2739. Total Distance Traveled

EasyMathSimulation
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Problem Description

In this problem, we have a truck with two fuel tanks: a main tank and an additional tank. The main tank's current fuel level is given by the integer mainTank, and the additional tank's fuel level is given by additionalTank, both measured in liters. The truck's fuel efficiency is fixed at 10 kilometers per liter.

Fuel transfer between tanks works under a specific rule: for every 5 liters consumed from the main tank, 1 liter is transferred from the additional tank to the main tank if there is at least 1 liter of fuel in the additional tank. This transfer is not continuous but occurs instantaneously every time the main tank's fuel level goes down by 5 liters.

We need to calculate the maximum distance the truck can travel given these conditions.

Intuition

The core idea behind the solution is to simulate the truck's fuel consumption while considering the rule for transferring fuel from the additional tank to the main tank.

We keep track of the total distance (ans) and the fuel consumption (cur) as we decrement fuel from the mainTank. For every liter of fuel used, the truck travels 10 km, so with each iteration, we increase the total distance by 10 km.

We need to track every time the main tank uses 5 liters of fuel to simulate the transfer from the additional tank. If the additionalTank has at least 1 liter, we transfer 1 liter to the mainTank. This is done by checking if cur is divisible by 5 and there is fuel in the additionalTank. If both conditions are met, we decrement 1 liter from the additionalTank and add 1 liter to the mainTank.

The loop continues until there is no more fuel in the mainTank. At this point, we have traveled the maximum possible distance, and we return the ans variable as the result.

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Solution Approach

To implement the solution, the Solution class defines a method distanceTraveled that takes mainTank and additionalTank as its parameters. This method does not use any complex data structures or algorithms but follows a straightforward iterative process.

Here's a step-by-step breakdown of the approach:

  1. Initialize ans to 0, which will keep track of the total distance traveled by the truck.
  2. Keep track of liters consumed from the main tank with cur, initializing it to 0.
  3. Use a while loop to iterate until the mainTank is empty (i.e., mainTank becomes 0).
  4. Inside the loop, increment cur by 1 for each iteration to count the fuel consumption.
  5. For every liter of fuel consumed, add 10 to ans, which is the distance traveled per liter of fuel.
  6. Decrement 1 from mainTank since a liter of fuel is used.
  7. Check if the value of cur is a multiple of 5 and that additionalTank has at least 1 liter.
  8. If both conditions are true, transfer 1 liter from the additionalTank to the mainTank by decrementing 1 from additionalTank and incrementing 1 to mainTank.

The key idea is to simulate the consumption and transfer of fuel from one tank to another while keeping track of the distance traveled every time the fuel is consumed. The loop halts when the main tank is empty, meaning no more distance can be traveled. At this point, the ans variable holds the maximum distance the truck can travel, so we return ans as the final result.

Note that no additional data structures are needed for this implementation, and the method uses simple arithmetic and conditional checks to achieve the required simulation.

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1int factorial(int n) {
2  if (n < 0) {
3    return -1;
4  } else if (n == 0) {
5    return 1;
6  } else {
7    return n * factorial(n - 1);
8  }
9}

Example Walkthrough

Let's assume the following scenario where mainTank has 12 liters of fuel and additionalTank has 2 liters of fuel. We want to calculate the maximum distance the truck can travel. Here's how the solution approach is applied in this example:

  1. We initialize ans to 0 to start counting the total distance traveled, and cur to 0 to track the liters consumed from the main tank.
  2. We enter the while loop since mainTank is not empty (contains 12 liters).
  3. The truck uses 1 liter of fuel, cur is incremented to 1, and mainTank is decremented to 11. We add 10 km to ans, for a total of 10 km traveled.
  4. This process repeats, with cur incrementing each time and mainTank decrementing for every liter used. Also, with every liter, another 10 km is added to ans.
  5. When cur becomes 5, indicating we have consumed 5 liters, and since additionalTank has at least 1 liter, we transfer 1 liter from the additionalTank to the mainTank. Now, mainTank has 7 liters (originally 6 before the transfer) and additionalTank has 1 liter left.
  6. The loop continues; we increment cur, decrement mainTank by 1, and add 10 km to ans for each liter of fuel.
  7. When cur reaches 10, we have consumed another batch of 5 liters from the main tank, but this time the additionalTank is empty, so no transfer occurs.
  8. The loop finally terminates when mainTank reaches 0, meaning there is no more fuel left to use.

By the end of the process:

  • The ans value reflects the total distance traveled. After consuming all 12 liters from the mainTank and the 1 liter transferred from the additionalTank, the truck would have traveled 130 km (12 liters + 1 transferred liter, each multiplied by 10 km per liter).
  • The while loop has terminated because the mainTank is now empty.

