<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) {
        // brute force:
            // generate all partitions
            // compute sums

        int target = 0;
        for(const int& n : nums) target += n;
        // if odd, partition impossible
        if(target % 2) return false;    
       
        target /= 2;

        unordered_set<int> possibleSums;
        possibleSums.insert(0);

        // for each num
        for(const int& num : nums) {
            if(num > target) continue;
            // need copy of set because cannot iterate through set that is being mutated
            unordered_set<int> newPossibleSums = possibleSums;
            // create new possible sums by adding to existing possible sums
            for(const int& sum : possibleSums) {
                int val = sum + num;
                if(val == target) return true;
                if(val < target) newPossibleSums.insert(sum + num);
            }
            possibleSums = newPossibleSums;
        }

        return false;
    }
};
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