<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>& arr) {
        ios::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);
        //average(A) = average(B) means sum of elements in A = i*totsum/n (a+b = totsum) (j*a = i*b) (i+j = n) i and j are no. of elements in A and B respectively and a and b are sum of elements in A and B
        int n = arr.size();
        int n1 = n/2;
        int n2 = n - n1;
        int totsum = accumulate(arr.begin(), arr.end(), 0);
        vector<vector<bool>> dp(n+1, vector<bool>(totsum+1, false));
        dp[0][0] = true;

        int l; int r; int sz;
        for(int mask = 1; mask<(1<<n1); mask++) {
            l = 0; sz = 0;
            for(int i = 0; i<n1; i++) {
                if(mask & (1<<i)){
                    sz++; 
                    l += arr[i];
                }
            }
            dp[sz][l] = true;
            if(1LL*n*l == 1LL*sz*totsum) return true;
        }

         for(int mask = 1; mask<(1<<n2); mask++) {
            r = 0; sz = 0;
            for(int i = 0; i<n2; i++) {
                if(mask & (1<<i)){
                    sz++; 
                    r += arr[i+n1];
                }
            }

            //Check if element from right + left is a valid sum or not starting by taking all elements from right to decreasing it 1 by 1
            for(int j = sz; j <= n/2; j++){
                if((j*totsum)%n != 0) continue;
                int s = j*totsum/n;
                if(s>=r && dp[j-sz][s-r]) return true; //j-sz is size of elements taken from l array
            }
        }

        return false;
    }
};
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