<p>You are given a <strong>0-indexed</strong> array of integers <code>nums</code>, and an integer <code>target</code>.</p>

<p>Return <em>the <strong>length of the longest subsequence</strong> of</em> <code>nums</code> <em>that sums up to</em> <code>target</code>. <em>If no such subsequence exists, return</em> <code>-1</code>.</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 = [1,2,3,4,5], target = 9
<strong>Output:</strong> 3
<strong>Explanation:</strong> There are 3 subsequences with a sum equal to 9: [4,5], [1,3,5], and [2,3,4]. The longest subsequences are [1,3,5], and [2,3,4]. Hence, the answer is 3.
</pre>

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

<pre>
<strong>Input:</strong> nums = [4,1,3,2,1,5], target = 7
<strong>Output:</strong> 4
<strong>Explanation:</strong> There are 5 subsequences with a sum equal to 7: [4,3], [4,1,2], [4,2,1], [1,1,5], and [1,3,2,1]. The longest subsequence is [1,3,2,1]. Hence, the answer is 4.
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,1,5,4,5], target = 3
<strong>Output:</strong> -1
<strong>Explanation:</strong> It can be shown that nums has no subsequence that sums up to 3.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 1000</code></li>
	<li><code>1 &lt;= target &lt;= 1000</code></li>
</ul>

class Solution:
    def lengthOfLongestSubsequence(self, nums: List[int], target: int) -> int:
        nums.sort()
        n = len(nums)
        
        memo = {}
        def dp(index: int, total: int) -> int:
            if (index, total) in memo:
                return memo[(index, total)]
            if total == 0:
                return 0
            if index == n or total < nums[index]:
                return -float("inf")
            res = max(1 + dp(index + 1, total - nums[index]), dp(index + 1, total))
            memo[(index, total)] = res
            return res
            
            
        res = dp(0, target)
        memo.clear() # absence of this would result in a MLE
        return res if res != -float("inf") else -1
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