<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>

#define ll long long
class Solution {
public:
    ll n;
    void f(ll ind, vector<ll> &v, ll sum, int c, unordered_map<int, set<double>> &dp) {
        if(ind>=v.size()) {
           if(c) dp[c].insert(sum);
           return;
        }
        f(ind+1, v, sum, c, dp);
        f(ind+1, v, sum+v[ind], c+1, dp);
    }
    bool splitArraySameAverage(vector<int>& nums) {
        this->n = nums.size();

        sort(nums.begin(), nums.end());

        vector<ll> v1, v2;

        int mid = n/2;
        double ts=0;

        for(ll i=0; i<mid; i++) v1.push_back(nums[i]), ts+=nums[i];
        for(ll i=mid; i<n; i++) v2.push_back(nums[i]), ts+=nums[i];

        unordered_map<int, set<double>> dp1, dp2;

        f(0, v1, 0, 0, dp1);
        f(0, v2, 0, 0, dp2);

        for(int cnt1=0; cnt1<=mid; cnt1++) {
            for(auto &sum1: dp1[cnt1]) {
                for(int cnt2=0; cnt2<=n-mid; cnt2++) {
                    if((cnt1+cnt2==n) || (cnt1+cnt2==0)) continue;
                    double sum2 = ts*(cnt1+cnt2)/n - sum1;

                    if(dp2[cnt2].count(sum2)) return true;
                }
            }
        }

        return false;
    }
};
36016