feat: add solutions to lc problem: No.3660 (#4663)

No.3660.Jump Game IX

Author: yanglbme

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

@@ -84,32 +84,124 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3600-3699/3660.Ju
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：动态规划
+
+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。
+
+具体步骤如下：
+
+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。
+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 $\textit{sufMin} = \infty$。
+3. 首先预处理 $\textit{preMax}$ 数组。
+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 $\textit{sufMin}$。
+5. 最后返回结果数组 $\textit{ans}$。
+
+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。
 
 <!-- tabs:start -->
 
 #### Python3
 
 ```python
-
+class Solution:
+    def maxValue(self, nums: List[int]) -> List[int]:
+        n = len(nums)
+        ans = [0] * n
+        pre_max = [nums[0]] * n
+        for i in range(1, n):
+            pre_max[i] = max(pre_max[i - 1], nums[i])
+        suf_min = inf
+        for i in range(n - 1, -1, -1):
+            ans[i] = ans[i + 1] if pre_max[i] > suf_min else pre_max[i]
+            suf_min = min(suf_min, nums[i])
+        return ans
 ```
 
 #### Java
 
 ```java
-
+class Solution {
+    public int[] maxValue(int[] nums) {
+        int n = nums.length;
+        int[] ans = new int[n];
+        int[] preMax = new int[n];
+        preMax[0] = nums[0];
+        for (int i = 1; i < n; ++i) {
+            preMax[i] = Math.max(preMax[i - 1], nums[i]);
+        }
+        int sufMin = 1 << 30;
+        for (int i = n - 1; i >= 0; --i) {
+            ans[i] = preMax[i] > sufMin ? ans[i + 1] : preMax[i];
+            sufMin = Math.min(sufMin, nums[i]);
+        }
+        return ans;
+    }
+}
 ```
 
 #### C++
 
 ```cpp
-
+class Solution {
+public:
+    vector<int> maxValue(vector<int>& nums) {
+        int n = nums.size();
+        vector<int> ans(n);
+        vector<int> preMax(n, nums[0]);
+        for (int i = 1; i < n; ++i) {
+            preMax[i] = max(preMax[i - 1], nums[i]);
+        }
+        int sufMin = 1 << 30;
+        for (int i = n - 1; i >= 0; --i) {
+            ans[i] = preMax[i] > sufMin ? ans[i + 1] : preMax[i];
+            sufMin = min(sufMin, nums[i]);
+        }
+        return ans;
+    }
+};
 ```
 
 #### Go
 
 ```go
+func maxValue(nums []int) []int {
+	n := len(nums)
+	ans := make([]int, n)
+	preMax := make([]int, n)
+	preMax[0] = nums[0]
+	for i := 1; i < n; i++ {
+		preMax[i] = max(preMax[i-1], nums[i])
+	}
+	sufMin := 1 << 30
+	for i := n - 1; i >= 0; i-- {
+		if preMax[i] > sufMin {
+			ans[i] = ans[i+1]
+		} else {
+			ans[i] = preMax[i]
+		}
+		sufMin = min(sufMin, nums[i])
+	}
+	return ans
+}
+```
 
+#### TypeScript
+
+```ts
+function maxValue(nums: number[]): number[] {
+    const n = nums.length;
+    const ans = Array(n).fill(0);
+    const preMax = Array(n).fill(nums[0]);
+    for (let i = 1; i < n; i++) {
+        preMax[i] = Math.max(preMax[i - 1], nums[i]);
+    }
+    let sufMin = 1 << 30;
+    for (let i = n - 1; i >= 0; i--) {
+        ans[i] = preMax[i] > sufMin ? ans[i + 1] : preMax[i];
+        sufMin = Math.min(sufMin, nums[i]);
+    }
+    return ans;
+}
 ```
 
 <!-- tabs:end -->

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

@@ -82,32 +82,124 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3600-3699/3660.Ju
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Dynamic Programming
+
+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.
+
+The specific steps are as follows:
+
+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.
+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 $\textit{sufMin} = \infty$.
+3. First preprocess the $\textit{preMax}$ array.
+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 $\textit{sufMin}$.
+5. Finally return the result array $\textit{ans}$.
+
+Time complexity $O(n)$, space complexity $O(n)$. Where $n$ is the length of the array $\textit{nums}$.
 
 <!-- tabs:start -->
 
 #### Python3
 
 ```python
-
+class Solution:
+    def maxValue(self, nums: List[int]) -> List[int]:
+        n = len(nums)
+        ans = [0] * n
+        pre_max = [nums[0]] * n
+        for i in range(1, n):
+            pre_max[i] = max(pre_max[i - 1], nums[i])
+        suf_min = inf
+        for i in range(n - 1, -1, -1):
+            ans[i] = ans[i + 1] if pre_max[i] > suf_min else pre_max[i]
+            suf_min = min(suf_min, nums[i])
+        return ans
 ```
 
 #### Java
 
 ```java
-
+class Solution {
+    public int[] maxValue(int[] nums) {
+        int n = nums.length;
+        int[] ans = new int[n];
+        int[] preMax = new int[n];
+        preMax[0] = nums[0];
+        for (int i = 1; i < n; ++i) {
+            preMax[i] = Math.max(preMax[i - 1], nums[i]);
+        }
+        int sufMin = 1 << 30;
+        for (int i = n - 1; i >= 0; --i) {
+            ans[i] = preMax[i] > sufMin ? ans[i + 1] : preMax[i];
+            sufMin = Math.min(sufMin, nums[i]);
+        }
+        return ans;
+    }
+}
 ```
 
