<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:
        n = len(nums)
        total_sum = sum(nums)
        if n == 1:
            return False
        if total_sum % 2 == 1:
            return False

        target = total_sum // 2
        
        # 0/1 Knapsack problem
        # cache[(k, s)]: using nums[:k+1], can we reach a sum of s?

        # base cases
        cache = defaultdict(lambda: False)
        for k in range(n):
            cache[(k, nums[k])] = True
            cache[(k, 0)] = True
        # for s in range(nums[0] + 1, target + 1):
        #     cache[(0, s)]= False

        for k in range(1, n):
            for s in range(target + 1):
                if nums[k] <= s:
                    # (do not take nums[k]) or (take nums[k])
                    cache[(k, s)] = cache[(k - 1, s)] or cache[(k - 1, s - nums[k])]
                else:
                    # do not take nums[k]
                    cache[(k, s)] = cache[(k - 1, s)]

        
        return cache[(n - 1, target)]
                

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