Coding tests are becoming mandatory for more and more companies, and JavaScript is one of the most popular languages in web development.
In this course, we will solve commonly occurring algorithm problems found in coding tests using JavaScript.
Today’s topic is the problem ‘Fast Travel with a Time Machine’.
Problem Description
You are a scientist who can operate a time machine. The time machine can move to a specific time, and
given two integers a and b, you need to calculate the minimum time (number of moves) it takes to move from a to b.
The time machine follows these rules:
- You can add 1 or subtract 1 from your current position.
- You can double your current position.
Write a function that calculates the minimum number of moves required to travel from a to b for given a and b.
Input Format
– Two integers a (0 ≤ a ≤ 105), b (0 ≤ b ≤ 105) are given.
Output Format
– Output the minimum number of moves required to travel from a to b as an integer.
Examples
Input: a = 2, b = 9
Output: 4
Input: a = 5, b = 22
Output: 7
Approach to Solve the Problem
To solve this problem, we will use BFS (Breadth-First Search). BFS is an algorithm suitable for finding the shortest path,
exploring all possible states from the given state to find the shortest move sequence to reach the target state.
In this case, each state represents a time the time machine could occupy.
Step 1: Prepare to Use BFS Algorithm
To implement BFS, we will use a queue. First, we add the starting position a to the queue.
Then, we will repeat adding all possible moves to the queue until we reach the target position b.
The possible moves from each position are as follows:
- Add 1 to the current position
- Subtract 1 from the current position
- Double the current position
We will keep track of the number of moves taken for each move until we reach the target position, and output the count when we reach the target.
Step 2: Implement the Code
function minimumMoves(a, b) {
if (a >= b) return a - b; // If a is greater than or equal to b, determine moves by simply subtracting the difference
const queue = [[a, 0]]; // [current position, number of moves]
const visited = new Set(); // Record visited positions
visited.add(a); // Mark the starting position as visited
while (queue.length > 0) {
const [current, moves] = queue.shift();
// Possible moves
const nextMoves = [current - 1, current + 1, current * 2];
for (let next of nextMoves) {
if (next === b) return moves + 1; // Return move count when the target position is reached
if (next < 0 || next > 100000 || visited.has(next)) continue; // Only for valid range and unvisited positions
visited.add(next); // Mark the next position as visited
queue.push([next, moves + 1]); // Add the next position and move count to the queue
}
}
}
Step 3: Analyze Algorithm Complexity
The time complexity of the algorithm is O(n).
In the worst case, we need to explore all possible positions, which corresponds to O(n),
and the space complexity is also O(n). This complexity arises from the space needed for the queue and visited records.
Step 4: Optimization Possibilities
This algorithm is already based on BFS, which explores the shortest path
and does not require additional optimization. However, depending on the situation,
DFS (Depth-First Search) could also be applied, but BFS is more effective for this problem.
Step 5: Conclusion
Through the ‘Fast Travel with a Time Machine’ problem, we have briefly understood the principle of BFS and learned how to solve a problem using JavaScript.
In this way, various problems can be solved, and it is crucial to master basic algorithms to achieve good results in coding tests.
Additional Resources
– To further study the BFS algorithm and practice various problem-solving, we recommend the following resources.