<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>& nums) {
        int n = nums.size();
        vector<unordered_set<int>>possibleSums((n/2)+1);
        possibleSums[0].insert(0);
        for(int num: nums){
            for(int i=n/2; i>=1; i--){
                for(int prevElem: possibleSums[i-1]){
                    possibleSums[i].insert(prevElem+num);
                }
            }
        }
        int sum = 0;
        for(int num: nums) sum+=num;
        for(int i=1; i<=n/2; i++){
            if((i*sum)%n == 0){
                if(possibleSums[i].find((i*sum)/n) != possibleSums[i].end()) return true;
            }
        }
        return false;
        
    }
};
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