<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, typed=False)
        def dfs(pos, sum):
            if 0 == sum:
                return True

            if len(nums) == pos:
                return False
            
           # if -1 != dp[pos][sum]:
           #     return dp[pos][sum]
            if dfs(pos + 1, sum):
                dp[pos][sum] = 1
                return True
            
            if sum >= nums[pos] and dfs(pos + 1, sum - nums[pos]):
                dp[pos][sum] = 1
                return True
            
            dp[pos][sum] = 0
            return False
        

        s = sum(nums)
        if 0 != s % 2:
            return False
        
        s //= 2
        nums = tuple(nums)

        dp = [[-1] * (s + 1) for _ in range(len(nums))]
        return dfs(0, s)
352586