<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 recursion(self, idx, target):
        if (idx, target) in self.dp_dict:
            return self.dp_dict[(idx, target)]
        
        if target == 0:
            return 0
        
        if idx >= self.n:
            return float("-inf")
        
        if self.nums[idx] > target:
            return float("-inf")

        #skip current eleement
        option1 = self.recursion(idx + 1, target)
        #use current element
        option2 = 1 + self.recursion(idx + 1, target - self.nums[idx])

        ans = max(option1, option2)
        self.dp_dict[(idx, target)] = ans
        return ans

    
    def lengthOfLongestSubsequence(self, nums: List[int], target: int) -> int:
        self.nums = sorted(nums)
        self.n = len(nums)
        self.dp_dict = {}
        self.t = target

        ans = self.recursion(0, target)
        if ans == float("-inf"):
            return -1

        return ans
7230