<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==1:
                return False
            

            dp=[[-1]*(sum(nums)//2+1) for _ in range(len(nums)+1)]

            def rec(i,x):

                if x==0:
                    return True
                if i<=0:
                    return False
                if dp[i][x]!=-1:
                    return  dp[i][x]
                
                if x>=nums[i-1]:
                    dp[i][x]= rec(i-1, x-nums[i-1]) or rec(i-1,x)
                else:
                    dp[i][x]=rec(i-1,x)
                return dp[i][x]
            return rec(len(nums), sum(nums)//2 )

            
42991