<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>& v) {
        int su=0,n=v.size(),m,ds,co;
        if(n==1) return false;
        for(int i=0;i<n;i++) su+=v[i];
        m=n/2;
        vector<vector<int>> v1(m, vector<int>((10000 * m) + 1, -1));
        vector<vector<int>> v2(n - m, vector<int>((10000 * (n - m)) + 1, -1));
        for(int i=1;i<(1<<m);i++){
            ds=0;co=0;
            for(int j=0;j<m;j++){
                if(i&(1<<j)) ds+=v[j],co++;
            }
            if(su*co==n*ds) return true;
            v1[co-1][ds]=1;
        }
        for(int i=1;i<(1<<(n-m));i++){
            ds=0;co=0;
            for(int j=0;j<(n-m);j++){
                if(i&(1<<j)) ds+=v[j+m],co++;
            }
            if(su*co==n*ds) return true;
            v2[co-1][ds]=1;
        }
        for(int i=0;i<m;i++){
            for(int j=0;j<=10000*(i+1);j++){
                if(v1[i][j]==1){
                    for(int k=0;k<(n-m);k++){
                        if((i+k+2)==n) break;
                        if((su*(i+k+2))%n!=0) continue;
                        co=(su*(i+k+2))/n;
                        if(co-j>=0&&v2[k][co-j]==1) return true;
                    }
                }
            }
        }
        return false;
    }
};
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