 #### C++
 
 ```cpp
-
+class Solution {
+public:
+    vector<int> maxValue(vector<int>& nums) {
+        int n = nums.size();
+        vector<int> ans(n);
+        vector<int> preMax(n, nums[0]);
+        for (int i = 1; i < n; ++i) {
+            preMax[i] = max(preMax[i - 1], nums[i]);
+        }
+        int sufMin = 1 << 30;
+        for (int i = n - 1; i >= 0; --i) {
+            ans[i] = preMax[i] > sufMin ? ans[i + 1] : preMax[i];
+            sufMin = min(sufMin, nums[i]);
+        }
+        return ans;
+    }
+};
 ```
 
 #### Go
 
 ```go
+func maxValue(nums []int) []int {
+	n := len(nums)
+	ans := make([]int, n)
+	preMax := make([]int, n)
+	preMax[0] = nums[0]
+	for i := 1; i < n; i++ {
+		preMax[i] = max(preMax[i-1], nums[i])
+	}
+	sufMin := 1 << 30
+	for i := n - 1; i >= 0; i-- {
+		if preMax[i] > sufMin {
+			ans[i] = ans[i+1]
+		} else {
+			ans[i] = preMax[i]
+		}
+		sufMin = min(sufMin, nums[i])
+	}
+	return ans
+}
+```
 
+#### TypeScript
+
+```ts
+function maxValue(nums: number[]): number[] {
+    const n = nums.length;
+    const ans = Array(n).fill(0);
+    const preMax = Array(n).fill(nums[0]);
+    for (let i = 1; i < n; i++) {
+        preMax[i] = Math.max(preMax[i - 1], nums[i]);
+    }
+    let sufMin = 1 << 30;
+    for (let i = n - 1; i >= 0; i--) {
+        ans[i] = preMax[i] > sufMin ? ans[i + 1] : preMax[i];
+        sufMin = Math.min(sufMin, nums[i]);
+    }
+    return ans;
+}
 ```
 
 <!-- tabs:end -->

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

@@ -0,0 +1,17 @@
+class Solution {
+public:
+    vector<int> maxValue(vector<int>& nums) {
+        int n = nums.size();
+        vector<int> ans(n);
+        vector<int> preMax(n, nums[0]);
+        for (int i = 1; i < n; ++i) {
+            preMax[i] = max(preMax[i - 1], nums[i]);
+        }
+        int sufMin = 1 << 30;
+        for (int i = n - 1; i >= 0; --i) {
+            ans[i] = preMax[i] > sufMin ? ans[i + 1] : preMax[i];
+            sufMin = min(sufMin, nums[i]);
+        }
+        return ans;
+    }
+};

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

@@ -0,0 +1,19 @@
+func maxValue(nums []int) []int {
+	n := len(nums)
+	ans := make([]int, n)
+	preMax := make([]int, n)
+	preMax[0] = nums[0]
+	for i := 1; i < n; i++ {
+		preMax[i] = max(preMax[i-1], nums[i])
+	}
+	sufMin := 1 << 30
+	for i := n - 1; i >= 0; i-- {
+		if preMax[i] > sufMin {
+			ans[i] = ans[i+1]
+		} else {
+			ans[i] = preMax[i]
+		}
+		sufMin = min(sufMin, nums[i])
+	}
+	return ans
+}

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

@@ -0,0 +1,17 @@
+class Solution {
+    public int[] maxValue(int[] nums) {
+        int n = nums.length;
+        int[] ans = new int[n];
+        int[] preMax = new int[n];
+        preMax[0] = nums[0];
+        for (int i = 1; i < n; ++i) {
+            preMax[i] = Math.max(preMax[i - 1], nums[i]);
+        }
+        int sufMin = 1 << 30;
+        for (int i = n - 1; i >= 0; --i) {
+            ans[i] = preMax[i] > sufMin ? ans[i + 1] : preMax[i];
+            sufMin = Math.min(sufMin, nums[i]);
+        }
+        return ans;
+    }
+}

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

@@ -0,0 +1,12 @@
+class Solution:
+    def maxValue(self, nums: List[int]) -> List[int]:
+        n = len(nums)
+        ans = [0] * n
+        pre_max = [nums[0]] * n
+        for i in range(1, n):
+            pre_max[i] = max(pre_max[i - 1], nums[i])
+        suf_min = inf
+        for i in range(n - 1, -1, -1):
+            ans[i] = ans[i + 1] if pre_max[i] > suf_min else pre_max[i]
+            suf_min = min(suf_min, nums[i])
+        return ans

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

@@ -0,0 +1,14 @@
+function maxValue(nums: number[]): number[] {
+    const n = nums.length;
+    const ans = Array(n).fill(0);
+    const preMax = Array(n).fill(nums[0]);
+    for (let i = 1; i < n; i++) {
+        preMax[i] = Math.max(preMax[i - 1], nums[i]);
+    }
+    let sufMin = 1 << 30;
+    for (let i = n - 1; i >= 0; i--) {
+        ans[i] = preMax[i] > sufMin ? ans[i + 1] : preMax[i];
+        sufMin = Math.min(sufMin, nums[i]);
+    }
+    return ans;
+}