<p>Given an <code>m x n</code> integers <code>matrix</code>, return <em>the length of the longest increasing path in </em><code>matrix</code>.</p>

<p>From each cell, you can either move in four directions: left, right, up, or down. You <strong>may not</strong> move <strong>diagonally</strong> or move <strong>outside the boundary</strong> (i.e., wrap-around is not allowed).</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/05/grid1.jpg" style="width: 242px; height: 242px;" />
<pre>
<strong>Input:</strong> matrix = [[9,9,4],[6,6,8],[2,1,1]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The longest increasing path is <code>[1, 2, 6, 9]</code>.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/27/tmp-grid.jpg" style="width: 253px; height: 253px;" />
<pre>
<strong>Input:</strong> matrix = [[3,4,5],[3,2,6],[2,2,1]]
<strong>Output:</strong> 4
<strong>Explanation: </strong>The longest increasing path is <code>[3, 4, 5, 6]</code>. Moving diagonally is not allowed.
</pre>

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

<pre>
<strong>Input:</strong> matrix = [[1]]
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>m == matrix.length</code></li>
	<li><code>n == matrix[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 200</code></li>
	<li><code>0 &lt;= matrix[i][j] &lt;= 2<sup>31</sup> - 1</code></li>
</ul>

class Solution {
public:
int longestIncreasingPath(vector<vector<int>>& matrix) {

    int m = matrix.size();
    int n = matrix[0].size();

    vector<vector<int>> G( m * n);
    vector<int> out(m * n);
    vector<int> ans(m * n);

    for(int i = 0 ; i < m;i++){
        for(int j=0;j < n;j++){
            if( j + 1 < n){
                if( matrix[i][j] > matrix[i][j+1] ){
                    G[i*n + j + 1].push_back(i*n + j);
                    out[i*n+j]++;
                }
                else if( matrix[i][j+1] > matrix[i][j] ){
                    G[i*n + j].push_back(i*n + j + 1);
                    out[i*n+j+1]++;
                }
            }
            if( i + 1 < m ){
                if( matrix[i + 1][j] > matrix[i][j] ){
                    G[i*n + j].push_back(( i + 1)*n + j);
                    out[(i + 1)*n+j]++;
                }
                else if( matrix[i][j] > matrix[i+1][j] ){
                    G[(i + 1)*n + j].push_back(i*n + j);
                    out[i*n+j]++;
                }
            }
        }
    }  

    set<pair<int,int>> s;
    for(int i = 0 ; i < m;i++){
        for(int j=0;j < n;j++){
            s.insert({out[i*n+j],i*n+j});
            ans[i*n+j] = 1;
        }
    }


    int a = 1;

    while(s.size() > 0 && (*s.begin()).first==0 ){
        int node = (*s.begin()).second;
        s.erase(*s.begin());
        for(auto it : G[node]){
            ans[it] = max(ans[it],1 + ans[node]);
            a = max(ans[it],a);
            s.erase({out[it],it});
            out[it]--;
            s.insert({out[it],it});
        }
    }

    return a;

}

};
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