<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();
        int sumt = 0;
        for (int i : nums) sumt += i;
        if (nums.size() == 1) return 0;
        if (sumt == 0) return 1;

        vector<unordered_set<int>> dp(n, unordered_set<int>());
        dp[0].insert(0);
        
        for (int len = 1; len < n; len++)
            for (int i = len - 1; i >= 0; i--)
                for (int x : dp[i]) dp[i + 1].insert(nums[len - 1] + x);


        for (int len = 1; len < n; len++) {
            int m = sumt * len / n;
            if (m * (n - len) == (sumt - m) * len) {
                if (dp[len].count(m)) return true;
            }
        }

        return false;
    }
};
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