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feat: add solutions to lc problem: No.2188 (#1630)
No.2188.Minimum Time to Finish the Race
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‎README.md

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</p>
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<p align="center">
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<a href="https://github.com/doocs/leetcode"><img src="https://img.shields.io/badge/langs-Java%20%7C%20Python%20%7C%20C%2B%2B%20%7C%20Go%20%7C%20TypeScript%20%7C%20...-red?style=flat-square&color=42b883" alt="languages"></a>
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<a href="https://github.com/doocs/leetcode/blob/main/LICENSE"><img src="https://img.shields.io/github/license/doocs/leetcode?color=42b883&style=flat-square" alt="LICENSE"></a><br>
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‎README_EN.md

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</p>
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<p align="center">
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<a href="https://github.com/doocs/leetcode"><img src="https://img.shields.io/badge/langs-Java%20%7C%20Python%20%7C%20C%2B%2B%20%7C%20Go%20%7C%20TypeScript%20%7C%20...-red?style=flat-square&color=42b883" alt="languages"></a>
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<a href="https://github.com/doocs/leetcode"><img src="https://img.shields.io/badge/langs-Python%20%7C%20Java%20%7C%20C++%20%7C%20Go%20%7C%20TypeScript%20%7C%20Rust%20%7C%20...-red?style=flat-square&color=42b883" alt="languages"></a>
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<a href="https://github.com/doocs/leetcode/blob/main/LICENSE"><img src="https://img.shields.io/github/license/doocs/leetcode?color=42b883&style=flat-square" alt="LICENSE"></a><br>
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‎solution/2100-2199/2188.Minimum Time to Finish the Race/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：预处理 + 动态规划**
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我们注意到，连续使用同一个轮胎 $(f, r)$ 跑 $i$ 圈，那么第 $i$ 圈的耗时不应该超过 $changeTime + f$，否则我们可以在第 $i$ 圈的时候换轮胎，这样总耗时会更少。即：
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$$
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f \times r^{i-1} \leq changeTime + f
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$$
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我们可以求出满足上式的最大的 $i$，要使得 $i$ 最大，那么 $f$ 和 $r$ 应该尽可能小，根据题目的数据范围，我们取 $f=1$, $r=2$，那么 $2^{i-1} \leq changeTime + 1$，即 $i \leq \log_2(changeTime + 1) + 1$。根据这个结论，以及题目中 $changeTime$ 的数据范围，我们可以知道 $i$ 最大为 $17$。
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我们定义 $cost[i]$ 表示使用同一个轮胎跑 $i$ 圈的最小耗时，那么我们可以预处理出 $cost$ 数组，然后使用动态规划求解即可。定义 $f[i]$ 表示跑 $i$ 圈的最小耗时，那么我们可以得到状态转移方程：
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$$
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f[i] = (\min_{1 \leq j \leq \min(17, i)} f[i-j] + cost[j]) + changeTime
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$$
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初始时 $f[0] = -changeTime$，最终答案为 $f[numLaps]$。
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时间复杂度 $O((n + numLaps) \times \log T_{max})$，空间复杂度 $O(n + \log T_{max})$，其中 $T_{max}$ 是题目中 $f_i$, $r_i$ 和 $changeTime$ 的最大值。本题中 $\log T_{max} \approx 17$。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def minimumFinishTime(
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self, tires: List[List[int]], changeTime: int, numLaps: int
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) -> int:
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cost = [inf] * 18
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for f, r in tires:
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i, s, t = 1, 0, f
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while t <= changeTime + f:
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s += t
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cost[i] = min(cost[i], s)
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t *= r
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i += 1
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f = [inf] * (numLaps + 1)
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f[0] = -changeTime
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for i in range(1, numLaps + 1):
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for j in range(1, min(18, i + 1)):
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f[i] = min(f[i], f[i - j] + cost[j])
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f[i] += changeTime
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return f[numLaps]
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int minimumFinishTime(int[][] tires, int changeTime, int numLaps) {
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final int inf = 1 << 30;
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int[] cost = new int[18];
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Arrays.fill(cost, inf);
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for (int[] e : tires) {
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int f = e[0], r = e[1];
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int s = 0, t = f;
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for (int i = 1; t <= changeTime + f; ++i) {
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s += t;
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cost[i] = Math.min(cost[i], s);
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t *= r;
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}
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}
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int[] f = new int[numLaps + 1];
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Arrays.fill(f, inf);
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f[0] = -changeTime;
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for (int i = 1; i <= numLaps; ++i) {
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for (int j = 1; j < Math.min(18, i + 1); ++j) {
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f[i] = Math.min(f[i], f[i - j] + cost[j]);
