818. Race Car
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:
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.
Approach #1: C++. [DFS]
class Solution {
public:
int racecar(int target) {
queue<pair<int, int>> q;
q.push({0, 1});
unordered_set<string> v;
v.insert("0_1");
v.insert("0_-1");
int steps = 0;
while (!q.empty()) {
int size = q.size();
while (size--) {
auto p = q.front(); q.pop();
int pos = p.first;
int speed = p.second;
{
int pos1 = pos + speed;
int speed1 = speed * 2;
pair<int, int> p1{pos1, speed1};
if (pos1 == target) return steps+1;
if (p1.first > 0 && p1.first < 2 * target)
q.push(p1);
}
{
int speed2 = speed > 0 ? -1 : 1;
pair<int, int> p2{pos, speed2};
string key2 = to_string(pos) + "_" + to_string(speed2);
if (!v.count(key2)) {
q.push(p2);
v.insert(key2);
}
}
}
steps++;
}
return -1;
}
};
Approach #2: Java. [DP]
class Solution {
private static int[][] m;
public int racecar(int target) {
if (m == null) {
final int kMaxT = 10000;
m = new int[kMaxT + 1][2];
for (int t = 1; t <= kMaxT; ++t) {
int n = (int)Math.ceil(Math.log(t + 1) / Math.log(2));
if (1 << n == t + 1) {
m[t][0] = n;
m[t][1] = n + 1;
continue;
}
int l = (1 << n) - 1 - t;
m[t][0] = n + 1 + Math.min(m[l][1], m[l][0] + 1);
m[t][1] = n + 1 + Math.min(m[l][0], m[l][1] + 1);
for (int i = 1; i < t; ++i) {
for (int d = 0; d <= 1; ++d) {
m[t][d] = Math.min(m[t][d], Math.min(
m[i][0] + 2 + m[t-i][d],
m[i][1] + 1 + m[t-i][d]));
}
}
}
}
return Math.min(m[target][0], m[target][1]);
}
}
Approach #3: Python. [DP]
class Solution(object):
def __init__(self): self.dp = {0: 0} def racecar(self, t):
"""
:type target: int
:rtype: int
"""
if t in self.dp: return self.dp[t]
n = t.bit_length()
if 2**n - 1 == t: self.dp[t] = n
else:
self.dp[t] = self.racecar(2**n - 1 - t) + n + 1
for m in range(n-1):
self.dp[t] = min(self.dp[t], self.racecar(t - 2**(n-1) + 2**m) + n + m + 1)
return self.dp[t]
Analysis:
http://zxi.mytechroad.com/blog/dynamic-programming/leetcode-818-race-car/
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