I think your explanations are already on par with NeetCode. However, NeetCode's diagramming tools are more professional. Your advantage is that you cover more topics.
Waiting for the M5 iPad Pro before getting one to do the drawings on in the future. Can't bring myself to buy one now since there is a refresh just around the corner haha
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
![LOWEST COMMON ANCESTOR OF A BINARY TREE III [PYTHON]](/img/n.gif)
The first example isn't an area of 2 though right b/c 4-directionally doesn't mean diagonal?
I lost my mind doing this problem, but I finally got it, thanks
In the problem statement it says 4-directional, why do you keep counting the diagonals to be a part of the island?
Probably just a mistake on my part. Update the code to use it without it, should be the same
very good explanation. Thank you
keep it up bro, great job!
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?
Doesn’t matter, just personal preference. Both are fine
@@crackfaang thank you
Very well explained, I made a lot of mistakes while coding this up!
Yea it's a tricky one for sure
Explained very well 🔥🔥
Great walk through, thanks!
great explanation and solution, thanks :)
Can you post the solution to the code?
Thank you very much , this is quite helpful.
I think your explanations are already on par with NeetCode. However, NeetCode's diagramming tools are more professional. Your advantage is that you cover more topics.
Waiting for the M5 iPad Pro before getting one to do the drawings on in the future. Can't bring myself to buy one now since there is a refresh just around the corner haha
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
Please cover this one:
249. Group Shifted Strings
thanks!:)
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