<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:

    const int target = 100001;

    bool splitArraySameAverage(vector<int>& nums) 
    {
        int n = nums.size();
        int sum = 0;
        int mx = 0;
        for(auto no : nums){
            sum += no;
            mx = max(mx, no);
        }
        bool poss[n+1][n+1][target];
        memset(poss, 0, sizeof(bool)*(n+1)*(n+1)*(target));
        for(int i=0; i<=n; i++)
            poss[i][0][0] = true;
        for(int i=1; i<=n; i++)
            for(int k=1; k<=i; k++)
                for(int t=0; t<min(k*mx+1,target); t++)
                {
                    poss[i][k][t] = poss[i-1][k][t];
                    if(t>=nums[i-1]){
                        poss[i][k][t] = poss[i][k][t] || poss[i-1][k-1][t-nums[i-1]];
                    }
                    // cout << i << "," << k << "," << t << ": " << poss[i][k][t] << "\n";
                }
        

        // iterate through all values
        for(int k=1; k<n; k++){
            int t=k*sum/n;
            if(t*n==k*sum && poss[n][k][t]){
                cout << k << " " << t << "\n";
                return true;
            }
        }
        
        return false;
    }
};
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