<p>You are given an integer array <code>nums</code> and an integer <code>k</code>.</p>

<p>Find the longest subsequence of <code>nums</code> that meets the following requirements:</p>

<ul>
	<li>The subsequence is <strong>strictly increasing</strong> and</li>
	<li>The difference between adjacent elements in the subsequence is <strong>at most</strong> <code>k</code>.</li>
</ul>

<p>Return<em> the length of the <strong>longest</strong> <strong>subsequence</strong> that meets the requirements.</em></p>

<p>A <strong>subsequence</strong> is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.</p>

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

<pre>
<strong>Input:</strong> nums = [4,2,1,4,3,4,5,8,15], k = 3
<strong>Output:</strong> 5
<strong>Explanation:</strong>
The longest subsequence that meets the requirements is [1,3,4,5,8].
The subsequence has a length of 5, so we return 5.
Note that the subsequence [1,3,4,5,8,15] does not meet the requirements because 15 - 8 = 7 is larger than 3.
</pre>

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

<pre>
<strong>Input:</strong> nums = [7,4,5,1,8,12,4,7], k = 5
<strong>Output:</strong> 4
<strong>Explanation:</strong>
The longest subsequence that meets the requirements is [4,5,8,12].
The subsequence has a length of 4, so we return 4.
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,5], k = 1
<strong>Output:</strong> 1
<strong>Explanation:</strong>
The longest subsequence that meets the requirements is [1].
The subsequence has a length of 1, so we return 1.
</pre>

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

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

class SegmentTree:
    def __init__(self, n):
        self.n = n
        self.tree = [0] * 4 * self.n

    def query(self, left, right, index, qleft, qright):
        if qright < left or qleft > right:
            return 0

        if qleft <= left and right <= qright:
            return self.tree[index]

        mid = (left + right) // 2
        resLeft = self.query(left, mid, 2*index+1, qleft, qright)
        resRight = self.query(mid+1, right, 2*index+2, qleft, qright)
        return max(resLeft, resRight)

    def update(self, left, right, index, pos, val):
        if left == right:
            self.tree[index] = val
            return

        mid = (left + right) // 2
        if pos <= mid:
            self.update(left, mid, 2*index+1, pos, val)
        else:
            self.update(mid+1, right, 2*index+2, pos, val)
        self.tree[index] = max(self.tree[2*index+1], self.tree[2*index+2])


class Solution:
    def lengthOfLIS(self, nums: List[int], k: int) -> int:
        mx = max(nums)
        segmentTree = SegmentTree(mx + 1)
        ans = 0
        for num in nums:
            subLongest = segmentTree.query(0, segmentTree.n-1, 0, num-k, num-1) + 1
            ans = max(ans, subLongest)
            segmentTree.update(0, segmentTree.n-1, 0, num, subLongest)
        return ans

# T: O(N log N)
# S: O(N)
29200