<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 {
    int m,n;
public:
    void solve(int i,int j,int s,int prev,vector<vector<int>>& matrix,vector<vector<int>>& ans){
        if(i>=0 && i<m && j>=0 && j<n && matrix[i][j]>prev){
            ans[i][j] = max(ans[i][j],s);
        }
    }
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        m = matrix.size();
        n = matrix[0].size();
        vector<vector<int>> ans(m,vector<int>(n,1));
        priority_queue<pair<int,pair<int,int>>,vector<pair<int,pair<int,int>>>,greater<pair<int,pair<int,int>>>>pq;
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                pq.push({matrix[i][j],{i,j}});
            }
        }
        while(!pq.empty()){
            pair<int,pair<int,int>> temp = pq.top();
            pq.pop();
            int i = temp.second.first;
            int j = temp.second.second;
            solve(i-1,j,ans[i][j]+1,matrix[i][j],matrix,ans);
            solve(i+1,j,ans[i][j]+1,matrix[i][j],matrix,ans);
            solve(i,j-1,ans[i][j]+1,matrix[i][j],matrix,ans);
            solve(i,j+1,ans[i][j]+1,matrix[i][j],matrix,ans);
        }
        int a = 0;
        for(auto i:ans){
            for(auto j:i){
                a = max(a,j);
            }
        }
        return a;
    }
};
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