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feat: add solutions to lc problem: No.3660 (#4663)
No.3660.Jump Game IX
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‎solution/3600-3699/3660.Jump Game IX/README.md

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@@ -84,32 +84,124 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3600-3699/3660.Ju
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<!-- solution:start -->
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### 方法一
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### 方法一：动态规划
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如果 $i = n - 1$，那么它可以跳到 $\textit{nums}$ 中的最大值，因此 $\textit{ans}[i] = \max(\textit{nums})$。对于其他位置 $i$，我们可以通过维护一个前缀最大值数组和一个后缀最小值变量来计算。
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具体步骤如下：
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1. 创建一个数组 $\textit{preMax}$，其中 $\textit{preMax}[i]$ 表示从左到右遍历时 $[0, i]$ 区间内的最大值。
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2. 创建一个变量 $\textit{sufMin}$，表示从右到左遍历时，当前元素右侧的最小值。初始时 $\textit{sufMin} = \infty$。
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3. 首先预处理 $\textit{preMax}$ 数组。
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4. 接下来，从右到左遍历数组，对于每个位置 $i$，如果 $\textit{preMax}[i] > \textit{sufMin}$，说明可以从 $i$ 跳到 $\textit{preMax}$ 所在的位置，再跳到 $\textit{sufMin}$ 所在的位置，最后跳到 $i + 1$。因此在 $i + 1$ 能跳到的数，在 $i$ 也能跳到，因此 $\textit{ans}[i] = \textit{ans}[i + 1]$；否则更新为 $\textit{preMax}[i]$。然后更新 $\textit{sufMin}$。
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5. 最后返回结果数组 $\textit{ans}$。
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时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是数组 $\textit{nums}$ 的长度。
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<!-- tabs:start -->
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#### Python3
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```python
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class Solution:
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def maxValue(self, nums: List[int]) -> List[int]:
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n = len(nums)
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ans = [0] * n
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pre_max = [nums[0]] * n
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for i in range(1, n):
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pre_max[i] = max(pre_max[i - 1], nums[i])
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suf_min = inf
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for i in range(n - 1, -1, -1):
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ans[i] = ans[i + 1] if pre_max[i] > suf_min else pre_max[i]
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suf_min = min(suf_min, nums[i])
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return ans
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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[] maxValue(int[] nums) {
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int n = nums.length;
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int[] ans = new int[n];
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int[] preMax = new int[n];
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preMax[0] = nums[0];
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for (int i = 1; i < n; ++i) {
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preMax[i] = Math.max(preMax[i - 1], nums[i]);
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}
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int sufMin = 1 << 30;
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for (int i = n - 1; i >= 0; --i) {
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ans[i] = preMax[i] > sufMin ? ans[i + 1] : preMax[i];
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sufMin = Math.min(sufMin, nums[i]);
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}
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return ans;
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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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vector<int> maxValue(vector<int>& nums) {
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int n = nums.size();
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vector<int> ans(n);
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vector<int> preMax(n, nums[0]);
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for (int i = 1; i < n; ++i) {
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preMax[i] = max(preMax[i - 1], nums[i]);
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}
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int sufMin = 1 << 30;
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for (int i = n - 1; i >= 0; --i) {
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ans[i] = preMax[i] > sufMin ? ans[i + 1] : preMax[i];
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sufMin = min(sufMin, nums[i]);
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}
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return ans;
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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 maxValue(nums []int) []int {
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n := len(nums)
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ans := make([]int, n)
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preMax := make([]int, n)
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preMax[0] = nums[0]
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for i := 1; i < n; i++ {
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preMax[i] = max(preMax[i-1], nums[i])
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}
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sufMin := 1 << 30
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for i := n - 1; i >= 0; i-- {
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if preMax[i] > sufMin {
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ans[i] = ans[i+1]
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} else {
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ans[i] = preMax[i]
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}
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sufMin = min(sufMin, nums[i])
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}
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return ans
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}
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```
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#### TypeScript
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```ts
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function maxValue(nums: number[]): number[] {
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const n = nums.length;
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const ans = Array(n).fill(0);
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const preMax = Array(n).fill(nums[0]);
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for (let i = 1; i < n; i++) {
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preMax[i] = Math.max(preMax[i - 1], nums[i]);
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}
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let sufMin = 1 << 30;
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for (let i = n - 1; i >= 0; i--) {
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ans[i] = preMax[i] > sufMin ? ans[i + 1] : preMax[i];
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sufMin = Math.min(sufMin, nums[i]);
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}
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return ans;
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}
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```
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<!-- tabs:end -->

