<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:
        # can break it into a target sum problem

        # calculate suffix sum
        size = len(nums)
        suffix_sum = [0] * (size+1)
        for i in range(size-1,-1,-1):
            suffix_sum[i] = nums[i] + suffix_sum[i+1]

        total = suffix_sum[0]
        if total % 2 == 1: return False
        goal = total//2
        self.save = dict()

        def solve(i, target): # sum to target with nums[i:]
            if (i, target) in self.save: return self.save[(i,target)]
            if target == 0:
                return True
            elif target < 0 or i == size:
                return False
            else:
                res = solve(i+1,target-nums[i]) or solve(i+1, target)
                self.save[(i,target)] = res
                self.save[(i, suffix_sum[i]-target)] = res
                return res

        return solve(0,goal)

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