Total Distance Traveled

Difficulty: Easy | Acceptance: 41.40% | Paid: No Topics: Math, Simulation

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

The truck consumes 1 liter of fuel for every kilometer it travels. If the fuel in the main tank drops to 0 liters while driving, the truck cannot move forward.

Whenever 5 liters of fuel are consumed from the main tank, if there is at least 1 liter of fuel in the additional tank, 1 liter of fuel will be transferred from the additional tank to the main tank.

Return the maximum distance the truck can travel.

• Examples
• Constraints
• Simulation
• Mathematical Optimization

Examples

Example 1:

Input: mainTank = 5, additionalTank = 10
Output: 60
Explanation:
- Start with 5 liters in main tank and 10 liters in additional tank.
- Travel 5 km consuming 5 liters from main tank. Transfer 1 liter from additional to main.
- Now main tank has 1 liter, additional tank has 9 liters.
- Travel 1 km consuming 1 liter from main tank.
- Total distance traveled is 5 + 1 = 60 km.

Example 2:

Input: mainTank = 1, additionalTank = 2
Output: 10
Explanation:
- Start with 1 liter in main tank and 2 liters in additional tank.
- Travel 1 km consuming 1 liter from main tank.
- No transfer happens because main tank is empty.
- Total distance traveled is 10 km.

Constraints

0 <= mainTank, additionalTank <= 100

Simulation

Intuition We can directly simulate the process of driving the truck. We iterate while there is fuel in the main tank, consuming 1 liter (10 km) at a time. After every 5 liters consumed, we check if we can transfer fuel from the additional tank to the main tank.

Steps

• Initialize distance to 0.
• Loop while mainTank is greater than 0.
• Decrement mainTank by 1 and increment distance by 10.
• Check if distance is a multiple of 50 (meaning 5 liters have been consumed) and if additionalTank is greater than 0.
• If true, increment mainTank by 1 and decrement additionalTank by 1.
• Return distance.

class Solution:
    def distanceTraveled(self, mainTank: int, additionalTank: int) -> int:
        distance = 0
        while mainTank > 0:
            mainTank -= 1
            distance += 10
            if distance % 50 == 0 and additionalTank > 0:
                mainTank += 1
                additionalTank -= 1
        return distance

Complexity

• Time: O(mainTank + additionalTank) - In the worst case, we iterate once for every liter of fuel available.
• Space: O(1) - We only use a few variables for tracking state.
• Notes: This approach is intuitive and perfectly acceptable given the small constraints (max 100).

Mathematical Optimization

Intuition We can calculate the result mathematically without simulation. To transfer 1 liter from the additional tank, we must burn 5 liters from the main tank. Effectively, using 1 liter from the additional tank “costs” 5 liters of consumption from the main tank to unlock. However, the transferred liter itself adds to the main tank. So, for every 1 liter transferred, the net decrease in the initial main tank fuel is 4 liters (5 burned - 1 transferred). Thus, the maximum amount of fuel we can transfer is limited by mainTank / 4 (integer division) and the amount available in additionalTank.

Steps

• Calculate the maximum fuel that can be transferred from the additional tank. This is the minimum of additionalTank and mainTank / 4.
• The total fuel used is the sum of the initial mainTank and the calculated transfer amount.
• Multiply the total fuel by 10 to get the distance.

class Solution:
    def distanceTraveled(self, mainTank: int, additionalTank: int) -> int:
        transfer = min(additionalTank, mainTank // 4)
        return (mainTank + transfer) * 10

Complexity

• Time: O(1) - Constant time arithmetic operations.
• Space: O(1) - No extra space used.
• Notes: This is the optimal solution, reducing the problem to a simple formula.