**Race to Finish Line - Google Top Interview Questions**

### Problem Statement :

You are driving a car in a one-dimensional line and are currently at position = 0 with speed = 1. You can make one of two moves: Accelerate: position += speed and speed *= 2 Reverse: speed = -1 if speed > 0 otherwise speed = 1. Return the minimum number of moves it would take to reach target. Constraints 1 ≤ target ≤ 100,000 Example 1 Input target = 7 Output 3 Explanation We can accelerate 3 times to reach 7. 0 -> 1 -> 3 -> 7 Example 2 Input target = 6 Output 5 Explanation We can accelerate 3 times to reach 7. 0 -> 1 -> 3 -> 7. Then we reverse to change our speed to -1. Then we accelerate to reach 6.

### Solution :

` ````
Solution in C++ :
int dp[100005];
// dp[i] is the minimum time needed to travel exactly a distance of i
int solve(int target) {
if (dp[target]) {
return dp[target];
}
int time = 0;
int speed = 1;
int position = 0;
// this is achievable by accelerating, then reversing twice, repeating "target" times until we
// get to the target
int besttime = 3 * target;
while (true) {
position += speed;
time++;
if (position == target) {
besttime = min(besttime, time);
break;
}
// reverse once, but since we're after the target we don't continue traveling further
// forward
if (position > target) {
besttime = min(besttime, time + 1 + solve(position - target));
break;
}
// reverse twice to reset speed to 1
besttime = min(besttime, time + 2 + solve(target - position));
// reverse in the middle, but don't reverse all the way
if (speed > 1) {
int candtime = time + 2;
int candposition = target - position;
for (int revspeed = 1; candposition + revspeed < target; revspeed *= 2) {
candposition += revspeed;
candtime++;
besttime = min(besttime, candtime + solve(candposition));
}
}
speed *= 2;
}
dp[target] = besttime;
return besttime;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(int target) {
if (target == 0)
return 0;
boolean[][] ff = new boolean[38][target * 2 + 1];
LinkedList<Integer> ll1 = new LinkedList(), ll2 = new LinkedList();
ll1.add(0);
ll2.add(0);
ff[0][0] = true;
int lay = 0, count = 1;
while (!ll1.isEmpty()) {
int p = ll1.poll(), s = ll2.poll();
int speed = s < 19 ? 1 << s : -(1 << (s - 19));
if (p + speed == target)
return lay + 1;
if (p + speed >= 0 && p + speed < ff[0].length && !ff[s + 1][p + speed]) {
ll1.add(p + speed);
ll2.add(s + 1);
ff[s + 1][p + speed] = true;
}
s = s < 19 ? 19 : 0;
if (!ff[s][p]) {
ll1.add(p);
ll2.add(s);
ff[s][p] = true;
}
--count;
if (count == 0) {
++lay;
count = ll1.size();
}
}
return -1;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, target):
self.ans = int(1e9)
hi = 1
while (1 << hi) < target:
hi += 1
self.dfs(hi, 0, 0, 0, target)
return self.ans
def dfs(self, digit, cost, pos, neg, target):
tot = cost + max(2 * (pos - 1), 2 * neg - 1)
if tot >= self.ans:
return
if target == 0:
self.ans = min(self.ans, tot)
return
step = (1 << digit) - 1
if step * 2 < abs(target):
return
self.dfs(digit - 1, cost, pos, neg, target)
self.dfs(digit - 1, cost + digit, pos + 1, neg, target - step)
self.dfs(digit - 1, cost + digit * 2, pos + 2, neg, target - step * 2)
self.dfs(digit - 1, cost + digit, pos, neg + 1, target + step)
self.dfs(digit - 1, cost + digit * 2, pos, neg + 2, target + step * 2)
```

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