818. Race Car

Problem

Your car starts at position 0 and speed +1 on an infinite number line. Your car can go into negative positions. Your car drives automatically according to a sequence of instructions 'A' (accelerate) and 'R' (reverse):

When you get an instruction 'A', your car does the following:

• position += speed

• speed *= 2

  When you get an instruction 'R', your car does the following:

• If your speed is positive then speed = -1

• otherwise speed = 1

  Your position stays the same.

For example, after commands "AAR", your car goes to positions 0 --> 1 --> 3 --> 3, and your speed goes to 1 --> 2 --> 4 --> -1.

Given a target position target, return the length of the shortest sequence of instructions to get there.

  Example 1:

Input: target = 3
Output: 2
Explanation: 
The shortest instruction sequence is "AA".
Your position goes from 0 --> 1 --> 3.

Example 2:

Input: target = 6
Output: 5
Explanation: 
The shortest instruction sequence is "AAARA".
Your position goes from 0 --> 1 --> 3 --> 7 --> 7 --> 6.

  Constraints:

• 1 <= target <= 10^4

Solution

class Solution {
    public int racecar(int target) {
        Queue<int[]> queue = new LinkedList<>();
        queue.add(new int[] {0, 1, 0});
        while (!queue.isEmpty()) {
            int[] arr = queue.poll();
            if (arr[0] == target) {
                return arr[2];
            }
            queue.add(new int[] {arr[0] + arr[1], 2 * arr[1], 1 + arr[2]});
            if ((arr[0] + arr[1]) > target && arr[1] > 0) {
                queue.add(new int[] {arr[0], -1, 1 + arr[2]});
            }
            if ((arr[0]) + arr[1] < target && (arr[1] < 0)) {
                queue.add(new int[] {arr[0], 1, 1 + arr[2]});
            }
        }
        return -1;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).