Playground: Shortest Bridge Between Islands

Clase 32 de 52 • Curso de Algoritmos Avanzados: Grafos y Árboles

John Eduard Meneses González

student•

My Slution!!

Carlos Hernandez

student•

Mi solución:

import math

def check_spot(islands, i, j):
    for index, islands_spots in islands.items():
        if (i - 1, j) in islands_spots or (i, j-1) in islands_spots or (i + 1, j) in islands_spots or (i, j + 1) in islands_spots:
            return index
    return -1

def get_islands(input_map):
    islands = {}

    for i, row in enumerate(input_map):
        for j, column in enumerate(row):
            if column == "0":
                continue
            else:
                # Check contiguous spots
                found_index = check_spot(islands, i, j)
                # Clock-wise order starting from left
                if found_index == -1:
                    next_index = len(islands.items())
                    islands[next_index] = [(i, j)]
                else:
                    islands[found_index].append((i, j))
    return islands

def generate_next_steps(islands, i, j, destination):
    next_steps = []

    if (i - 1) >= 0 and islands[i-1][j] == "0" or (i-1,j) == destination:
        next_steps.append((i-1, j))
    if (j + 1) < len(islands[i]) and islands[i][j+1] == "0" or (i,j+1) == destination:
        next_steps.append((i, j+1))
    if (i + 1) < len(islands) and islands[i+1][j] == "0" or (i+1,j) == destination:
        next_steps.append((i+1, j))
    if (j - 1) >= 0 and islands[i][j-1] == "0" or (i,j-1) == destination:
        next_steps.append((i, j-1))
    
    return next_steps


def get_shortest_bridge_between_islands(input_map):
    islands = get_islands(input_map)
    
    shortest_path_data = {
        "spot_1" : None,
        "spot_2" : None,
        "distance": 1000000
    }

    for i1, j1 in islands[0]:
        for i2, j2 in islands[1]:
            spots_distance = math.dist([i1, j1], [i2, j2])
            if spots_distance < shortest_path_data["distance"]:
                shortest_path_data['spot_1'] = (i1, j1)
                shortest_path_data['spot_2'] = (i2, j2)
                shortest_path_data['distance'] = spots_distance
    
    origin, destination = shortest_path_data["spot_1"], shortest_path_data["spot_2"]
    print(origin, destination)

    bridge_amount = -1
    steps = [origin]


    while steps:
        if destination in steps: 
            break
        for _ in range(len(steps)):
            current_step = steps.pop(0)
            i, j = current_step
            next_steps = generate_next_steps(input_map, i, j, destination)
            for next_step in next_steps:
                if not next_step in steps:
                    steps.append(next_step)
        bridge_amount += 1
        print(steps)
        
    return bridge_amount
    


input_map = [
    ["1","1","1","1","1"],
    ["1","0","0","0","1"],
    ["1","0","1","0","1"],
    ["1","0","0","0","1"],
    ["1","1","1","1","1"],
]
response = get_shortest_bridge_between_islands(input_map)
print(response)

Miguel Angel Reyes Moreno

student•

comparto una solución con algunos comentarios:

from typing import List

from collections import deque



class Solution:

&#x20;   def shortestBridge(self, grid: List\[List\[int]]) -> int:

&#x20;       \# Step 1: Find the first island and mark it, adding boundary cells to the queue

&#x20;       found = False

&#x20;       queue = deque()  # Boundary of the first island for expansion

&#x20;       for i in range(len(grid)):

&#x20;           for j in range(len(grid\[i])):

&#x20;               if grid\[i]\[j] == 1:

&#x20;                   self.mark\_first\_island(grid, i, j, queue)  # Mark first island and add boundaries to queue

&#x20;                   found = True

&#x20;                   break

&#x20;           if found:

&#x20;               break



&#x20;       \# Step 2: Expand BFS from the first island to reach the second island

&#x20;       return self.expand(grid, queue)



&#x20;   def mark\_first\_island(self, grid: List\[List\[int]], x: int, y: int, queue: deque):

&#x20;       directions = \[(1, 0), (-1, 0), (0, 1), (0, -1)]

&#x20;       grid\[x]\[y] = 2  # Mark the island as visited

&#x20;       queue\_for\_marking = deque(\[(x, y)])



&#x20;       while queue\_for\_marking:

&#x20;           x, y = queue\_for\_marking.popleft()

&#x20;           for direction in directions:

&#x20;               new\_x, new\_y = x + direction\[0], y + direction\[1]

&#x20;               if 0 <= new\_x < len(grid) and 0 <= new\_y < len(grid\[0]):

&#x20;                   if grid\[new\_x]\[new\_y] == 1:

&#x20;                       grid\[new\_x]\[new\_y] = 2  # Mark connected land

&#x20;                       queue\_for\_marking.append((new\_x, new\_y))

&#x20;                   elif grid\[new\_x]\[new\_y] == 0:

&#x20;                       \# Add boundary water cells to queue for BFS expansion

&#x20;                       queue.append((x, y))



&#x20;   def expand(self, grid: List\[List\[int]], queue: deque) -> int:

&#x20;       directions = \[(1, 0), (-1, 0), (0, 1), (0, -1)]

&#x20;       steps = 0



&#x20;       \# Start BFS to expand towards the second island

&#x20;       while queue:

