<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:
        if sum (nums) % 2 != 0:
            return False
        
        target = sum(nums) //2
        memo = {}
        
        # run dfs to find if target is in nums, if so, return True, else return False
        
        def dfs(idx, currSum):
            
            # returns True if currSum is in nums
            if (idx, currSum) in memo:
                return memo[(idx, currSum)]
            
            if currSum == target:
                return True
            if idx >= len(nums) or currSum > target:
                return False
            
            memo[(idx, currSum)] = dfs(idx + 1, currSum) or dfs(idx + 1, currSum + nums[idx])
            
            return memo[(idx, currSum)]
            
            
            
        
        return dfs(0, 0)
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