<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 hashVector(vector<int>pos){
        long long val=0;
        long long M=1e9+7;
        for(int j=0;j<pos.size();j++){
            val=((val*10)%M+pos[j])%M;
        } 
        return val;
    }

    int kthSmallest(vector<vector<int>>& mat, int k) {
        priority_queue<pair<long, vector<int>>, vector<pair<long, vector<int>>>, greater<pair<long, vector<int>>>> pq;
        int m=mat.size();
        int n=mat[0].size();
        unordered_set<long long> set;
        vector<int> pos(m, 0);
        long S=0;
        for(int i=0;i<m;i++){
            S+=mat[i][0];
        }
        pair p1={S, pos};
        pq.push(p1);
        int place=1;
        while(place<k){
            pair top=pq.top();
            pq.pop();
            vector<int> pos_=top.second;
            long long vectorHash=hashVector(pos_);
            //cout<<vectorHash<<endl;
            if(set.find(vectorHash)!=set.end()){
                continue;
            } 
            set.insert(vectorHash);
            long sum=top.first;
            
            for(int i=0;i<m;i++){
                
                pos_[i]++;
                if(pos_[i]<n){
                    pq.push({sum+mat[i][pos_[i]]-mat[i][pos_[i]-1], pos_});
                    // cout<<sum<<" "<<mat[i][pos_[i]-1]<<"->"<<mat[i][pos_[i]]<<endl;
                    // cout<<sum+mat[i][pos_[i]]-mat[i][pos_[i]-1]<<"<-pushed"<<endl;
                    // for(int j:pos_){
                    //     cout<<j<<" ";
                    // }
                }
                // cout<<endl;
                pos_[i]--;

            }
            place++;
        }
        // cout<<"Final\n";
        // vector<int> pos_=pq.top().second;
        // for(int j:pos_){
        //             cout<<j<<" ";
        //         }
        //         cout<<endl;
        while(set.find(hashVector(pq.top().second))!=set.end()){
            pq.pop();
        }
        return pq.top().first;
    }
};
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