# 1696. Jump Game VI

#### Medium

***

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:**

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

```python
class Solution:
    def maxResult(self, nums: List[int], k: int) -> int:
        #.
        dq = deque([0])
        for i in range(1, len(nums)):
            # Remove old indexes
            while dq and dq[0] < (i-k):
                dq.popleft()
            # Notice that Line 10 causes dq to be in decreasing order,
            # So [0] element in dq will be max
            nums[i] += nums[dq[0]]
            # Make dq in decreasing order
            while dq and nums[dq[-1]] <= nums[i]:
                dq.pop()
            dq.append(i)
        return nums[-1]
```