<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 m, n;
    short path[200][200];
int dfs(int i, int j, vector<vector<int>>& mat) {
        if(path[i][j] >   0) return path[i][j];
        if(path[i][j] == -1) return 0;
        int max_next = 0;
        path[i][j] = -1;
        if(i > 0   && mat[i][j] < mat[i-1][j]) max_next = max(max_next, dfs(i-1, j, mat));
        if(j > 0   && mat[i][j] < mat[i][j-1]) max_next = max(max_next, dfs(i, j-1, mat));
        if(i < m-1 && mat[i][j] < mat[i+1][j]) max_next = max(max_next, dfs(i+1, j, mat));
        if(j < n-1 && mat[i][j] < mat[i][j+1]) max_next = max(max_next, dfs(i, j+1, mat));
        return path[i][j] = 1 + max_next;
    }
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        m = matrix.size();
        n = matrix[0].size();

        memset(path, 0, sizeof(path));
    
        int max_path = 1;
        for(int i = 0; i < m; i++)
            for(int j = 0; j < n; j++)
                max_path = max(max_path, dfs(i, j, matrix));

        return max_path;
    }
};
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