<p>You are given an integer array <code>nums</code>.</p>

<p>You should move each element of <code>nums</code> into one of the two arrays <code>A</code> and <code>B</code> such that <code>A</code> and <code>B</code> are non-empty, and <code>average(A) == average(B)</code>.</p>

<p>Return <code>true</code> if it is possible to achieve that and <code>false</code> otherwise.</p>

<p><strong>Note</strong> that for an array <code>arr</code>, <code>average(arr)</code> is the sum of all the elements of <code>arr</code> over the length of <code>arr</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,4,5,6,7,8]
<strong>Output:</strong> true
<strong>Explanation:</strong> We can split the array into [1,4,5,8] and [2,3,6,7], and both of them have an average of 4.5.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,1]
<strong>Output:</strong> false
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 30</code></li>
	<li><code>0 &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution {
public:

    int dp[31][31];



    bool splitArraySameAverage(vector<int>& nums) {
        int sum=0;
        for(auto it:nums)sum+=it;
        int n=nums.size();
        set<pair<int,int>>s;
        s.insert({0,0});
        for(auto x:nums)
        {
            vector<pair<int,int>>v;
            for(auto it:s)
            {
                v.push_back({it.first+x,it.second+1});
            }

            for(auto it:v)
            {
                long long sum1=it.first;
                long long sum2=sum-it.first;
                sum1*=n-it.second;
                sum2*=it.second;
                if(sum1==sum2&&(it.second!=n&&it.second!=0))return 1;
                s.insert(it);
            }
        }
        return 0;
    }
};
378438