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}
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f[i] += changeTime;
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}
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return f[numLaps];
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int minimumFinishTime(vector<vector<int>>& tires, int changeTime, int numLaps) {
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int cost[18];
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memset(cost, 0x3f, sizeof(cost));
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for (auto& e : tires) {
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int f = e[0], r = e[1];
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int s = 0;
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long long t = f;
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for (int i = 1; t <= changeTime + f; ++i) {
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s += t;
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cost[i] = min(cost[i], s);
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t *= r;
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}
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}
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int f[numLaps + 1];
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memset(f, 0x3f, sizeof(f));
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f[0] = -changeTime;
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for (int i = 1; i <= numLaps; ++i) {
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for (int j = 1; j < min(18, i + 1); ++j) {
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f[i] = min(f[i], f[i - j] + cost[j]);
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}
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f[i] += changeTime;
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}
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return f[numLaps];
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}
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};
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```
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### **Go**
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```go
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func minimumFinishTime(tires [][]int, changeTime int, numLaps int) int {
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const inf = 1 << 30
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cost := [18]int{}
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for i := range cost {
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cost[i] = inf
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}
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for _, e := range tires {
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f, r := e[0], e[1]
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s, t := 0, f
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for i := 1; t <= changeTime+f; i++ {
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s += t
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cost[i] = min(cost[i], s)
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t *= r
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}
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}
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f := make([]int, numLaps+1)
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for i := range f {
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f[i] = inf
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}
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f[0] = -changeTime
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for i := 1; i <= numLaps; i++ {
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for j := 1; j < min(18, i+1); j++ {
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f[i] = min(f[i], f[i-j]+cost[j])
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}
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f[i] += changeTime
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}
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return f[numLaps]
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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```
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### **TypeScript**
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```ts
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function minimumFinishTime(tires: number[][], changeTime: number, numLaps: number): number {
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const cost: number[] = Array(18).fill(Infinity);
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for (const [f, r] of tires) {
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let s = 0;
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let t = f;
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for (let i = 1; t <= changeTime + f; ++i) {
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s += t;
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cost[i] = Math.min(cost[i], s);
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t *= r;
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}
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}
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const f: number[] = Array(numLaps + 1).fill(Infinity);
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f[0] = -changeTime;
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for (let i = 1; i <= numLaps; ++i) {
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for (let j = 1; j < Math.min(18, i + 1); ++j) {
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f[i] = Math.min(f[i], f[i - j] + cost[j]);
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}
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f[i] += changeTime;
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}
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return f[numLaps];
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}
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```
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### **...**

