<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:
        weight = sum(nums) / 2
        if not weight.is_integer():
            return False
        weight = int(weight)
        # dp: dp[i][j]: select num from [0, i], such that subset can fit j perfectly
        # return dp[len(nums)-1][weight]
        # dp[i][j] = dp[i-1][j] or dp[i-1][j-nums[i]]
        import numpy as np
        # dp = np.zeros((len(nums), weight+1))
        # # dp[i][0]: select num from [0,i] such that subset can fit 0
        # # dp[0][j]: select only nums[0], such that subset can fit 0
        # dp[:,0] = True

        # for i in range(len(nums)):
        #     for j in range(1, weight+1):
        #         if j >= nums[i]:
        #             dp[i][j] = dp[i-1][j] or dp[i-1][j-nums[i]]
        #         else:   # size smaller than new number
        #             dp[i][j] = dp[i-1][j]

        # because i only depends on i-1, can optimize
        dp = [0] * (weight+1)
        dp[0] = 1

        for num in nums:
            for j in range(weight, num-1, -1):
                dp[j] = dp[j] or dp[j-num]

        return dp[weight]

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