<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 {
    int dp[201][201];
    int dfs(int r, int c, int par, vector<vector<int>>&matrix) {
        int dir[] = {0, -1, 0, 1, 0};
        int len = 1;
        if(dp[r][c] != -1) return dp[r][c];

        for(int k=0; k<4; ++k) {
            int nr = r + dir[k];
            int nc = c + dir[k+1];

            if(nr >=0 && nr<matrix.size() && nc>=0 &&
             nc<matrix[0].size() && par < matrix[nr][nc]) {

                len = max(len, 1 + dfs(nr, nc, matrix[nr][nc], matrix));
            }
        }
        return dp[r][c] = len;
    }
public:
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        int n = matrix.size();
        int m = matrix[0].size();
        memset(dp, -1, sizeof(dp));

        int maxLen = 1;

        for(int i=0; i<n; ++i) {
            for(int j=0; j<m; ++j) {
                int len = dfs(i, j,matrix[i][j], matrix);
                maxLen = max(maxLen, len);
            }
        }
        return maxLen;
    }
};
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