<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) {
        unordered_set<int> sums = {0};
        int sum = accumulate(nums.begin(), nums.end(), 0);
        if (sum % 2)
            return false;
        
        int target = sum / 2;
        for (int n : nums)
        {
            vector<int> to_add;

            // You can't be adding elements while you're iterating over this
            for (int s : sums)
            {
                if (n + s == target)
                    return true;
                to_add.push_back(n + s);
            }

            copy(to_add.begin(), to_add.end(), inserter(sums, sums.end()));
        }

        return sums.find(target) != sums.end();
    }
};
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