<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 isPossible(int ind, int len, int target, vector<int>& nums, vector<vector<vector<bool>>>& dp){
        if(len==0) return target==0;
        if(ind>=nums.size()) return false;
        if(dp[ind][len][target]==false) return false;

        bool take=false;
        if(nums[ind]<=target) 
            take=isPossible(ind+1,len-1,target-nums[ind],nums,dp);
        bool notTake=isPossible(ind+1,len,target,nums,dp);

        return dp[ind][len][target]=(take || notTake);
    }

    bool splitArraySameAverage(vector<int>& nums) {
        int n=nums.size();
        long sum=accumulate(nums.begin(),nums.end(),0);
        vector<vector<vector<bool>>> dp(n,vector<vector<bool>>(n+1,vector<bool>(sum+1,1)));
        //s1+s2=s
        //n1+n2=n
        //s1/n1=s2/n2 --> s2=(n-n1)s1/n1
        //s1=s*n1/n
        for(int len=1;len<n;len++){
            //for reqSum to be an integer (sum*len)%n==0
            if((sum*len)%n) continue;
            int reqSum=sum*len;
            reqSum/=n;
            if(isPossible(0,len,reqSum,nums,dp))
                return true;
        }
        return false;
    }
};
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