<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:
        @lru_cache(maxsize=None)
        def dfs(i, subset_sum):
            if subset_sum == 0 and i == 0:
                return True
            
            if i == 0 or subset_sum == 0:
                return False

            # true if you can sum up with i or without it
            return (
                dfs(i - 1, subset_sum) or dfs(i - 1, subset_sum - nums[i-1])
            )
        
        total_sum = sum(nums)
        if total_sum % 2 == 1:
            return False

        return dfs(len(nums), total_sum // 2)
            
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