<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 longestIncreasingPath(vector<vector<int>>& matrix) {
        int m = matrix.size();
        int n = matrix[0].size();
        vector<vector<int>> memo(m, vector<int>(n, 0));
        vector<vector<bool>> visited(m, vector<bool>(n, false));

        int result = 0;
        for (int i = 0; i < m; i++)
        {
            for (int j = 0; j < n; j++)
            {
                // visited[i][j] = true;
                result = max(result, dfs(matrix, i, j, memo, visited));
                // visited[i][j] = false;
            }
        }

        return result;
    }

    int dfs(vector<vector<int>>& matrix, int row, int col, 
            vector<vector<int>>& memo, vector<vector<bool>>& visited)
    {
        if (memo[row][col] > 0)
        {
            return memo[row][col];
        }

        int result = 1;
        vector<pair<int, int>> dirs{{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
        for (int i = 0; i < dirs.size(); i++)
        {
            int new_row = row + dirs[i].first;
            int new_col = col + dirs[i].second;
            if (new_row >= 0 && new_row < matrix.size() 
                && new_col >= 0 && new_col < matrix[0].size() 
                && visited[new_row][new_col] == false
                && matrix[new_row][new_col] > matrix[row][col])
            {
                visited[new_row][new_col] = true;
                result = max(result, 1 + dfs(matrix, new_row, new_col, memo, visited));
                visited[new_row][new_col] = false;
            }
        }

        memo[row][col] = result;
        return result;
    }
};
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