<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 x[4]={0,0,1,-1};
    int y[4]={1,-1,0,0};
    vector<vector<int>>dp;
    int dfs(vector<vector<int>>& matrix,int r,int c,int m,int n){
        if(dp[r][c]!=-1){
            return dp[r][c];
        }
        int ans=0;
        for(int i=0;i<4;i++){
            int r1=r+x[i],c1=c+y[i];
            if(r1>=0&&r1<m&&c1>=0&&c1<n&&matrix[r][c]<matrix[r1][c1]){
                ans=max(ans,dfs(matrix,r1,c1,m,n));
            }
        }

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

        return ans;
    }
};
20271