• For example, if fi = 3 and ri = 2, then the tire would finish its 1st lap in 3 seconds, its 2nd lap in 3 * 2 = 6 seconds, its 3rd lap in 3 * 22 = 12 seconds, etc.

You are also given an integer changeTime and an integer numLaps.

The race consists of numLaps laps and you may start the race with any tire. You have an unlimited supply of each tire and after every lap, you may change to any given tire (including the current tire type) if you wait changeTime seconds.

Return the minimum time to finish the race.

Example 1:

Input: tires = [[2,3],[3,4]], changeTime = 5, numLaps = 4
Output: 21
Explanation: 
Lap 1: Start with tire 0 and finish the lap in 2 seconds.
Lap 2: Continue with tire 0 and finish the lap in 2 * 3 = 6 seconds.
Lap 3: Change tires to a new tire 0 for 5 seconds and then finish the lap in another 2 seconds.
Lap 4: Continue with tire 0 and finish the lap in 2 * 3 = 6 seconds.
Total time = 2 + 6 + 5 + 2 + 6 = 21 seconds.
The minimum time to complete the race is 21 seconds.

Example 2:

Input: tires = [[1,10],[2,2],[3,4]], changeTime = 6, numLaps = 5
Output: 25
Explanation: 
Lap 1: Start with tire 1 and finish the lap in 2 seconds.
Lap 2: Continue with tire 1 and finish the lap in 2 * 2 = 4 seconds.
Lap 3: Change tires to a new tire 1 for 6 seconds and then finish the lap in another 2 seconds.
Lap 4: Continue with tire 1 and finish the lap in 2 * 2 = 4 seconds.
Lap 5: Change tires to tire 0 for 6 seconds then finish the lap in another 1 second.
Total time = 2 + 4 + 6 + 2 + 4 + 6 + 1 = 25 seconds.
The minimum time to complete the race is 25 seconds. 

Constraints:

• 1 <= tires.length <= 105
• tires[i].length == 2
• 1 <= fi, changeTime <= 105
• 2 <= ri <= 105
• 1 <= numLaps <= 1000

Solution: DP

Observation: since ri >= 2, we must change tire within 20 laps, otherwise it will be slower.

pre-compute the time to finish k laps using each type of tire (k < 20), find min for each lap.

dp[i] = best[i], i < 20,
dp[i] = min{dp[i – j] + changeTime + best[j]}, i > 20

Time complexity: O(n)
Space complexity: O(n) -> O(20)

// Author: Huahua
class Solution {
public:
  int minimumFinishTime(vector<vector<int>>& tires, int changeTime, int numLaps) {
    constexpr int kMax = 20;
    vector<long> best(kMax, INT_MAX);
    for (const auto& t : tires)
      for (long i = 1, l = t[0], s = t[0]; i < kMax && l < t[0] + changeTime; ++i, l *= t[1], s += l)
        best[i] = min(best[i], s);
    vector<long> dp(numLaps + 1, INT_MAX);    
    for (int i = 1; i <= numLaps; ++i)
      for (int j = 1; j <= min(i, kMax - 1); ++j)
        dp[i] = min(dp[i], best[j] + (i > j) * (changeTime + dp[i - j]));
    return dp[numLaps];
  }
};

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