‎solution/3600-3699/3660.Jump Game IX/README_EN.md

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Dynamic Programming
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If $i = n - 1$, then it can jump to the maximum value in $\textit{nums}$, so $\textit{ans}[i] = \max(\textit{nums})$. For other positions $i$, we can calculate by maintaining a prefix maximum array and a suffix minimum variable.
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The specific steps are as follows:
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1. Create an array $\textit{preMax}$, where $\textit{preMax}[i]$ represents the maximum value in the interval $[0, i]$ when traversing from left to right.
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2. Create a variable $\textit{sufMin}$, which represents the minimum value to the right of the current element when traversing from right to left. Initially $\textit{sufMin} = \infty$.
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3. First preprocess the $\textit{preMax}$ array.
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4. Next, traverse the array from right to left. For each position $i$, if $\textit{preMax}[i] > \textit{sufMin}$, it means we can jump from $i$ to the position where $\textit{preMax}$ is located, then jump to the position where $\textit{sufMin}$ is located, and finally jump to $i + 1$. Therefore, the numbers that can be reached from $i + 1$ can also be reached from $i$, so $\textit{ans}[i] = \textit{ans}[i + 1]$; otherwise update to $\textit{preMax}[i]$. Then update $\textit{sufMin}$.
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5. Finally return the result array $\textit{ans}$.
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Time complexity $O(n)$, space complexity $O(n)$. Where $n$ is the length of the array $\textit{nums}$.
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<!-- tabs:start -->
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#### Python3
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```python
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class Solution:
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def maxValue(self, nums: List[int]) -> List[int]:
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n = len(nums)
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ans = [0] * n
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pre_max = [nums[0]] * n
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for i in range(1, n):
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pre_max[i] = max(pre_max[i - 1], nums[i])
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suf_min = inf
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for i in range(n - 1, -1, -1):
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ans[i] = ans[i + 1] if pre_max[i] > suf_min else pre_max[i]
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suf_min = min(suf_min, nums[i])
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return ans
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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[] maxValue(int[] nums) {
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int n = nums.length;
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int[] ans = new int[n];
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int[] preMax = new int[n];
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preMax[0] = nums[0];
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for (int i = 1; i < n; ++i) {
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preMax[i] = Math.max(preMax[i - 1], nums[i]);
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}
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int sufMin = 1 << 30;
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for (int i = n - 1; i >= 0; --i) {
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ans[i] = preMax[i] > sufMin ? ans[i + 1] : preMax[i];
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sufMin = Math.min(sufMin, nums[i]);
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}
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return ans;
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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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vector<int> maxValue(vector<int>& nums) {
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int n = nums.size();
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vector<int> ans(n);
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vector<int> preMax(n, nums[0]);
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for (int i = 1; i < n; ++i) {
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preMax[i] = max(preMax[i - 1], nums[i]);
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}
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int sufMin = 1 << 30;
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for (int i = n - 1; i >= 0; --i) {
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ans[i] = preMax[i] > sufMin ? ans[i + 1] : preMax[i];
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sufMin = min(sufMin, nums[i]);
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}
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return ans;
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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 maxValue(nums []int) []int {
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n := len(nums)
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ans := make([]int, n)
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preMax := make([]int, n)
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preMax[0] = nums[0]
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for i := 1; i < n; i++ {
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preMax[i] = max(preMax[i-1], nums[i])
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}
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sufMin := 1 << 30
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for i := n - 1; i >= 0; i-- {
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if preMax[i] > sufMin {
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ans[i] = ans[i+1]
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} else {
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ans[i] = preMax[i]
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}
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sufMin = min(sufMin, nums[i])
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}
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return ans
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}
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```
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#### TypeScript
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```ts
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function maxValue(nums: number[]): number[] {
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const n = nums.length;
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const ans = Array(n).fill(0);
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const preMax = Array(n).fill(nums[0]);
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for (let i = 1; i < n; i++) {
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preMax[i] = Math.max(preMax[i - 1], nums[i]);
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}
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let sufMin = 1 << 30;
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for (let i = n - 1; i >= 0; i--) {
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ans[i] = preMax[i] > sufMin ? ans[i + 1] : preMax[i];
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sufMin = Math.min(sufMin, nums[i]);
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}
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return ans;
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}
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```
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<!-- tabs:end -->

