<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 dfs(int r, int c, vector<vector<int>>& matrix, vector<vector<int>> &dp) {
        
        int m = matrix.size();
        int n = matrix[0].size();
        vector<int> dr = {0, +1, 0, -1};
        vector<int> dc = {+1, 0, -1, 0};

        if(dp[r][c] != -1){
            return dp[r][c];
        }

        int ans = 1;
        for(int i=0; i<4; i++){
            int rr = r + dr[i];
            int cc = c + dc[i];
            if(rr>=0 && cc>=0 && rr<m && cc<n && matrix[rr][cc] > matrix[r][c]){
                ans = max(ans, dfs(rr, cc, matrix, dp) + 1);
            }
        }

        return dp[r][c] = ans;
    }

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

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

        for(int i=0; i<m; i++){
            for(int j=0; j<n; j++){
                maxi = max(maxi, dfs(i, j, matrix, dp));
            }
        }

        return maxi;
    }
};
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