<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:
        dp = [[-1 for _ in range((sum(nums)//2)+1)] for _ in range(len(nums)+1)]
        def recurr(idx, currTotal):
            if idx < 0:
                return False
            
            if currTotal == 0:
                return True

            if dp[idx][currTotal] != -1:
                return dp[idx][currTotal]
            
            option1 = recurr(idx-1, currTotal - nums[idx])
            option2 = recurr(idx-1, currTotal)

            dp[idx][currTotal] = option1 or option2

            return dp[idx][currTotal]
            
        total = sum(nums)
        if total % 2 == 0:
            return recurr(len(nums)-1, total//2)
        else:
            return False
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