<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:
        sumofnums=sum(nums)
        if sumofnums%2!=0:
            return False
        dp=[[-1]*(sumofnums+1) for i in range(len(nums)+1)]
        def dfs(index,target):
            if dp[index][target]!=-1:
                    return dp[index][target]
            if index==len(nums) or target<=0 or (index,target) in dp:
                if target==0:
                    
                    return True
                
                return False
            dp[index][target]=dfs(index+1,target) or dfs(index+1,target-nums[index])
            return dp[index][target]


        ## base cases
        

        return dfs(0,sumofnums//2)
        
66806