<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:
    //int dp[32][300009];
    bool splitArraySameAverage(vector<int>& nums) {
        int n=nums.size();
        
        int sum=0;
        for(int i=0;i<n;i++)
        {
            sum+=nums[i];
        }
        vector<vector<int>> dp(n,vector<int>(sum+1,-1));
       
        //memset(dp,-1,sizeof(dp));
        sort(nums.begin(),nums.end(),greater<int>());
        return solve(nums,0,0,n,0,sum,dp);
    }

    bool solve(vector<int>&nums,int i,int cnt,int n,float sum,float tot,vector<vector<int>> &dp)
    {
        if(i==n )  
        {
            return sum/cnt==(tot-sum)/(n-cnt);
        }
        if(sum/cnt<(tot-sum)/(n-cnt)) 
            return false; 
        
        if(dp[i][sum]!=-1) 
            return dp[i][sum];

        return dp[i][sum]=solve(nums,i+1,cnt,n,sum,tot,dp) || solve(nums,i+1,cnt+1,n,sum+nums[i],tot,dp);
    }
};
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