<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:
        s=sum(nums)
        if s%2!=0:
            return False
        
        s//=2

        dp={}
        def fun(i,t):
            if t==0:
                return True
            if i==0:
                if nums[i]==t:
                    return True
                else:
                    return False
            if (i,t) in dp:
                return dp[(i,t)]
            notTake=fun(i-1,t)
            take=False
            if nums[i]<=t:
                take=fun(i-1,t-nums[i])
            dp[(i,t)]=notTake or take
            return dp[(i,t)]
        
        dp = [[False for _ in range(len(nums) + 1)] for _ in range(s + 1)]
        for i in range(len(nums) + 1):
            dp[0][i] = True

        for i in range(1,s+1):
            for j in range(1,len(nums)+1):
                if nums[j-1]>i:
                    dp[i][j] = dp[i][j - 1]
                else:
                    dp[i][j] = dp[i - nums[j - 1]][j - 1] or dp[i][j - 1]
        return dp[s][len(nums)]
72760