<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:
        # 1. calculate sum of equal: sums // 2
        # 2. write DFS to check if there is any a pair equals
            # interate through nums, enter the nums[i], check dfs(i), exit nums[i]
            # if reaching the last nums i == len(nums), it means found the equals, return True

        @cache
        def dfs(i: int, target: int) -> bool: 
            if target == 0: return True
            if i == len(nums): return False
            if dfs(i+1, target): return True
            if target < nums[i]: return False
            target -= nums[i]
            if dfs(i+1, target): return True
            target += nums[i]
            return False

        total = sum(nums)
        if total % 2: return False
        nums.sort()
        return dfs(0, total // 2)

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