<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) {
        int m=matrix.size();
        int n=matrix[0].size();
        map<pair<int,int>,vector<pair<int,int>>> adj;
        int dx[4]={-1,0,1,0};
        int dy[4]={0,-1,0,1};
        map<pair<int,int>,int> indegree;
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                for(int k=0;k<4;k++){
                    int ni=i+dx[k];
                    int nj=j+dy[k];
                    if(ni>=0 && ni<m && nj>=0 && nj<n && matrix[ni][nj] > matrix[i][j]){
                        adj[{i,j}].push_back({ni,nj});
                        indegree[{ni,nj}]++;
                    }
                }
            }
        }
        queue<pair<int, int>> q;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (indegree[{i, j}] == 0) {
                    q.push({i, j});
                }
            }
        }
        
        int ans=0;
        map<pair<int,int>,int> dist;
        while(!q.empty()){
            int i=q.front().first;
            int j=q.front().second;
            q.pop();
            int curr=dist[{i,j}]+1;
            ans=max(ans,curr);
            for(auto it : adj[{i,j}]){
                int ni=it.first;
                int nj=it.second;
                indegree[{ni,nj}]--;
                if(indegree[{ni,nj}]==0)
                    q.push({ni,nj});
                dist[{ni,nj}]=max(dist[{ni,nj}],curr);
            }
        }
        return ans;
    }
};
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