<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:
    # Time complexity: O(n*sum(nums))
    # Space complexity: O(n*sum(nums))
    def canPartition(self, nums: List[int]) -> bool:
        self.nums = nums
        # If the sum of the list items is not divisible by two, then the list cannot be split into two parts with the
        # same sum
        if sum(nums) % 2 != 0:
            return False

        n = len(self.nums)
        
        self.target = sum(self.nums) // 2
        self.cache = [None for _ in range(n)]

        return self.target in self.find_sum(0, 0)

    def find_sum(self, i: int, current_sum: int) -> Optional[set]:
        n = len(self.nums)
        # If we get a sum greater than the target, we don't need to store it because it will certainly result in a
        # unbalance the partitions
        if current_sum > self.target:
            return set()

        if i >= n:
            return {0}
        # Checking for cached results
        if self.cache[i] is not None:
            return self.cache[i]
        # Ignoring the ith item
        sums_1 = self.find_sum(i + 1, current_sum)
        # Adding the ith item to the partition
        tmp = self.find_sum(i + 1, current_sum + self.nums[i])
        sums_2 = set()

        for sum in tmp:
            if sum + self.nums[i] <= self.target:
                sums_2.add(sum + self.nums[i])
        # Cache the result
        self.cache[i] = sums_1.union(sums_2)
        return self.cache[i]
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