<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) {
        vector<pair<int,pair<int,int>>> v;
        vector<vector<int>> visit;
        for (int i = 0; i < matrix.size(); i ++) {
            visit.push_back({});
            for (int j = 0; j < matrix[i].size(); j ++) {
                v.push_back({matrix[i][j],{i,j}});
                visit[i].push_back(0);
            }
        }
        sort (v.begin(), v.end());
        for (int i = 0; i < v.size(); i ++) {
            int a = v[i].second.first;
            int b = v[i].second.second;
            int dx[4] = {0,0,1,-1};
            int dy[4] = {-1,1,0,0};
            int value = 0;
            for (int i = 0; i < 4; i ++) {
                int c = a + dx[i];
                int d = b + dy[i];
                if (c >= 0 && c < matrix.size() && d >= 0 && d < matrix[0].size() 
                    && matrix[c][d] < matrix[a][b]) {
                    value = max (value, visit[c][d]);
                }
            }
            visit[a][b] = value+1;
        }
        int maxim = 0;
        for (int i = 0; i < matrix.size(); i ++) {
            for (int j = 0; j < matrix[i].size(); j ++) {
                maxim = max (maxim,visit[i][j]);
            }
        }
        return maxim;
    }
};
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