<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>

from bisect import bisect_right

class Solution:
    def lengthOfLIS(self, nums: List[int], k: int) -> int:
        memo = [1] * len(nums)
        n2 = []

        index = {}
        for i in range(len(nums) - 1, -1, -1):
            n = nums[i]
            index[n] = i
            opt = 0
            idx = bisect_right(n2, n)
            if idx == len(n2):
                n2.append(n)
                idx += 1
            elif n2[idx] != n:
                n2.insert(idx, n)
                idx += 1
            while idx < len(n2):
                if n2[idx] - n <= k:
                    j = index.get(n2[idx])
                    if j is None:
                        idx += 1
                        continue
                    opt = max(memo[j], opt)
                    idx += 1
                    if j == i + 1:
                        break      
                else:
                    break
            memo[i] = opt + 1
        return max(memo)
35400