forked from lzl124631x/LeetCode
-
Notifications
You must be signed in to change notification settings - Fork 0
Commit
This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository.
- Loading branch information
1 parent
164bd6b
commit b9cfdd1
Showing
2 changed files
with
98 additions
and
0 deletions.
There are no files selected for viewing
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,93 @@ | ||
| # [2188. Minimum Time to Finish the Race (Hard)](https://leetcode.com/problems/minimum-time-to-finish-the-race/) | ||
|
|
||
| <p>You are given a <strong>0-indexed</strong> 2D integer array <code>tires</code> where <code>tires[i] = [f<sub>i</sub>, r<sub>i</sub>]</code> indicates that the <code>i<sup>th</sup></code> tire can finish its <code>x<sup>th</sup></code> successive lap in <code>f<sub>i</sub> * r<sub>i</sub><sup>(x-1)</sup></code> seconds.</p> | ||
|
|
||
| <ul> | ||
| <li>For example, if <code>f<sub>i</sub> = 3</code> and <code>r<sub>i</sub> = 2</code>, then the tire would finish its <code>1<sup>st</sup></code> lap in <code>3</code> seconds, its <code>2<sup>nd</sup></code> lap in <code>3 * 2 = 6</code> seconds, its <code>3<sup>rd</sup></code> lap in <code>3 * 2<sup>2</sup> = 12</code> seconds, etc.</li> | ||
| </ul> | ||
|
|
||
| <p>You are also given an integer <code>changeTime</code> and an integer <code>numLaps</code>.</p> | ||
|
|
||
| <p>The race consists of <code>numLaps</code> laps and you may start the race with <strong>any</strong> tire. You have an <strong>unlimited</strong> supply of each tire and after every lap, you may <strong>change</strong> to any given tire (including the current tire type) if you wait <code>changeTime</code> seconds.</p> | ||
|
|
||
| <p>Return<em> the <strong>minimum</strong> time to finish the race.</em></p> | ||
|
|
||
| <p> </p> | ||
| <p><strong>Example 1:</strong></p> | ||
|
|
||
| <pre><strong>Input:</strong> tires = [[2,3],[3,4]], changeTime = 5, numLaps = 4 | ||
| <strong>Output:</strong> 21 | ||
| <strong>Explanation:</strong> | ||
| 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. | ||
| </pre> | ||
|
|
||
| <p><strong>Example 2:</strong></p> | ||
|
|
||
| <pre><strong>Input:</strong> tires = [[1,10],[2,2],[3,4]], changeTime = 6, numLaps = 5 | ||
| <strong>Output:</strong> 25 | ||
| <strong>Explanation:</strong> | ||
| 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. | ||
| </pre> | ||
|
|
||
| <p> </p> | ||
| <p><strong>Constraints:</strong></p> | ||
|
|
||
| <ul> | ||
| <li><code>1 <= tires.length <= 10<sup>5</sup></code></li> | ||
| <li><code>tires[i].length == 2</code></li> | ||
| <li><code>1 <= f<sub>i</sub>, changeTime <= 10<sup>5</sup></code></li> | ||
| <li><code>2 <= r<sub>i</sub> <= 10<sup>5</sup></code></li> | ||
| <li><code>1 <= numLaps <= 1000</code></li> | ||
| </ul> | ||
|
|
||
|
|
||
| **Similar Questions**: | ||
| * [Minimum Skips to Arrive at Meeting On Time (Hard)](https://leetcode.com/problems/minimum-skips-to-arrive-at-meeting-on-time/) | ||
|
|
||
| ## Solution 1. DP | ||
|
|
||
| The `best[i]` is the least time we need to finish `i+1` laps **using a single tire**. For each tire, we try to update the `best` values with it. | ||
|
|
||
| The `dp` part is doing knapsack using the `best` values to get the total `numLaps` laps. | ||
|
|
||
| ```cpp | ||
| // OJ: https://leetcode.com/problems/minimum-time-to-finish-the-race/ | ||
| // Author: github.com/lzl124631x | ||
| // Time: O(N * numLaps) | ||
| // Space: O(numLaps) | ||
| class Solution { | ||
| public: | ||
| int minimumFinishTime(vector<vector<int>>& A, int change, int numLaps) { | ||
| int N = A.size(), len = 0; | ||
| vector<long> best(numLaps, LONG_MAX), dp(numLaps + 1, INT_MAX); | ||
| for (int i = 0; i < N; ++i) { | ||
| long f = A[i][0], r = A[i][1], sum = change, p = 1; // We assume we also need `change` time to use the first tire so that we don't need to care the first tire as a special case | ||
| for (int j = 0; j < numLaps; ++j) { | ||
| sum += f * p; | ||
| if (f * p >= f + change) break; // If using the same tire takes no less time than changing the tire, stop further using the current tire | ||
| best[j] = min(best[j], sum); | ||
| len = max(len, j + 1); | ||
| p *= r; | ||
| } | ||
| } | ||
| dp[0] = 0; // dp[i + 1] is the minimum time to finish `numLaps` laps | ||
| for (int i = 0; i < numLaps; ++i) { | ||
| for (int j = 0; j < len && i - j >= 0; ++j) { // try using the same tire in the last `j+1` laps | ||
| dp[i + 1] = min(dp[i + 1], dp[i - j] + best[j]); | ||
| } | ||
| } | ||
| return dp[numLaps] - change; // minus the `change` we added to the first tire | ||
| } | ||
| }; | ||
| ``` |