<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:
        n = len(nums)
        dp = [[-1] * (target + 1) for _ in range(n + 1)]
        for i in range(n + 1):
            dp[i][0] = 0

        for i in range(1, n + 1):
            for s in range(1, target + 1):
                dp[i][s] = dp[i - 1][s]
                
                if s - nums[i - 1] >= 0 and dp[i - 1][s - nums[i - 1]] != -1:
                    dp[i][s] = max(dp[i][s], dp[i - 1][s - nums[i - 1]] + 1)

        return dp[n][target]

        

        # n = len(nums)
        # memo = dict()


        # def backtracking(i: int, current_sum: int, current_length: int) -> int:
        #     nonlocal n
            
        #     if current_sum == target:
        #         return current_length

        #     if (i, current_sum, current_length) in memo:
        #         return memo[(i, current_sum, current_length)]
            
        #     result = 0
            
        #     for j in range(i, n):
        #         if current_sum + nums[j] <= target:
        #             t = backtracking(j + 1, current_sum + nums[j], current_length + 1)
        #             result = max(result, t)
            
        #     memo[(i, current_sum, current_length)] = result
        #     return result

        # result = backtracking(0, 0, 0)
        # return result if result > 0 else -1
        


        

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