<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 S = accumulate(nums.begin(), nums.end(), 0);
        vector<vector<int>> dp(N + 1, vector<int>(S + 1, false));

        dp[0][0] = true;

        for(int x : nums) {
            for(int i = N; i > 0; i--) {
                for(int j = S - x; j >= 0; j--) {
                    if(dp[i - 1][j])
                        dp[i][j + x] = true;
                }
            }
        }

        for(int i = 1; i < N; i++) {
            for(int j = 0; j <= S; j++) {
                if(dp[i][j]) {
                    int ASum = j, BSum = S - j;
                    int ANum = i, BNum = N - i;
                    if(ASum * BNum == BSum * ANum)
                        return true;
                }
            }
        }

        return false;
    }
};
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