<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 dp[201][201];
    int fun(int x,int y,vector<vector<int>>& arr){
        int n = arr.size(); 
        int m = arr[0].size();

        if( dp[x][y] != -1 )return dp[x][y];

        int maxi = 0;
        if( x+1 < n && arr[x][y] > arr[x+1][y] )maxi = fun(x+1,y,arr);
        if( y+1 < m && arr[x][y] > arr[x][y+1] )maxi = max(fun(x,y+1,arr),maxi);
        if( x-1 >= 0 && arr[x][y] > arr[x-1][y] )maxi = max(fun(x-1,y,arr),maxi);
        if( y-1 >= 0 && arr[x][y] > arr[x][y-1] )maxi = max(fun(x,y-1,arr),maxi);

        return dp[x][y]=maxi+1;
    }

    int longestIncreasingPath(vector<vector<int>>& matrix) {
        
        int maxi = 0; 
        int n = matrix.size(); 
        int m = matrix[0].size();

        memset(dp,-1,sizeof dp);

        for(int i = 0; i < n; i++)
            for(int j = 0; j < m; j++ )maxi = max(maxi,fun(i,j,matrix));

        return maxi;
    }
};
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