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Easy
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Weekly Contest 350 Q1
Math
Simulation

中文文档

Description

A truck has two fuel tanks. You are given two integers, mainTank representing the fuel present in the main tank in liters and additionalTank representing the fuel present in the additional tank in liters.

The truck has a mileage of 10 km per liter. Whenever 5 liters of fuel get used up in the main tank, if the additional tank has at least 1 liters of fuel, 1 liters of fuel will be transferred from the additional tank to the main tank.

Return the maximum distance which can be traveled.

Note: Injection from the additional tank is not continuous. It happens suddenly and immediately for every 5 liters consumed.

 

Example 1:

Input: mainTank = 5, additionalTank = 10
Output: 60
Explanation: 
After spending 5 litre of fuel, fuel remaining is (5 - 5 + 1) = 1 litre and distance traveled is 50km.
After spending another 1 litre of fuel, no fuel gets injected in the main tank and the main tank becomes empty.
Total distance traveled is 60km.

Example 2:

Input: mainTank = 1, additionalTank = 2
Output: 10
Explanation: 
After spending 1 litre of fuel, the main tank becomes empty.
Total distance traveled is 10km.

 

Constraints:

  • 1 <= mainTank, additionalTank <= 100

Solutions

Solution 1: Simulation

We can simulate the process of the truck's movement. Each time, it consumes 1 liter of fuel from the main fuel tank and travels 10 kilometers. Whenever the fuel in the main fuel tank is consumed by 5 liters, if there is fuel in the auxiliary fuel tank, 1 liter of fuel is transferred from the auxiliary fuel tank to the main fuel tank. The simulation continues until the fuel in the main fuel tank is exhausted.

The time complexity is $O(n + m)$, where $n$ and $m$ are the amounts of fuel in the main and auxiliary fuel tanks, respectively. The space complexity is $O(1)$.

Python3

class Solution:
    def distanceTraveled(self, mainTank: int, additionalTank: int) -> int:
        ans = cur = 0
        while mainTank:
            cur += 1
            ans += 10
            mainTank -= 1
            if cur % 5 == 0 and additionalTank:
                additionalTank -= 1
                mainTank += 1
        return ans

Java

class Solution {
    public int distanceTraveled(int mainTank, int additionalTank) {
        int ans = 0, cur = 0;
        while (mainTank > 0) {
            cur++;
            ans += 10;
            mainTank--;
            if (cur % 5 == 0 && additionalTank > 0) {
                additionalTank--;
                mainTank++;
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int distanceTraveled(int mainTank, int additionalTank) {
        int ans = 0, cur = 0;
        while (mainTank > 0) {
            cur++;
            ans += 10;
            mainTank--;
            if (cur % 5 == 0 && additionalTank > 0) {
                additionalTank--;
                mainTank++;
            }
        }
        return ans;
    }
};

Go

func distanceTraveled(mainTank int, additionalTank int) (ans int) {
	cur := 0
	for mainTank > 0 {
		cur++
		ans += 10
		mainTank--
		if cur%5 == 0 && additionalTank > 0 {
			additionalTank--
			mainTank++
		}
	}
	return
}

Rust

impl Solution {
    pub fn distance_traveled(mut main_tank: i32, mut additional_tank: i32) -> i32 {
        let mut cur = 0;
        let mut ans = 0;

        while main_tank > 0 {
            cur += 1;
            main_tank -= 1;
            ans += 10;

            if cur % 5 == 0 && additional_tank > 0 {
                additional_tank -= 1;
                main_tank += 1;
            }
        }

        ans
    }
}

JavaScript

var distanceTraveled = function (mainTank, additionalTank) {
    let ans = 0,
        cur = 0;
    while (mainTank) {
        cur++;
        ans += 10;
        mainTank--;
        if (cur % 5 === 0 && additionalTank) {
            additionalTank--;
            mainTank++;
        }
    }
    return ans;
};