<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:
        # basically we wanna find a subset that is equal to half the sum of the
        # entire list

        total = sum(nums)

        if total % 2 != 0:
            return False
        total = int(total/2)
        @cache
        def dp(t, i):
            if t == 0:
                return True
            if t < 0:
                return False
            if i == len(nums):
                if t == 0:
                    return True
                else:
                    return False
            else:
                return dp(t - nums[i], i + 1) or dp(t, i + 1)

        arr = [[None]* (len(nums) + 1) for i in range((total) + 1)]
        for t in range((total) + 1):
            for i in range(len(nums)):
                arr[t][i] = dp(t, i)

        return arr[total][0]



                

                

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