<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:
        # you can partition into 2 lists, l1 and l2
        # each element has to be apart of one of the lists
        #
        # at d[-1], want to check if there are any list sums that are different by d[-1]
        # so each at d[i], we should store all the possible difference in list sums
        # assume a list of no elements has a sum of zero

        d = [set() for _ in range(len(nums))]

        # base case: 
        #   1. l1 = nums[0], l2 = 0
        #   2. l1 = 0, l2 = nums[0]
        d[0].add(abs(nums[0]))

        for i in range(1, len(nums)):
            for diff in d[i-1]:
                d[i].add(abs(diff+nums[i]))
                d[i].add(abs(diff-nums[i]))
            
        return 0 in d[-1]




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