Longest Increasing Path in a matrix

'''
In this code, we maintain a 2D DP table visited where visited[i][j] stores the length of the longest increasing path starting from cell (i, j). For each cell, if it hasn’t been computed yet, we use DFS (fun) to explore all 4 directions and update visited[i][j] by taking the maximum of 1 + path from a valid increasing neighbor. The initial value of visited[i][j] is set to 1 to count the cell itself. Passing row-1 and col-1 ensures boundary checks in is_feasible work correctly. After processing all cells, the answer is the maximum value in visited, representing the longest increasing path in the matrix.
'''

def is_feasible(new_i,new_j,row,col,arr,i,j):
    if new_i>=0 and new_i<=row and new_j>=0 and new_j<=col:
        return True
    return False
def fun(i,j,row,col,arr,visited):
    if visited[i][j]:
        return visited[i][j]
    visited[i][j]=1 #atleast length is gonna be 1

    x=[-1,0,1,0]
    y=[0,1,0,-1]

    for k in range(4):
        new_i=i+x[k]
        new_j=j+y[k]

        if is_feasible(new_i,new_j,row,col,arr,i,j) and arr[new_i][new_j]>arr[i][j]:
            visited[i][j]=max(visited[i][j],1+ fun(new_i,new_j,row,col,arr,visited))
    return visited[i][j]

class Solution:
    def longestIncreasingPath(self, arr: List[List[int]]) -> int:
        row=len(arr)
        col=len(arr[0])
        visited=[[0]*col for i in range(row)]
        for i in range(row):
            for j in range(col):
                if visited[i][j]==0:
                    fun(i,j,row-1,col-1,arr,visited)
        return max(max(row) for row in visited)