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107 lines (99 loc) · 3.08 KB
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package breadth_first_search;
import java.util.*;
/**
* Created by gouthamvidyapradhan on 19/08/2018
* 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.
*
* Now for some target position, say the length of the shortest sequence of instructions to get there.
*
* Example 1:
* Input:
*>
* Output: 2
* Explanation:
* The shortest instruction sequence is "AA".
* Your position goes from 0->1->3.
* Example 2:
* Input:
*>
* Output: 5
* Explanation:
* The shortest instruction sequence is "AAARA".
* Your position goes from 0->1->3->7->7->6.
*
*
* Note:
*
* 1 <= target <= 10000.
*
* Solution: O(n log n) Do a BFS and visit every possible state. Prune the search space by avoiding negative vertices
* and keep a boundary target of approximately (target * 2) - beyond this boundary target the race car should not
* progress in the forward direction.
*/
public class RaceCar {
/**
* Main method
* @param args
*/
public static void main(String[] args) {
System.out.println(new RaceCar().racecar(1000));
}
private class Node{
int v, s, d;
Node(int v, int s, int d){
this.v = v;
this.s = s;
this.d = d;
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (!(o instanceof Node)) return false;
Node node = (Node) o;
return v == node.v &&
s == node.s;
}
@Override
public int hashCode() {
return Objects.hash(v, s);
}
}
public int racecar(int target) {
if(target == 0) return 0;
Queue<Node> queue = new ArrayDeque<>();
Set<Node> done = new HashSet<>();
Node start = new Node(0, 1, 0);
done.add(start);
queue.offer(start);
while(!queue.isEmpty()){
Node curr = queue.poll();
if(curr.v < (target * 2)){
Node c1 = new Node(curr.v + curr.s, curr.s * 2, curr.d + 1);
if(c1.v >= 0){
if(!done.contains(c1)){
queue.offer(c1);
done.add(c1);
if(target == c1.v){
return c1.d;
}
}
}
}
Node c2 = new Node(curr.v, curr.s < 0 ? 1 : -1, curr.d + 1);
if(!done.contains(c2)){
done.add(c2);
queue.offer(c2);
}
}
return -1;
}
}