LeetCode #2739 — EASY

Total Distance Traveled

Build confidence with an intuition-first walkthrough focused on math fundamentals.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math

Example 1

5
10

Example 2

1
2

Step 02

Core Insight

What unlocks the optimal approach

Avoid calculations in decimal to prevent precision errors.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2739: Total Distance Traveled
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;
    }
}

// Accepted solution for LeetCode #2739: Total Distance Traveled
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
}

# Accepted solution for LeetCode #2739: Total Distance Traveled
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

// Accepted solution for LeetCode #2739: Total Distance Traveled
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
    }
}

// Accepted solution for LeetCode #2739: Total Distance Traveled
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2739: Total Distance Traveled
// 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;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.