<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:
    vector<vector<int>>dp;
    int n,m;
    vector<pair<int,int>>find_smaller(int i,int j,vector<vector<int>>& matrix)
    {
        vector<pair<int,int>>a;
        if(i>0 && matrix[i][j]>matrix[i-1][j])
            a.push_back({i-1,j});
        if(j>0 && matrix[i][j]>matrix[i][j-1])
            a.push_back({i,j-1});
        if(i<n-1 && matrix[i][j]>matrix[i+1][j])
            a.push_back({i+1,j});
        if(j<m-1 && matrix[i][j]>matrix[i][j+1])
            a.push_back({i,j+1});
        return a;
    }
    int rec(int i,int j,vector<vector<int>>& matrix)
    {
        vector<pair<int,int>>smaller=find_smaller(i,j,matrix);
        if(smaller.empty())
            return 1;
        if(dp[i][j]!=-1)
            return dp[i][j];
        int ans=1;
        for(auto v:smaller)
        {
            ans=max(ans,rec(v.first,v.second,matrix)+1);
        }
        return dp[i][j]=ans;
    }
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        n=matrix.size();
        m=matrix[0].size();

        dp.assign(n,vector<int>(m,-1));
        int ans=0;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(dp[i][j]==-1)
                {
                    ans=max(ans,rec(i,j,matrix));
                }
            }
        }
        return ans;
    }
};
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