<p>You are given an <code>m x n</code> matrix <code>mat</code> that has its rows sorted in non-decreasing order and an integer <code>k</code>.</p>

<p>You are allowed to choose <strong>exactly one element</strong> from each row to form an array.</p>

<p>Return <em>the </em><code>k<sup>th</sup></code><em> smallest array sum among all possible arrays</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> mat = [[1,3,11],[2,4,6]], k = 5
<strong>Output:</strong> 7
<strong>Explanation:</strong> Choosing one element from each row, the first k smallest sum are:
[1,2], [1,4], [3,2], [3,4], [1,6]. Where the 5th sum is 7.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> mat = [[1,3,11],[2,4,6]], k = 9
<strong>Output:</strong> 17
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> mat = [[1,10,10],[1,4,5],[2,3,6]], k = 7
<strong>Output:</strong> 9
<strong>Explanation:</strong> Choosing one element from each row, the first k smallest sum are:
[1,1,2], [1,1,3], [1,4,2], [1,4,3], [1,1,6], [1,5,2], [1,5,3]. Where the 7th sum is 9.  
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == mat.length</code></li>
	<li><code>n == mat.length[i]</code></li>
	<li><code>1 &lt;= m, n &lt;= 40</code></li>
	<li><code>1 &lt;= mat[i][j] &lt;= 5000</code></li>
	<li><code>1 &lt;= k &lt;= min(200, n<sup>m</sup>)</code></li>
	<li><code>mat[i]</code> is a non-decreasing array.</li>
</ul>

class Solution {
public:
   long long  dp[200001][41];
   long long check(int sum,int row,vector<vector<int>>&mat){
      if(row==mat.size()) {
        if(sum==0) return 1;
        return 0;
      }
      if(sum<0) return 0;
      if(dp[sum][row]!=-1) return dp[sum][row];
       int n=mat.size(),m=mat[0].size();
      long long ans=0;
      for(int col=0;col<m;col++){
         ans=ans+check(sum-mat[row][col],row+1,mat);
         if(ans>=200) return ans;
      }
      return dp[sum][row]=ans;
   }
    int kthSmallest(vector<vector<int>>& mat, int k) {
        int n=mat.size(),m=mat[0].size();
        memset(dp,-1,sizeof(dp));
       // cout<<check(7,0,mat)<<endl;
        for(int i=1;i<=200000;i++){
            long long cnt=check(i,0,mat);
            if(cnt>0){
                k-=cnt;
                if(k<=0) return i;
            }
        }
        return -1;
    }
};
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