<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 maxResult(self, nums: List[int], k: int) -> int:
        """
        - check if greedy
            k=2, [1,-1,-5,-6,-7,-8] <- greedy will not work
        - dynamic programming <- constraint
            dp[i] be "the maximum score to reach the end starting at index i". 
            The answer for dp[i] is nums[i] + max{dp[i+j]} for 1 <= j <= k.
            time-complexity: O(N*k)
        - base-case: dp[i] for n-k<i<n: sum of all positive integers nums[j] for i<=j<n
        - can use sliding max window or heap approach to find max{dp[i+j]} for 1<=j<=k in O(1)
        """
        n = len(nums)
        dp = [0]*n
        # base-case:
        dp[n-1] = nums[n-1]
        pq = []
        heappush(pq, (-dp[n-1], n-1))
        for i in range(n-2, -1, -1):
            val, j = heappop(pq)
            while j>i+k:
                val, j = heappop(pq)
            heappush(pq, (val, j))
            dp[i] = nums[i]-val
            heappush(pq, (-dp[i], i))
        return dp[0]
# TC: O(N*log(K))
# SC: O(N)
38800