<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:
        ####### Top-down DP
        setSum = sum(nums)
        if setSum & 1:
            return False

        subSetSum = setSum // 2

        @cache
        def backtrack(i, subSetSum):
            # print(i, subSetSum)
            if subSetSum == 0:
                return True

            if i == 0:
                if subSetSum == nums[0]:
                    return True
                else:
                    return False

            # if i < 0 or subSetSum < 0:
            #     return False

            return backtrack(i - 1, subSetSum - nums[i]) or backtrack(i - 1, subSetSum)

        return backtrack(len(nums) - 1, subSetSum)

        ####### Brute force: O(2^n * n) -> TLE ########
        # n = len(nums)
        # for i in range(1 << n):
        #     left_sum = 0
        #     right_sum = 0
        #     for j in range(n):
        #         if i & (1 << j):
        #             right_sum += nums[j]
        #         else:
        #             left_sum += nums[j]

        #     if left_sum == right_sum:
        #         return True

        # return False

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