<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:
        i_max = 0
        d = [[nums.pop()]]
        for n in reversed(nums):
            for i in range(i_max, -1, -1):
                a = d[i]
                j = bisect.bisect_right(a, n)
                if j == len(a):
                    j -= 1
                if n < a[j] <= n + b:
                    i += 1
                    if i == len(d):
                        d.append([])
                    a = d[i]
                    j = bisect.bisect_left(a, n)
                    if j == len(a):
                        a.append(n)
                    elif a[j] != n:
                        d[i] = a[:j] + [n] + a[j:]
                    i_max = max(i_max, i)
                    break
            else:
                a = d[0]
                j = bisect.bisect_left(a, n)
                if j == len(a):
                    a.append(n)
                elif a[j] != n:
                    d[0] = a[:j] + [n] + a[j:]
        return i_max + 1

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