<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>

/*
    Given matrix, return length of longest increasing path
    Ex. matrix = [[9,9,4],[6,6,8],[2,1,1]] -> 4, [1,2,6,9]

    DFS + memo, cache on indices, compare to prev for increasing check

    Time: O(m x n)
    Space: O(m x n)
*/

class Solution {
public:
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        int m = matrix.size();
        int n = matrix[0].size();
        
        int result = 0;
        
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                result = max(result, dfs(matrix, -1, i, j, m, n));
            }
        }
        
        return result;
    }
private:
    // {(i, j) -> longest increasing path at (i, j)}
    map<pair<int, int>, int> dp;
    
    int dfs(vector<vector<int>>& matrix, int prev, int i, int j, int m, int n) {
        if (i < 0 || i >= m || j < 0 || j >= n || matrix[i][j] <= prev) {
            return 0;
        }
        if (dp.find({i, j}) != dp.end()) {
            return dp[{i, j}];
        }
        
        int result = 1;
        result = max(result, 1 + dfs(matrix, matrix[i][j], i - 1, j, m, n));
        result = max(result, 1 + dfs(matrix, matrix[i][j], i + 1, j, m, n));
        result = max(result, 1 + dfs(matrix, matrix[i][j], i, j - 1, m, n));
        result = max(result, 1 + dfs(matrix, matrix[i][j], i, j + 1, m, n));
        dp[{i, j}] = result;
        
        return dp[{i, j}];
    }
};

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