<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>

int dfs(int i, vector<vector<int>> &v, vector<int> &visit, map<int,int> &arr){
    if(visit[i]){
        return arr[i];
    }
    visit[i] = 1;
    int ans = 0;
    for(auto j:v[i]){
        ans = max(ans, dfs(j,v,visit,arr));
    }
    ans++;
    arr[i] = max(arr[i], ans);
    return ans;
}
int dfs2(int i, vector<vector<int>> &v, vector<int> &visit){
    if(visit[i]){
        return 0;
    }
    visit[i] = 1;
    int ans = 0;
    for(auto j:v[i]){
        ans = max(ans, dfs2(j,v,visit));
    }
    ans++;
    cout << i << "-" << ans << endl;
    return ans;
}

class Solution {
public:
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        int i,j,n,m,i1,j1;
        n = matrix.size(), m = matrix[0].size();
        vector<vector<int>> v(n*m);
        vector<pair<int,int>>  v1 = {{1,0}, {-1,0}, {0,1}, {0,-1}};
        for(i = 0; i<n; i++){
            for(j = 0; j<m; j++){
                for(auto p:v1){
                    i1 = i+p.first, j1=j+p.second;
                    if((i1>=0) && (i1<n) && (j1>=0) && (j1<m) && (matrix[i][j]<matrix[i1][j1])){
                        v[i*m+j].push_back(i1*m+j1);
                    }
                }
            }
        }
        n = n*m;
        vector<int> visit(n,0), visit2(n,0), v2;
        map<int,int> arr;
        int ans = 0;
        for(i = 0; i<n; i++){
            if(visit[i]==0){
                ans = max(ans, dfs(i,v,visit,arr));
            }
        }
        // reverse(v2.begin(), v2.end());
        
        // for(auto i: v2){
        //     cout << i << " " << visit2[i] << endl;
        //     if(visit2[i]==0){
        //         ans = max(ans, dfs2(i,v,visit2));
        //     }
        //     cout << endl;
        // }
        return ans;

    }
};
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