<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 func(vector<vector<int>> &mm,int x,int y,vector<vector<int>> &dp,int m,int n){
    vector<int> dir={-1,0,1,0,-1};
    if(dp[x][y]!=-1)
    return dp[x][y];

    int ans=0;
    for(int i=0;i<4;i++){
        int a=min(m-1,max(0,x+dir[i]));
        int b=min(n-1,max(0,y+dir[i+1]));

        if(mm[a][b]>mm[x][y])
        ans=max(ans,func(mm,a,b,dp,m,n)+1);
    }
    return dp[x][y]=ans;
   }

public:
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        int m=matrix.size();
        int n=matrix[0].size();
        priority_queue<pair<int,pair<int,int>>,vector<pair<int,pair<int,int>>>, greater<pair<int,pair<int,int>>>> q;
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                q.push({matrix[i][j],{i,j}});
            }
        }
        int ans=0;

        vector<vector<int>> dp(m,vector<int>(n,-1));
        while(!q.empty()){
            int x=q.top().second.first;
            int y=q.top().second.second;
            q.pop();
            ans=max(ans,func(matrix,x,y,dp,m,n));
        }
        return ans+1;
    }
};
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