<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(), SUM = accumulate(nums.begin(), nums.end(), 0);

        //we dont want floating point here
        // a - SUM/N => a*N - SUM -> would not affect the average
        for(auto & a : nums){
            a = a * N - SUM;
        }
        if(N == 1)return 0;

        vector<int> arrA;
        vector<int> arrB;
        for(int i = 0; i < N; i++){
            if(i < N/2)arrA.push_back(nums[i]);
            else arrB.push_back(nums[i]);
        } 
        set<int> s;
        for(int bitmask = 1; bitmask < (1 << arrB.size()); bitmask++){
            int sum = 0;
            for(int i = 0; i < arrB.size(); i++){
                if((1 << i) & bitmask)sum+=arrB[i];
            } 
            if(sum == 0)return 1;
            s.insert(sum);
        }
        int lsum = accumulate(arrA.begin(), arrA.end(), 0);
        for(int bitmask = 1; bitmask < (1 << arrA.size()); bitmask++){
            int sum = 0;
            for(int i = 0; i < arrA.size(); i++){
                if((1 << i) & bitmask)sum+=arrA[i];
            }
            if(sum == 0 || (lsum != sum && s.find(-sum) != s.end())){
                return 1;
            }
        }
    
        return 0;
    }
};
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