<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>>& dp,vector<vector<int>>& visited){
        if(dp[i][j]!=-1){
            return dp[i][j];
        }
        if(visited[i][j]==1){
            return 0;
        }
        
        visited[i][j]=1;
        
        int l=0;
        int r=0;
        int u=0;
        int d=0;
        
        if(j-1>=0 && matrix[i][j-1]>matrix[i][j]){
            l=1+dfs(i,j-1,matrix,dp,visited);
        }
        if(j+1<matrix[0].size() && matrix[i][j+1]>matrix[i][j]){
            r=1+dfs(i,j+1,matrix,dp,visited);
        }
        if(i-1>=0 && matrix[i-1][j]>matrix[i][j]){
            u=1+dfs(i-1,j,matrix,dp,visited);
        }
        if(i+1<matrix.size() && matrix[i+1][j]>matrix[i][j]){
            d=1+dfs(i+1,j,matrix,dp,visited);
        }
        visited[i][j]=0;
        return dp[i][j]=max({l,r,u,d});
    }
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        int n=matrix.size();
        int m=matrix[0].size();
        
        vector<vector<int>> dp(n,vector<int>(m,-1));
        vector<vector<int>> visited(n,vector<int>(m,0));
        int ans=0;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(dp[i][j]==-1){
                    
                    dfs(i,j,matrix,dp,visited);
                }
                ans=max(ans,dp[i][j]);
            }
        }
        
        return ans+1;
    }
};
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