, cnt = 0;

        while (cnt < len) {
            int j = (i + cnt) % len;
            sumOfGas += gas[j];
            sumOfCost += cost[j];
            if (sumOfCost > sumOfGas) {
                break;
            }
            cnt++;
        }

        if (cnt == len) {
            return i;
        } else {
            i += cnt + 1;
        }
    }

    return -1;
}

func canCompleteCircuit(gas []int, cost []int) int {
    size, i := len(gas), 0

    for i < size {
        sumOfGas, sumOfCost, cnt := 0, 0, 0

        for cnt < size {
            j := (i + cnt) % size
            sumOfGas += gas[j]
            sumOfCost += cost[j]
            if sumOfCost > sumOfGas {
                break
            }
            cnt++
        }

        if cnt == size {
            return i
        } else {
            i += cnt + 1
        }
    }

    return -1
}