<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 count(int i,int j,int row,int col,int prevVal,vector<vector<int>>& matrix,vector<vector<int>> &dp){
   if(i<0 || i>=row || j<0 || j>=col || matrix[i][j]<=prevVal)return 0;
   if(dp[i][j]!=-1)return dp[i][j];
   int up=1+count(i-1,j,row,col,matrix[i][j],matrix,dp);
   int left=1+count(i,j-1,row,col,matrix[i][j],matrix,dp);
   int right=1+count(i,j+1,row,col,matrix[i][j],matrix,dp);
   int down=1+count(i+1,j,row,col,matrix[i][j],matrix,dp);
   return dp[i][j]=max(up,max(down,max(left,right)));
}
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        int row=matrix.size();
        int col=matrix[0].size();
        int maxi=0;
        vector<vector<int>> dp(row,vector<int>(col,-1));
        for(int i=0;i<row;i++){
            for(int j=0;j<col;j++){
                maxi=max(maxi,count(i,j,row,col,-1,matrix,dp));
            }
        }
     return maxi;
            }
};
20271