Constrained Subsequence Sum
<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>
Queue; Array; Dynamic Programming; Sliding Window; Monotonic Queue; Heap (Priority Queue)
class Solution {
    public int constrainedSubsetSum(int[] nums, int k) {
        Deque<Integer> q = new ArrayDeque<>();
        q.offer(0);
        int n = nums.length;
        int[] f = new int[n];
        int ans = -(1 << 30);
        for (int i = 0; i < n; ++i) {
            while (i - q.peekFirst() > k) {
                q.pollFirst();
            }
            f[i] = Math.max(0, f[q.peekFirst()]) + nums[i];
            ans = Math.max(ans, f[i]);
            while (!q.isEmpty() && f[q.peekLast()] <= f[i]) {
                q.pollLast();
            }
            q.offerLast(i);
        }
        return ans;
    }
}
