<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 dfs(int i, int j, vector<vector<int>>& matrix, vector<vector<int>>& lip, vector<vector<bool>>& visited){
        vector<pair<int,int>> dirs={{0,1},{0,-1},{1,0},{-1,0}};
        if(i<0 || i>=matrix.size() || j<0 || j>=matrix[0].size()){
            return 0;
        }
        if(visited[i][j]==true){
            return lip[i][j];
        }
        visited[i][j]=true;
        for(auto p:dirs){
            int r=i+p.first;
            int c=j+p.second;
            if(r>=0 && r<matrix.size() && c>=0 && c<matrix[0].size() && matrix[r][c]>matrix[i][j]){
                lip[i][j]=max(lip[i][j],1+dfs(r,c,matrix,lip,visited));
            }
        }
        return lip[i][j];
    }
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        vector<vector<int>> lip(matrix.size(),vector<int>(matrix[0].size(),1));
        vector<vector<bool>> visited(matrix.size(),vector<bool>(matrix[0].size(),false));
        int maxval=0;
        for(int i=matrix.size();i>=0;i--){
            for(int j=matrix[0].size();j>=0;j--){
                maxval=max(maxval,dfs(i,j,matrix,lip,visited));
            }
        }
        return maxval;
    }
};
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