<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 row[4] = {0,0,-1,1};
    int col[4] = {1,-1,0,0};

    bool valid(int i, int j, int n, int m){
        return i>=0 && j>=0 && i<n && j<m;
    }

    int dfs(vector<vector<int>>& matrix, int i, int j, int n, int m, vector<vector<bool>> &visited, vector<vector<int>> &dp){

        if(dp[i][j] != -1){
            return dp[i][j];
        }

        visited[i][j] = true;
        int ans = 1;

        for(int k=0; k<4; k++){
            int newR = i+row[k];
            int newC = j+col[k];

            if(valid(newR,newC,n,m) && matrix[i][j] < matrix[newR][newC]){
                ans = max(ans,1 + dfs(matrix,newR,newC,n,m,visited,dp));
            }
        }

        return dp[i][j] = ans;

    }

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

        vector<vector<int>> dp(n,vector<int>(m,-1));
        vector<vector<bool>> visited(n,vector<bool>(m,0));

        int ans = 1;

        for(int i=0; i<n; i++){
            for(int j=0; j<m; j++)
            if(!visited[i][j])        
            ans = max(ans,dfs(matrix,i,j,n,m,visited,dp)); 
        }
        return ans;
    }
};
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