<p>You are given an integer array <code>nums</code> and an integer <code>k</code>. You may partition <code>nums</code> into one or more <strong>subsequences</strong> such that each element in <code>nums</code> appears in <strong>exactly</strong> one of the subsequences.</p>

<p>Return <em>the <strong>minimum </strong>number of subsequences needed such that the difference between the maximum and minimum values in each subsequence is <strong>at most</strong> </em><code>k</code><em>.</em></p>

<p>A <strong>subsequence</strong> is a sequence that can be derived from another sequence 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 = [3,6,1,2,5], k = 2
<strong>Output:</strong> 2
<strong>Explanation:</strong>
We can partition nums into the two subsequences [3,1,2] and [6,5].
The difference between the maximum and minimum value in the first subsequence is 3 - 1 = 2.
The difference between the maximum and minimum value in the second subsequence is 6 - 5 = 1.
Since two subsequences were created, we return 2. It can be shown that 2 is the minimum number of subsequences needed.
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,2,3], k = 1
<strong>Output:</strong> 2
<strong>Explanation:</strong>
We can partition nums into the two subsequences [1,2] and [3].
The difference between the maximum and minimum value in the first subsequence is 2 - 1 = 1.
The difference between the maximum and minimum value in the second subsequence is 3 - 3 = 0.
Since two subsequences were created, we return 2. Note that another optimal solution is to partition nums into the two subsequences [1] and [2,3].
</pre>

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

<pre>
<strong>Input:</strong> nums = [2,2,4,5], k = 0
<strong>Output:</strong> 3
<strong>Explanation:</strong>
We can partition nums into the three subsequences [2,2], [4], and [5].
The difference between the maximum and minimum value in the first subsequences is 2 - 2 = 0.
The difference between the maximum and minimum value in the second subsequences is 4 - 4 = 0.
The difference between the maximum and minimum value in the third subsequences is 5 - 5 = 0.
Since three subsequences were created, we return 3. It can be shown that 3 is the minimum number of subsequences needed.
</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>0 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= k &lt;= 10<sup>5</sup></code></li>
</ul>

# [2,2,4,5], k=0
# min = 2
# max = 4
# ans = 2
class Solution:
    def partitionArray(self, nums: List[int], k: int) -> int:
        nums.sort()
        curr_min = nums[0]
        ans = 1
        for i in range(1, len(nums)):
            if nums[i] - curr_min > k:
                ans += 1
                curr_min = nums[i]
        return ans

# Started considering min and max of each subset separately
# but becasue it's sorted the newest number will always be the max, 
# and the first number in the subset will be the min, so you can just
# update the max at each step instead of both the min and the max
718