<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 mod=1e9+7;
int dp[1001][1001];
vector<int>delr={-1,0,1,0};
vector<int>delc={0,1,0,-1};
vector<vector<int>>v;
int n,m;
int dfs(int i,int j){
    int r=1;
    if(dp[i][j]!=-1) return dp[i][j];
    for(int l=0; l<4; l++){
        int nr=i+delr[l];
        int nc=j+delc[l];
        if(nr>=0 && nc>=0 && nr<n && nc<m && v[i][j]<v[nr][nc]){
            r=max(r,1+dfs(nr,nc));
        }
    }
    return dp[i][j]=r;
}
    int longestIncreasingPath(vector<vector<int>>& mat) {
         n=mat.size(); m=mat[0].size(); v=mat;
         memset(dp,-1,sizeof(dp));
         int ans=0;
         for(int i=0; i<n; i++){
            for(int j=0; j<m; j++){
                 ans=max(ans,dfs(i,j));
            }
         }
         return ans;
    }
};
23803