<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:
    bool splitArraySameAverage(vector<int>& a) {
        int n = a.size();
        int sum = accumulate(a.begin(), a.end(), 0);
        // vector<vector<bool>> dp(n + 1, vector<bool>(sum + 1, 0));
        vector<int> dp(sum+1,0);
        dp[0]=0;
        vector<pair<int,int>> v,x;
        v.push_back({0,0});
        for(int i=1;i<=n;i++)
        {
            x.clear();
            for(auto &[f,s] : v)
            {
                if(s+1>n-1 || ((dp[f+a[i-1]]>>(s+1))&1)==1)continue;
                if((f+a[i-1])*n==sum*(s+1))return true;
                x.push_back({f+a[i-1],s+1});
                dp[f+a[i-1]]+=1<<(s+1);
            }
            v.insert(v.end(),x.begin(),x.end());
        }
        return false;
    }
};
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