<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();

        //dp[r][c] represents len of longest incr path starting at r, c
        vector<vector<int>> dp(m, vector<int>(n, 0));

        //fill in dp for results
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (dp[i][j] == 0)
                    backtrack(matrix, i, j, dp, m, n);
            }
        }

        //get max of dp
        int res = 1;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                res = max(res, dp[i][j]);
            }
        }

        return res;

    }

    int backtrack(vector<vector<int>>& matrix, int i, int j, vector<vector<int>>& dp, int m, int n) {
        //prereq: i, j valid coords in matrix, dp

        if (dp[i][j] != 0) 
            return dp[i][j];

        //assert: dp[i][j] == 0
        vector<int> neighbors {1}; //min longest will now be 0
        //neighbors will be only 1 if is local max

        for (auto d : dirs) {
            if (i+d[0] < 0 || i+d[0] >= m || j+d[1] < 0 || j+d[1] >= n) continue;
            if (matrix[i+d[0]][j+d[1]] > matrix[i][j])
                neighbors.push_back(backtrack(matrix, i+d[0], j+d[1], dp, m, n)+1);
        }

        int res = *max_element(neighbors.begin(), neighbors.end());
        dp[i][j] = res;

        //postreq: returns longest incr path starting at i, j (>= 1)
        return res;
    }

    vector<vector<int>> dirs {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
};
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