‎solution/3600-3699/3660.Jump Game IX/Solution.cpp

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class Solution {
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public:
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vector<int> maxValue(vector<int>& nums) {
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int n = nums.size();
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vector<int> ans(n);
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vector<int> preMax(n, nums[0]);
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for (int i = 1; i < n; ++i) {
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preMax[i] = max(preMax[i - 1], nums[i]);
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}
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int sufMin = 1 << 30;
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for (int i = n - 1; i >= 0; --i) {
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ans[i] = preMax[i] > sufMin ? ans[i + 1] : preMax[i];
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sufMin = min(sufMin, nums[i]);
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}
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return ans;
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}
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};

‎solution/3600-3699/3660.Jump Game IX/Solution.go

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func maxValue(nums []int) []int {
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n := len(nums)
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ans := make([]int, n)
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preMax := make([]int, n)
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preMax[0] = nums[0]
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for i := 1; i < n; i++ {
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preMax[i] = max(preMax[i-1], nums[i])
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}
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sufMin := 1 << 30
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for i := n - 1; i >= 0; i-- {
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if preMax[i] > sufMin {
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ans[i] = ans[i+1]
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} else {
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ans[i] = preMax[i]
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}
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sufMin = min(sufMin, nums[i])
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}
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return ans
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}

‎solution/3600-3699/3660.Jump Game IX/Solution.java

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class Solution {
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public int[] maxValue(int[] nums) {
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int n = nums.length;
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int[] ans = new int[n];
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int[] preMax = new int[n];
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preMax[0] = nums[0];
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for (int i = 1; i < n; ++i) {
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preMax[i] = Math.max(preMax[i - 1], nums[i]);
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}
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int sufMin = 1 << 30;
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for (int i = n - 1; i >= 0; --i) {
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ans[i] = preMax[i] > sufMin ? ans[i + 1] : preMax[i];
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sufMin = Math.min(sufMin, nums[i]);
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}
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return ans;
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}
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}

‎solution/3600-3699/3660.Jump Game IX/Solution.py

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class Solution:
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def maxValue(self, nums: List[int]) -> List[int]:
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n = len(nums)
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ans = [0] * n
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pre_max = [nums[0]] * n
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for i in range(1, n):
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pre_max[i] = max(pre_max[i - 1], nums[i])
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suf_min = inf
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for i in range(n - 1, -1, -1):
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ans[i] = ans[i + 1] if pre_max[i] > suf_min else pre_max[i]
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suf_min = min(suf_min, nums[i])
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return ans

‎solution/3600-3699/3660.Jump Game IX/Solution.ts

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function maxValue(nums: number[]): number[] {
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const n = nums.length;
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const ans = Array(n).fill(0);
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const preMax = Array(n).fill(nums[0]);
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for (let i = 1; i < n; i++) {
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preMax[i] = Math.max(preMax[i - 1], nums[i]);
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}
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let sufMin = 1 << 30;
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for (let i = n - 1; i >= 0; i--) {
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ans[i] = preMax[i] > sufMin ? ans[i + 1] : preMax[i];
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sufMin = Math.min(sufMin, nums[i]);
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}
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return ans;
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}

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