<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)
        dp = [[0]* (2*s+1) for _ in range(len(nums)+1)]
        # initial condition
        dp[0][s] = 1 
        for row in range(len(nums)):
            for col in range(len(dp[row])):
                if dp[row][col] != 0:
                    # print("row, col: ", row, col)
                    
                    dp[row+1][col-nums[row]] += dp[row][col]
                    dp[row+1][col+nums[row]] += dp[row][col]
        # print("dp[-1]: ", dp[-1])
        return dp[-1][s] != 0
            
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