<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 i, int j, vector<vector<int>>& matrix, vector<vector<int>>& dp, int& n, int& m) {
        if (dp[i][j] != -1) return dp[i][j];
        vector<pair<int, int>> directions = {{-1, 0}, {0, -1}, {1, 0}, {0, 1}};
        int maxLength = 1;  
        for (auto dir : directions) {
            int nx = i + dir.first;
            int ny = j + dir.second;
            if (nx >= 0 && nx < n && ny >= 0 && ny < m && matrix[nx][ny] > matrix[i][j]) {  // matrix[nx][ny] > matrix[i][j] prevent coming to par so no infinte loop
                maxLength = max(maxLength, 1 + dfs(nx, ny, matrix, dp, n, m));
            }
        }
        dp[i][j] = maxLength; 
        return maxLength;
    }
    
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        if (matrix.empty()) return 0;
        int n = matrix.size();
        int m = matrix[0].size();
        vector<vector<int>> dp(n, vector<int>(m, -1)); 
        int maxPath = 1;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                maxPath = max(maxPath, dfs(i, j, matrix, dp, n, m));
            }
        }
        
        return maxPath;
    }
};

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