‎solution/2100-2199/2188.Minimum Time to Finish the Race/README_EN.md

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### **Python3**
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```python
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class Solution:
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def minimumFinishTime(
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self, tires: List[List[int]], changeTime: int, numLaps: int
70+
) -> int:
71+
cost = [inf] * 18
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for f, r in tires:
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i, s, t = 1, 0, f
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while t <= changeTime + f:
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s += t
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cost[i] = min(cost[i], s)
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t *= r
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i += 1
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f = [inf] * (numLaps + 1)
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f[0] = -changeTime
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for i in range(1, numLaps + 1):
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for j in range(1, min(18, i + 1)):
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f[i] = min(f[i], f[i - j] + cost[j])
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f[i] += changeTime
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return f[numLaps]
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```
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### **Java**
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```java
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class Solution {
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public int minimumFinishTime(int[][] tires, int changeTime, int numLaps) {
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final int inf = 1 << 30;
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int[] cost = new int[18];
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Arrays.fill(cost, inf);
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for (int[] e : tires) {
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int f = e[0], r = e[1];
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int s = 0, t = f;
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for (int i = 1; t <= changeTime + f; ++i) {
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s += t;
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cost[i] = Math.min(cost[i], s);
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t *= r;
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}
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}
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int[] f = new int[numLaps + 1];
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Arrays.fill(f, inf);
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f[0] = -changeTime;
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for (int i = 1; i <= numLaps; ++i) {
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for (int j = 1; j < Math.min(18, i + 1); ++j) {
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f[i] = Math.min(f[i], f[i - j] + cost[j]);
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}
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f[i] += changeTime;
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}
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return f[numLaps];
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int minimumFinishTime(vector<vector<int>>& tires, int changeTime, int numLaps) {
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int cost[18];
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memset(cost, 0x3f, sizeof(cost));
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for (auto& e : tires) {
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int f = e[0], r = e[1];
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int s = 0;
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long long t = f;
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for (int i = 1; t <= changeTime + f; ++i) {
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s += t;
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cost[i] = min(cost[i], s);
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t *= r;
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}
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}
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int f[numLaps + 1];
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memset(f, 0x3f, sizeof(f));
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f[0] = -changeTime;
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for (int i = 1; i <= numLaps; ++i) {
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for (int j = 1; j < min(18, i + 1); ++j) {
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f[i] = min(f[i], f[i - j] + cost[j]);
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}
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f[i] += changeTime;
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}
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return f[numLaps];
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}
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};
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```
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### **Go**
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```go
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func minimumFinishTime(tires [][]int, changeTime int, numLaps int) int {
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const inf = 1 << 30
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cost := [18]int{}
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for i := range cost {
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cost[i] = inf
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}
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for _, e := range tires {
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f, r := e[0], e[1]
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s, t := 0, f
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for i := 1; t <= changeTime+f; i++ {
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s += t
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cost[i] = min(cost[i], s)
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t *= r
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}
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}
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f := make([]int, numLaps+1)
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for i := range f {
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f[i] = inf
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}
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f[0] = -changeTime
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for i := 1; i <= numLaps; i++ {
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for j := 1; j < min(18, i+1); j++ {
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f[i] = min(f[i], f[i-j]+cost[j])
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}
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f[i] += changeTime
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}
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return f[numLaps]
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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```
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### **TypeScript**
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```ts
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function minimumFinishTime(tires: number[][], changeTime: number, numLaps: number): number {
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const cost: number[] = Array(18).fill(Infinity);
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for (const [f, r] of tires) {
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let s = 0;
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let t = f;
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for (let i = 1; t <= changeTime + f; ++i) {
200+
s += t;
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cost[i] = Math.min(cost[i], s);
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t *= r;
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}
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}
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const f: number[] = Array(numLaps + 1).fill(Infinity);
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f[0] = -changeTime;
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for (let i = 1; i <= numLaps; ++i) {
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for (let j = 1; j < Math.min(18, i + 1); ++j) {
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f[i] = Math.min(f[i], f[i - j] + cost[j]);
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}
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f[i] += changeTime;
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}
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return f[numLaps];
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}
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```
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### **...**

‎solution/2100-2199/2188.Minimum Time to Finish the Race/Solution.cpp

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class Solution {
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public:
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int minimumFinishTime(vector<vector<int>>& tires, int changeTime, int numLaps) {
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int cost[18];
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memset(cost, 0x3f, sizeof(cost));
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for (auto& e : tires) {
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int f = e[0], r = e[1];
8+
int s = 0;
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long long t = f;
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for (int i = 1; t <= changeTime + f; ++i) {
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s += t;
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cost[i] = min(cost[i], s);
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t *= r;
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}
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}
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int f[numLaps + 1];
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memset(f, 0x3f, sizeof(f));
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f[0] = -changeTime;
19+
for (int i = 1; i <= numLaps; ++i) {
20+
for (int j = 1; j < min(18, i + 1); ++j) {
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f[i] = min(f[i], f[i - j] + cost[j]);
22+
}
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f[i] += changeTime;
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}
25+
return f[numLaps];
26+
}
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};

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