<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 {
public:
    bool canPartition(vector<int>& nums) {
        // dp
        // observe: both halfs have to equal HALF of the total sum
            // if total sum is odd, return false
        // bottom up: start at back and add up sums
            // if one sum equals target, return true;
        // top down - 
        int sum = 0;
        for(int i : nums)
            sum += i;
        if(sum % 2 != 0)
            return false;
        int target = sum / 2;
        unordered_set<int> sums;
        sums.insert(0);
        for(int i = nums.size() - 1; i >= 0; i--)
        {
            if(sums.count(target))
                return true;
            unordered_set<int> newsums = sums;
            for(int n : sums)
            {
                newsums.insert(n + nums[i]);
            }
            sums = newsums;
        }
        return false;
    }
};
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