So, by plugging in the actual liters and applying the approach outlined in the Problem Description, we've concluded that the truck can travel a maximum distance of 130 kilometers with the given amounts of fuel in the main and additional tanks.

Solution Implementation

1class Solution:
2    def distanceTraveled(self, main_tank: int, additional_tank: int) -> int:
3        # Initialize variables to track the distance and the current step count
4        distance = current_step = 0
5      
6        # Continue the loop as long as there is fuel in the main tank
7        while main_tank:
8            current_step += 1  # Increment the step count
9            distance += 10      # Increase the distance traveled by 10 (assumed unit of distance per unit of fuel)
10            main_tank -= 1      # Decrease the main tank fuel by 1 unit
11          
12            # Check if a unit from the additional tank should be transferred to the main tank
13            if current_step % 5 == 0 and additional_tank:
14                additional_tank -= 1  # Remove 1 unit from the additional tank
15                main_tank += 1        # Add 1 unit to the main tank (refueling from the additional tank)
16      
17        # Return the total distance traveled
18        return distance
19
1class Solution {
2    public int distanceTraveled(int mainTank, int additionalTank) {
3        int distance = 0;  // Initialize the total distance travelled to 0
4        int moves = 0;     // Counter to keep track of moves made
5
6        // Loop until the main tank is empty
7        while (mainTank > 0) {
8            moves++;                 // Increment moves count
9            distance += 10;          // Increase distance by 10 for each move
10            mainTank--;              // Use one unit of fuel from the main tank
11          
12            // Every 5 moves, if there is fuel in the additional tank, transfer it to the main tank
13            if (moves % 5 == 0 && additionalTank > 0) {
14                additionalTank--;    // Use one unit of fuel from the additional tank
15                mainTank++;          // Add one unit of fuel to the main tank
16            }
17        }
18
19        return distance; // Return the total distance travelled
20    }
21}
22
1class Solution {
2public:
3    // Function to calculate the distance traveled given the amount of fuel in the main tank and the additional tank
4    int distanceTraveled(int mainTank, int additionalTank) {
5        int distance = 0; // Initialize distance traveled
6        int steps = 0;    // Initialize steps taken
7      
8        // Loop runs as long as there is fuel in the main tank
9        while (mainTank > 0) {
10            steps++;             // Increment steps
11            distance += 10;      // Increase distance by 10 for each step
12            mainTank--;          // Decrease main tank fuel by 1
13          
14            // Every 5 steps, if there is fuel in the additional tank, transfer 1 unit to the main tank
15            if (steps % 5 == 0 && additionalTank > 0) {
16                additionalTank--; // Use 1 unit of fuel from the additional tank
17                mainTank++;       // Add 1 unit of fuel to the main tank
18            }
19        }
20      
21        // Return the total distance traveled
22        return distance;
23    }
24};
25
1// Function to calculate the distance traveled given the amount of fuel in the main tank and the additional tank
2function distanceTraveled(mainTank: number, additionalTank: number): number {
3    let distance: number = 0; // Variable to store the total distance traveled
4    let steps: number = 0;    // Variable to count the number of steps taken
5
6    // Loop runs as long as there is fuel in the main tank
7    while (mainTank > 0) {
8        steps++;              // Increment the step count by 1
9        distance += 10;       // Increase distance by 10 for each step
10        mainTank--;           // Decrement the main tank fuel by 1
11
12        // On every 5th step:
13        if (steps % 5 === 0 && additionalTank > 0) {
14            additionalTank--; // Decrease the additional tank fuel by 1
15            mainTank++;       // Transfer 1 unit of fuel to the main tank
16        }
17    }
18  
19    // Return the calculated distance traveled
20    return distance;
21}
22
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1int sqrt(int n) {
2  for (int guess = 1; guess * guess <= n; guess++) {
3    if (guess * guess == n) {
4      return guess;
5    }
6  }
7  return -1;
8}

Time and Space Complexity

Time Complexity

The time complexity of the code is primarily determined by the while loop that runs as long as mainTank is not empty. For each iteration in the loop, we perform a constant amount of work: increment cur, decrement mainTank, add 10 to ans, and conditionally transfer fuel from the additionalTank to mainTank. The loop could run for as many as mainTank iterations, and in the worst case, we do not add any fuel from additionalTank to mainTank. Thus, the worst-case time complexity is O(mainTank).

However, it's worth noting that every 5th iteration, if additionalTank has fuel, mainTank is incremented, which can happen up to additionalTank times. This effectively adds extra iterations to the loop, but since this action only happens after every 5th decrement of mainTank, the additional loop iterations are bounded by additionalTank / 5. Therefore, the precise worst-case time complexity of the code would be O(mainTank + additionalTank/5).

Space Complexity

The space complexity of the code is O(1) because there is only a fixed number of variables used (ans, cur, mainTank, additionalTank), and no additional space is required that grows with the size of the input.

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