<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:
        # 1. state: (idx, subset_sum)
        # 2. output: T/F can be partition
        # 3. transition: (idx - 1, subset_sum) # include nums[idx - 1]
        # or (idx - 1, subset_sum - nums[idx - 1]) # not include nums[idx - 1]
        total = sum(nums)

        if total % 2 != 0:
            return False

        subset_sum = total // 2
        n = len(nums) - 1
        cache = {}

        def dfs(idx, subset_sum):
            if subset_sum == 0:
                return True

            if idx == 0 or subset_sum < 0:
                return False

            if (idx, subset_sum) in cache:
                return cache[(idx, subset_sum)]

            cache[(idx, subset_sum)] = dfs(idx - 1, subset_sum) or dfs(idx - 1, subset_sum - nums[idx - 1])
            return cache[(idx, subset_sum)]
        
        return dfs(n, subset_sum)
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