<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:
bool isvalid(int i, int j, int n, int m){
    return (((i>=0)&&(i<n))&&((j>=0)&&(j<m)));
}
    int dfs(int i, int j,vector<vector<int>>& matrix,vector<vector<int>>&dp){
          int ans=0;
          int n=matrix.size();
          int m=matrix[0].size();
          if(dp[i][j]!=-1) return dp[i][j];
          vector<int>vec={0,-1,0,1,0};
          for(int k=1;k<5;k++){
            int new_row=i+vec[k-1];
            int new_col=j+vec[k];
            if(isvalid(new_row,new_col,n,m)&&(matrix[new_row][new_col]>matrix[i][j])){
                   ans=max(ans,dfs(new_row,new_col,matrix,dp));
            }
          }
          return dp[i][j]=1+ans;
    }
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        int n=matrix.size();
        int m=matrix[0].size();
        queue<pair<int,int>>q;
        vector<vector<int>>dp(n,vector<int>(m,-1));
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                vector<int>vec={0,-1,0,1,0};
                bool flag=true;
                for(int k=1;k<5;k++){
                    int new_row=i+vec[k-1];
                    int new_col=j+vec[k];
                    if(isvalid(new_row,new_col,n,m)){
                        if(matrix[new_row][new_col]<matrix[i][j]){
                             flag=false;
                        }
                    }
                }
                if(flag){
                    q.push({i,j});
                }
            }
        }
        int ans=0;
        while(!q.empty()){
            pair<int,int>p=q.front();
            ans=max(ans,dfs(p.first,p.second,matrix,dp));
            q.pop();
        }
        return ans;
        
    }
};
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