<p>Given an integer array <code>nums</code>, return <code>true</code> <em>if you can partition the array into two subsets such that the sum of the elements in both subsets is equal or </em><code>false</code><em> otherwise</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,5,11,5]
<strong>Output:</strong> true
<strong>Explanation:</strong> The array can be partitioned as [1, 5, 5] and [11].
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,2,3,5]
<strong>Output:</strong> false
<strong>Explanation:</strong> The array cannot be partitioned into equal sum subsets.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 200</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 100</code></li>
</ul>

class Solution:
    def canPartition(self, nums: List[int]) -> bool:

        # find 1 subset u can partpartion 
        # def dp .... repeat work?
        # or backtracking rather

        # ask yourself, can you partition this? do u need sum?
        # should it just be array splicing and take sum then go next?
        # ohhh sub set

        # ok it's BT
        # it's include or not include this current sum

        # mf it's just a 2 equal subset thing lol

        tot = sum(nums)

        if tot % 2 != 0:
            return False
        
        target = tot / 2 # need this exact amount

        dp = {}

        def dfs(i, cur): # idx, cur = cur sum

            if (i, cur) in dp:
                return dp[(i, cur)]
            
            if cur == target:
                return True
            
            if cur > target:
                return False
            
            if i == len(nums):
                return False
            # include

            dp[(i, cur)] = False # default

            cur += nums[i]
            check = dfs(i + 1, cur)

            if check == True:
                dp[(i, cur)] = True
                return True
            
            cur -= nums[i]

            # exclude
            check = dfs(i + 1, cur) 
            if check == True:
                dp[(i, cur)] = True
                return True
            
            return dp[(i, cur)]
        
        return dfs(0, 0)
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