<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 {
private: 
    int m,n;
    vector<vector<int>> dir = {{0, 1},{0, -1},{1, 0},{-1, 0}};
    vector<vector<int>> memo;
    int dfs(vector<vector<int>>& matrix, int i, int j) {
        if(memo[i][j] != -1) return memo[i][j];
        int ans=1; // start with 1 length;
        for(auto d: dir) {
            int x=i+d[0];
            int y=j+d[1];
            if(x<n and y<m and x>=0 and y>=0 and matrix[i][j]<matrix[x][y]) {
                ans = max(ans, 1+dfs(matrix, x, y));
            }
        }
        memo[i][j] = ans;
        return ans;
        
    }
public:
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        int ans=0;
        n=matrix.size();
        if(n==0) return 0;
        m=matrix[0].size();
        memo = vector<vector<int>>(n, vector<int>(m, -1));
        for(int i=0;i<n;i++) {
            for(int j=0;j<m;j++) {
                ans = max(ans, dfs(matrix, i, j));
            }
        }
        return ans;
    }
};
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