<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:
        summ = sum(nums)
        if summ % 2 != 0:
            return False

        half_sum = summ // 2
        if max(nums) > half_sum:
            return False
        
        mycock = set()
        def backtrack(i, subsum):
            if (i, subsum) in mycock:
                return False
            if subsum > half_sum:
                return False
            if subsum == half_sum:
                return True
            if i == len(nums):
                return False
            mycock.add((i, subsum))
            return backtrack(i + 1, subsum) or backtrack(i + 1, subsum + nums[i])
        
        return backtrack(0, 0)
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