<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 helper(i: int, partial_sum: int) -> bool:            
            if i < 0:
                return False
            if nums[i] == target_sum or partial_sum-nums[i] == 0:
                dp[i] = True
                return dp[i]
            
            if not dp[i]:
                dp[i] = helper(i-1, partial_sum-nums[i]) or helper(i-1, partial_sum) 
            return dp[i]

        if sum(nums) % 2 != 0:
            return False
        
        target_sum = sum(nums) // 2
        n = len(nums) - 1
        dp = [0] * len(nums)
        return helper(n, target_sum)
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