<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 Solution:
    def lengthOfLIS(self, nums: List[int], b: int) -> int:
        k = 0
        n = len(nums)
        s = [-1]
        d = [0]*(n + 1)
        o = sorted(range(n), key=lambda x: (nums[x], -x))
        for i in o:
            m = nums[i] - b
            while nums[o[k]] < m:
                j = bisect_left(s, o[k])
                if j < len(s) and s[j] == o[k]:
                    s = s[:j] + s[j+1:]
                k += 1
            j = bisect_left(s, i)
            if j == len(s):
                s.append(i)
            else:
                s[j] = i
            d[i] = d[s[j-1]] + 1
        return max(d)
        
36900