<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:
    vector<int>dx{0,0,-1,1};
    vector<int>dy{1,-1,0,0};
    int n;
    int m;
    bool isValid(int x, int y){
        return (x >= 0 && x < n && y >= 0 && y < m);
    }
    int solve(vector<vector<int>>&matrix, vector<vector<int>>&dp, vector<vector<bool>>&vis, int x, int y) {
        if (dp[x][y] != -1){
            return dp[x][y];
        }
        vis[x][y] = true;
        int cur = matrix[x][y];

        int ans = 1;

        for(int k=0;k<4;k++){
            int newx = dx[k] + x;
            int newy = dy[k] + y;
            if (isValid(newx, newy)){
                if (!vis[newx][newy] && matrix[newx][newy] > cur) {
                    ans = max(ans, 1 + solve(matrix, dp, vis, newx, newy));
                }
            }
        }

        vis[x][y] = false;
        dp[x][y] = ans;

        return ans;
    }
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        n = matrix.size();
        m = matrix[0].size();


        vector<vector<bool>>vis(n+1, vector<bool>(m+1));
        vector<vector<int>>dp(n+1, vector<int>(m+1, -1));

        int ans = 1;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                ans = max(ans, solve(matrix, dp, vis, i, j));
            }
        }

        return ans;
    }
};
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