<p>Given an integer array <code>nums</code> and an integer <code>k</code>, return the maximum sum of a <strong>non-empty</strong> subsequence of that array such that for every two <strong>consecutive</strong> integers in the subsequence, <code>nums[i]</code> and <code>nums[j]</code>, where <code>i &lt; j</code>, the condition <code>j - i &lt;= k</code> is satisfied.</p>

<p>A <em>subsequence</em> of an array is obtained by deleting some number of elements (can be zero) from the array, leaving the remaining elements in their original order.</p>

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

<pre>
<strong>Input:</strong> nums = [10,2,-10,5,20], k = 2
<strong>Output:</strong> 37
<b>Explanation:</b> The subsequence is [10, 2, 5, 20].
</pre>

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

<pre>
<strong>Input:</strong> nums = [-1,-2,-3], k = 1
<strong>Output:</strong> -1
<b>Explanation:</b> The subsequence must be non-empty, so we choose the largest number.
</pre>

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

<pre>
<strong>Input:</strong> nums = [10,-2,-10,-5,20], k = 2
<strong>Output:</strong> 23
<b>Explanation:</b> The subsequence is [10, -2, -5, 20].
</pre>

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

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

import heapq

class Solution:
    def constrainedSubsetSum(self, arr: List[int], k: int) -> int:
        # max heap (-value, index).
        windows_maxs = [[-arr[0], 0]]
        res = arr[0]
        n = len(arr)

        for i in range(1, n):
            # check if the index of top element in heap is out of window
            while i - k > windows_maxs[0][1]:
                heapq.heappop(windows_maxs)
            
            # dp[i] = arr[i] + max(0, dp[i-1], dp[i-2], .... , dp[i-k])
            curr = arr[i] + max(0, -windows_maxs[0][0])
            res = max(res, curr)
            
            # update the window
            heapq.heappush(windows_maxs, [-curr, i])

        return res
38200