<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) {
        sort(nums.begin(), nums.end());
        N = nums.size();
        ng = std::move(nums);
        int total = accumulate(ng.begin(),ng.end(), 0);
        if((total % 2) == 1)return false;
        return helper(0, total/2);
    }
    vector<int> ng;
    int N;
    unordered_map<int,unordered_map<int,bool>> d;
    bool helper(int i, int target){
        int initialtarget = target;
        if(d.contains(i) && d[i].contains(target))return d[i][target];
        if(target == 0) return true;
        else if(target < 0 || i >= N)return false;

        int curnum = ng[i];
        int nextn = i;
        while(nextn < N && ng[nextn] == curnum) nextn++;
        for(int k = 0; k < nextn - i + 1;k++){
            if(helper(nextn,target)){
                d[i][initialtarget] = true;
                return true;
            }
            target -= curnum;
        }
        d[i][initialtarget] = false;
        return false;
    }


};





    


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