MAKING A LARGE ISLAND | LEETCODE # 827 | PYTHON SOLUTION

[@thezendawg]

The first example isn't an area of 2 though right b/c 4-directionally doesn't mean diagonal?

[@rsKayiira]

Excellent explanation. Only question is in second for loop of both double for loops why use len(grid[m]) when len(grid[0]) also works?

[@crackfaang]

Doesn’t matter, just personal preference. Both are fine

[@rsKayiira]

@@crackfaang thank you

[@saadkhan4049]

In the problem statement it says 4-directional, why do you keep counting the diagonals to be a part of the island?

[@crackfaang]

Probably just a mistake on my part. Update the code to use it without it, should be the same

[@syafzal273]

Very well explained, I made a lot of mistakes while coding this up!

[@crackfaang]

Yea it's a tricky one for sure

[@WaldoTheWombat]

This is my version: class Solution: def calculate_potential_connection_size(self, i, j,): indices_of_starting_cells = set() # up if i-1 > -1 and self.grid[i-1][j] != 0: # can be switched with checking if it's a tupple indices_of_starting_cells.add( self.grid[i-1][j]) # right if j+1 < len(self.grid[i]) and self.grid[i][j+1] != 0: # can be switched with checking if it's a tupple indices_of_starting_cells.add( self.grid[i][j+1]) # down if i+1 < len(self.grid) and self.grid[i+1][j] != 0: # can be switched with checking if it's a tupple indices_of_starting_cells.add( self.grid[i+1][j]) # left if j-1 > -1 and self.grid[i][j-1] != 0: # can be switched with checking if it's a tupple indices_of_starting_cells.add( self.grid[i][j-1]) return sum( [self.starting_cell_indices_to_island_size[pair] for pair in indices_of_starting_cells] ) + 1 # because of 0 the we would change to a 1 def get_island_size(self, i, j, start_i, start_j): size = 0 if self.grid[i][j] == 1: size += 1 self.grid[i][j] = (start_i,start_j) if i-1 > -1: # up size += self.get_island_size(i-1, j, start_i, start_j) if j+1 < len(self.grid[i]): # right size +=self.get_island_size(i, j+1, start_i, start_j) if i+1 < len(self.grid): # down size +=self.get_island_size(i+1, j, start_i, start_j) if j-1 > -1: # left size += self.get_island_size(i, j-1, start_i, start_j) else: # self.grid[i][j] must be 0 self.surrounding_zeroes.add((i,j)) return size def largestIsland(self, grid: List[List[int]]) -> int: self.grid = grid # print(f"Before : ") # self.print_grid() self.surrounding_zeroes = set() self.starting_cell_indices_to_island_size = dict() max_area = 1 # because of 0 the we would change to for i in range(len(self.grid)): for j in range(len(self.grid[i])): if self.grid[i][j] == 1: self.starting_cell_indices_to_island_size[i, j] = self.get_island_size(i,j, start_i=i, start_j=j) # print(f"{self.starting_cell_indices_to_island_size[i, j] = }") max_area = max([max_area, self.starting_cell_indices_to_island_size[i, j]]) # print(f"After : ") # self.print_grid() for i, j in self.surrounding_zeroes: if self.grid[i][j] == 0: # print(f"the 0 at {i}, {j} = {self.calculate_potential_connection_size(i, j)}") max_area = max([max_area, self.calculate_potential_connection_size(i, j)]) return max_area

[@nikethdonthula2123]

Explained very well 🔥🔥

[@SatinderSingh71]

Can you post the solution to the code?

[@mohammadkareem1187]

keep it up bro, great job!

[@nithinsastrytellapuri291]

very good explanation. Thank you

[@trueinviso1]

Great walk through, thanks!

[@AmolGautam]

Thank you very much , this is quite helpful.

[@strawberriesandcream2863]

great explanation and solution, thanks :)

[@Ankit-hs9nb]

Please cover this one: 249. Group Shifted Strings thanks!:)

[@crackfaang]

Only if you subscribe to the channel 😉

[@Ankit-hs9nb]

Already subscribed a month ago and notification bell also turned on!😄