---
description: https://leetcode.com/problems/jump-game-vi/
---

# 1696. Jump Game VI

You are given a **0-indexed** integer array `nums` and an integer `k`.

You are initially standing at index `0`. In one move, you can jump at most `k` steps forward without going outside the boundaries of the array. That is, you can jump from index `i` to any index in the range `[i + 1, min(n - 1, i + k)]` **inclusive**.

You want to reach the last index of the array (index `n - 1`). Your **score** is the **sum** of all `nums[j]` for each index `j` you visited in the array.

Return _the **maximum score** you can get_.

**Example 1:**

```
Input: nums = [1,-1,-2,4,-7,3], k = 2
Output: 7
Explanation: You can choose your jumps forming the subsequence [1,-1,4,3] (underlined above). The sum is 7.
```

**Example 2:**

```
Input: nums = [10,-5,-2,4,0,3], k = 3
Output: 17
Explanation: You can choose your jumps forming the subsequence [10,4,3] (underlined above). The sum is 17.
```

**Example 3:**

```
Input: nums = [1,-5,-20,4,-1,3,-6,-3], k = 2
Output: 0
```

**Constraints:**

* &#x20;`1 <= nums.length, k <= 105`
* `-104 <= nums[i] <= 104`

给定数组nums，从第一个元素开始，每次能往前最远跳k步且收益为所跳到格子的值，问跳到最后一个位置时最大的收益是多少。虽然有一点不同，但思路基本和1425. Constrained Subsequence Sum一致：subseq + argmax => dp\[i] == 此时最大序列和：argmax(dp\[i - k], dp\[i -k +1], ... dp\[i - 1]) + nums\[i]。找max(dp\[i - k: i - 1])的过程相当于维持一个大小为k的在dp数组上的滑动窗口，且每次找窗口内最大元素=>单调栈。和1425区别在于必须从起点跳到终点，因此结果不再是max(dp)而是dp\[-1]，且dp\[i - k: i - 1]必须要选一个，而不能因为是负数而不选。



```python
class Solution:
    def maxResult(self, nums: List[int], k: int) -> int:
        N = len(nums)
        dp, q = [0] * N, deque([nums[0]])
        dp[0] = nums[0]
        for i in range(1, N):
            f = i - k
            r = q[0] + nums[i]
            dp[i] = r
            if f >= 0 and dp[f] == q[0]:
                q.popleft()
            while q and r > q[-1]:
                q.pop()
            q.append(r)
        return dp[-1]
        
```
