<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:
        summ = sum(nums)

        if summ % 2 == 1:
            return False
        nums.sort(reverse=True)
        half = summ//2
        n = len(nums)

        stack = deque([(0, 0)])
        visited = set((0, 0))
        while stack:
            idx, st = stack.popleft()
            if st==half:
                return True
            
            if idx < n:
                if (idx+1, st+nums[idx]) not in visited:
                    stack.append((idx+1, st+nums[idx]))
                    visited.add((idx+1, st+nums[idx]))
                if (idx+1, st) not in visited:
                    stack.append((idx+1, st))
                    visited.add((idx+1, st))
        
        return False

        return dp(0, 0)

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