<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 __init__(self):
        self.cache = {}
    def canPartition(self, nums: List[int], left: int = 0, right: int = 0) -> bool:
        if len(nums) <= 0:
            # base case: all numbrers allocated
            return left == right        
        key = (min(left, right), max(left, right))
        
        if key in self.cache:
            # cached case: return memorised answer
            return self.cache[key]

        # recursively search for answer
        self.cache[key] = any(
            [self.canPartition(nums[1:], left + nums[0], right),
            self.canPartition(nums[1:], left, right + nums[0])]
        )
        return self.cache[key]
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