<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>

#include <iostream>
#include <algorithm>
#include <numeric>
#include <cstring>

using namespace std;

class Solution {
public:
    Solution() {
        ios_base::sync_with_stdio(false);
        cin.tie(nullptr);
    }

    bool splitArraySameAverage(vector<int>& nums) {
        int n = nums.size();
        int sum = accumulate(nums.begin(), nums.end(), 0);
        int mx = sum;

        // Create a 2D array dynamically using malloc
        bool* possible = (bool*)malloc((n + 1) * (mx + 1) * sizeof(bool));
        memset(possible, false, (n + 1) * (mx + 1) * sizeof(bool));
        possible[0 * (mx + 1) + 0] = true;

        for (int i = 0; i < n; i++) {
            for (int j = mx; j >= nums[i]; j--) {
                for (int cnt = 0; cnt <= i; cnt++) {
                    if (possible[cnt * (mx + 1) + (j - nums[i])]) {
                        possible[(cnt + 1) * (mx + 1) + j] = true;
                    }
                }
            }
        }

        for (int i = 1; i < n; i++) {
            if ((sum * i) % n == 0) {
                int req = (sum * i) / n;
                if (possible[i * (mx + 1) + req]) {
                    free(possible);
                    return true;
                }
            }
        }

        free(possible);
        return false;
    }
};

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