<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 {
    #define ll long long
public:
    int kthSmallest(vector<vector<int>>& mat, int k) {
        ll n = mat.size();
        ll m = mat[0].size();

        priority_queue<vector<ll>, vector<vector<ll>>, greater<vector<ll>>> pq;

        vector<ll> temp(n+1, 0);
        for(ll i=0; i<n; i++) temp[0] += mat[i][0];

        pq.push(temp);
        // k--;
        ll ans = temp[0];
        map<vector<ll>,ll> mp;
        mp[temp] = 1;
        while(k--){
            vector<ll> v = pq.top();
            ans = v[0];
            pq.pop();

            // cout << ans << endl;
            // for(int i=0; i<=n; i++) cout << v[i] << " "; cout << endl;

            for(ll i=1; i<=n; i++){
                if (v[i] >= m-1) continue;
                v[i]++;
                v[0] += mat[i-1][v[i]] - mat[i-1][v[i]-1];
                if (mp.find(v) == mp.end()){
                    pq.push(v);
                    mp[v] = 1;
                }        
                v[i]--;
                v[0] = ans;
            }
        }

        return ans;
    }
};
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