<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) {
        int target = reduce(nums.begin(), nums.end());

        if (target & 0x1)
            return false;
        
        target /= 2;

        // bitset<20000> b;

        // for (const auto &el: nums) {
        //     b |= b << el;
        //     b[el - 1] = true;
        // }

        // return b[target - 1];
        unordered_set<int> dp{};
        dp.insert(0);

        for (int n: nums) {
            vector<int> newDp{};
            for (int i: dp) {
                newDp.push_back(n + i);
            }
            dp.insert(newDp.begin(), newDp.end());
        }

        return dp.contains(target);
    }
};
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