<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>

static int dp[2][31][300002];
class Solution {
public:
    bool splitArraySameAverage(vector<int>& nums) {
        
        int n = nums.size();
        nums.insert(nums.begin(),0);
        int sum = 0;
        for(int i=1;i<=n;i++){
            sum+=nums[i];
        }
        // vector<vector<int>> prev(vector<vector<int>>(n+1,vector<int>(sum+1,0)));
        // vector<vector<int>> cur(vector<vector<int>>(n+1,vector<int>(sum+1,0)));
        //vector<vector<vector<int>>> dp(2,vector<vector<int>>(n+1,vector<int>(sum+1,0)));
        
         for(int i=0;i<2;i++){
            for(int j=0;j<=n;j++){
                for(int k=0;k<=sum;k++) dp[i][j][k]=0;
            }
         }
        dp[0][0][0]=1;
    //  prev[0][0] = 1;
    //  prev[1][nums[0]] = 1;
     for(int i=1;i<=n;i++){
        for(int j=0;j<=i;j++){
            for(int k=0;k<=sum;k++){
                if(j>0){
                if(k>=nums[i]){
                dp[i%2][j][k]=dp[(i-1)%2][j][k] | dp[(i-1)%2][j-1][k-nums[i]];
                }
                else dp[i%2][j][k]=dp[(i-1)%2][j][k];
            }
            else dp[i%2][j][k]=(k==0);
                
                if(dp[i%2][j][k]==1 && j>0 && j<n && (n-j)*(k)==(j*(sum-k))){
                    return 1;
                }
            }
        }
     }
     return 0;
    
      
    //  for(int i=2;i<=n;i++){
    //     for(int j=0;j<=i;j++){
    //         if(j==0){
    //            cur[j][0] = 1;
    //         }  
    //         else{
    //           for(int k=0;k<=sum;k++){
    //             if(k>=nums[i-1])
    //              cur[j][k] = (prev[j][k]|prev[j-1][k-nums[i-1]]);
    //             else
    //                cur[j][k] = prev[j][k];
    //             if(i==n&&j!=0&&j!=n){
    //                 if(cur[j][k]){
    //                     if((n-j)*(k)==(j)*(sum-k)){
    //                         return 1;
    //                     }
    //                   }
    //                 }
    //            }
    //         }
           
    //     }
    //      prev=cur;
    //  }
    //  return 0;

    }
};
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