<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 check(int idx,int n,int target,vector<int>& nums,vector<vector<vector<bool>>>& dp){
        // base case: at arr's end or n==0
        if(n==0){
            return target==0;
        }
        if(idx==nums.size()) return false;
            
        if(dp[idx][n][target]==false) return false;
            
        //take
        bool take = false;
        if(target-nums[idx]>=0) take = check(idx+1,n-1,target-nums[idx],nums,dp);
        if(take) return true;
        //not take
        bool nTake = check(idx+1,n,target,nums,dp);
        if(nTake) return true;

        return dp[idx][n][target]=false;
    }
    
    bool splitArraySameAverage(vector<int>& nums) {
        int N = nums.size();
        if(N<2) return false; // single elem case
        int S=0;
        for(int e:nums) S+=e;
        if(S==0) return true; // all 0's case
        vector<vector<vector<bool>>> dp(N,vector<vector<bool>>(N,vector<bool>(S,true))); // dp[idx][n][target]
        
        for(int n=1; n<N; n++){
            if((S*n)%N==0){
                int s = (S*n)/N;
                if(check(0,n,s,nums,dp)) return true; // checking if 
// there exist a subarray starting from idx=0 till end of size=n with sum=s exists in nums or not
            }
        }
        return false;
    }
};
// reason for marking dp vector true: 
// agar nhi mila to will mark false for ensuring ki future me fir check na kare but agar true hai to check karlo as there can be 2 cases ya to mil jayega ya fir pehli bar check ho rha..
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