<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:


    # Neetcode 150 9/04/2024
    # Time O(2^n) SpaceO(n)
    def canPartition(self, nums: List[int]) -> bool:
        total = sum(nums)
        if total % 2 != 0:
            return False 
        target = total // 2 

        dp = {}

        def dfs(i, runningSum):
            if runningSum == target:
                return True 
            if i == len(nums) or runningSum > target:
                return False 
            if (i, runningSum) in dp:
                return dp[(i, runningSum)]
            dp[(i, runningSum)] = dfs(i + 1, runningSum) or dfs(i + 1, runningSum + nums[i])
            return dp[(i, runningSum)]

        return dfs(0, 0)

    # # Neetcode 150 9/04/2024
    # # Time O(2^n) SpaceO(n)
    # def canPartition(self, nums: List[int]) -> bool:
    #     total = sum(nums)
    #     if total % 2 != 0:
    #         return False 
    #     target = total // 2 

    #     def dfs(i, runningSum):
    #         if runningSum == target:
    #             return True 
    #         if i == len(nums) or runningSum > target:
    #             return False 
    #         return dfs(i + 1, runningSum) or dfs(i + 1, runningSum + nums[i])

        
    #     return dfs(0, 0)
                

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