&#x20;           for \_ in range(len(queue)):

&#x20;               x, y = queue.popleft()

&#x20;               for direction in directions:

&#x20;                   new\_x, new\_y = x + direction\[0], y + direction\[1]

&#x20;                   if 0 <= new\_x < len(grid) and 0 <= new\_y < len(grid\[0]):

&#x20;                       if grid\[new\_x]\[new\_y] == 1:  # Found the second island

&#x20;                           return steps

&#x20;                       if grid\[new\_x]\[new\_y] == 0:  # Expand to water, turn it into land

&#x20;                           grid\[new\_x]\[new\_y] = 2

&#x20;                           queue.append((new\_x, new\_y))

&#x20;           steps += 1  # Increase the step count as we expand one layer outwards



&#x20;       return -1  # Should never reach here if there are two islands

import math

def check_spot(islands, i, j):
    for index, islands_spots in islands.items():
        if (i - 1, j) in islands_spots or (i, j-1) in islands_spots or (i + 1, j) in islands_spots or (i, j + 1) in islands_spots:
            return index
    return -1

def get_islands(input_map):
    islands = {}

    for i, row in enumerate(input_map):
        for j, column in enumerate(row):
            if column == "0":
                continue
            else:
                # Check contiguous spots
                found_index = check_spot(islands, i, j)
                # Clock-wise order starting from left
                if found_index == -1:
                    next_index = len(islands.items())
                    islands[next_index] = [(i, j)]
                else:
                    islands[found_index].append((i, j))
    return islands

def generate_next_steps(islands, i, j, destination):
    next_steps = []

    if (i - 1) >= 0 and islands[i-1][j] == "0" or (i-1,j) == destination:
        next_steps.append((i-1, j))
    if (j + 1) < len(islands[i]) and islands[i][j+1] == "0" or (i,j+1) == destination:
        next_steps.append((i, j+1))
    if (i + 1) < len(islands) and islands[i+1][j] == "0" or (i+1,j) == destination:
        next_steps.append((i+1, j))
    if (j - 1) >= 0 and islands[i][j-1] == "0" or (i,j-1) == destination:
        next_steps.append((i, j-1))
    
    return next_steps


def get_shortest_bridge_between_islands(input_map):
    islands = get_islands(input_map)
    
    shortest_path_data = {
        "spot_1" : None,
        "spot_2" : None,
        "distance": 1000000
    }

    for i1, j1 in islands[0]:
        for i2, j2 in islands[1]:
            spots_distance = math.dist([i1, j1], [i2, j2])
            if spots_distance < shortest_path_data["distance"]:
                shortest_path_data['spot_1'] = (i1, j1)
                shortest_path_data['spot_2'] = (i2, j2)
                shortest_path_data['distance'] = spots_distance
    
    origin, destination = shortest_path_data["spot_1"], shortest_path_data["spot_2"]
    print(origin, destination)

    bridge_amount = -1
    steps = [origin]


    while steps:
        if destination in steps: 
            break
        for _ in range(len(steps)):
            current_step = steps.pop(0)
            i, j = current_step
            next_steps = generate_next_steps(input_map, i, j, destination)
            for next_step in next_steps:
                if not next_step in steps:
                    steps.append(next_step)
        bridge_amount += 1
        print(steps)
        
    return bridge_amount
    


input_map = [
    ["1","1","1","1","1"],
    ["1","0","0","0","1"],
    ["1","0","1","0","1"],
    ["1","0","0","0","1"],
    ["1","1","1","1","1"],
]
response = get_shortest_bridge_between_islands(input_map)
print(response)

from typing import List

from collections import deque

class Solution:

&#x20;   def shortestBridge(self, grid: List\[List\[int]]) -> int:

&#x20;       \# Step 1: Find the first island and mark it, adding boundary cells to the queue

&#x20;       found = False

&#x20;       queue = deque()  # Boundary of the first island for expansion

&#x20;       for i in range(len(grid)):

&#x20;           for j in range(len(grid\[i])):

&#x20;               if grid\[i]\[j] == 1:

&#x20;                   self.mark\_first\_island(grid, i, j, queue)  # Mark first island and add boundaries to queue

&#x20;                   found = True

&#x20;                   break

&#x20;           if found:

&#x20;               break

&#x20;       \# Step 2: Expand BFS from the first island to reach the second island

&#x20;       return self.expand(grid, queue)

&#x20;   def mark\_first\_island(self, grid: List\[List\[int]], x: int, y: int, queue: deque):

&#x20;       directions = \[(1, 0), (-1, 0), (0, 1), (0, -1)]

&#x20;       grid\[x]\[y] = 2  # Mark the island as visited

&#x20;       queue\_for\_marking = deque(\[(x, y)])

&#x20;       while queue\_for\_marking:

&#x20;           x, y = queue\_for\_marking.popleft()

&#x20;           for direction in directions:

&#x20;               new\_x, new\_y = x + direction\[0], y + direction\[1]

&#x20;               if 0 <= new\_x < len(grid) and 0 <= new\_y < len(grid\[0]):

&#x20;                   if grid\[new\_x]\[new\_y] == 1:

&#x20;                       grid\[new\_x]\[new\_y] = 2  # