<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 ans=0;
    vector<vector<int>> dp;

    int dfs(int i,int j,vector<vector<int>>& matrix,int& n,int& m){
        int res=1;
        if(dp[i][j]!=-1) return dp[i][j];
        
        int x[] = { -1 , 1 , 0 , 0 };
        int y[] = { 0 , 0 , -1 , 1 };

        for(int l=0;l<4;l++){
            int h= i + x[l];
            int k = j+y[l];
            if( h<0 || h>=n || k>=m || k<0 ) continue;
            if(matrix[h][k] > matrix[i][j] ){
                    res = max(res , 1 + dfs(h,k,matrix,n,m));
            }
            

        }

        return dp[i][j]=res;

    }

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