<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> and an integer <code>k</code>.</p>

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

<p>You want to reach the last index of the array (index <code>n - 1</code>). Your <strong>score</strong> is the <strong>sum</strong> of all <code>nums[j]</code> for each index <code>j</code> you visited in the array.</p>

<p>Return <em>the <strong>maximum score</strong> you can get</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

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

<p><strong class="example">Example 2:</strong></p>

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

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,-5,-20,4,-1,3,-6,-3], k = 2
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length, k &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution:
    def maxResult2(self, nums: List[int], k: int) -> int:
        
        n = len(nums)

        dp = [-math.inf] * n
        def helper(i):
            if i == n-1:
                return 0
            
            if dp[i] > -math.inf:
                return dp[i]

            r1 = -math.inf
            for j in range(i+1, min(i+k,n-1)+1):
                r1 = max(r1, helper(j) + nums[j])

            dp[i] = r1
            return r1


        return helper(0) + nums[0]
    

    def maxResult3(self, nums: List[int], k: int) -> int:

        n = len(nums)
        heap = [(-nums[n-1], n-1)]
        result = nums[n-1]
        for i in range(n-2,-1,-1):
            while heap[0][1] > i + k:
                heappop(heap)
            result = -heap[0][0] + nums[i]
            heappush(heap, (-result, i))

        return result

    def maxResult(self, nums: List[int], k: int) -> int:
        n = len(nums)
        dp = [0] * n
        dp[0] = nums[0]
        dq = deque([0])
        result = nums[0]
        for i in range(1,n):
            while len(dq) > 0 and dq[0] < i - k:
                dq.popleft()

            result = dp[dq[0]] + nums[i]
            dp[i] = result
            
            while len(dq) > 0 and dp[dq[-1]] < result:
                dq.pop()

            dq.append(i)

        return result
704