<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>

from collections import Counter
from functools import cache


class Solution:

    def canPartition(self, nums: list[int]) -> bool:

        s = sum(nums)

        if s % 2 == 1:
            return False

        nums_count = list(Counter(nums).items())
        n_uniq_vals = len(nums_count)

        @cache
        def exist_subset(i: int, s_sub: int):
            # return true of if we can select a subset of the nums in nums_count[i:] with sum s1.
            nonlocal nums_count
            if i == n_uniq_vals:
                return s_sub == 0

            if s_sub < 0:
                # not possible subset - since no negative numbers
                return False
            x, x_cnt = nums_count[i]
            # we need to check all possible number of times we include x
            x_sum_in_sub = 0
            # The sum from i, after we omit the i-th possible numer from nums_count
            for _ in range(x_cnt + 1):
                if exist_subset(i=i + 1, s_sub=s_sub - x_sum_in_sub):
                    return True
                x_sum_in_sub += x
            return False

        return exist_subset(i=0, s_sub=s // 2)

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