<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 sum = 0;
        for (auto & num : nums) {
            sum += num;
        }

        if (sum % 2 == 1) {
            return false;
        }

        sum /= 2;
        vector<bool> dp(sum+1, false);
        dp[0] = true;

        for (int i = 0; i < nums.size(); i++) {
            vector<int> new_nums;
            for (int j = 0; j <= sum; j++) {
                if (nums[i]+j > sum) {
                    break;
                }
                if (dp[j] && nums[i]+j <= sum) {
                    new_nums.push_back(nums[i]+j);
                    if (nums[i]+j == sum) {
                        return true;
                    }
                }
            }
            for (auto & new_num : new_nums) {
                dp[new_num] = true;
            }
        }

        return dp[sum];
    }
};
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