<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:
        # 2 5 6 7 8
        sum_nums: int = sum(nums)
        
        if sum_nums % 2 == 1:
            return False
        
        half_sum = sum_nums // 2
        
        nums.sort()
        
        @cache
        def backtrack(i, s):
            if s == half_sum:
                # print(i, s, 'a')
                return True
            for j in range(i + 1, len(nums)):
                s += nums[j]
                if backtrack(j, s):
                    # print(i, s, 'b')
                    return True
                s -= nums[j]
            # print(i, s, 'c')
            return False
        
        return backtrack(-1, 0)

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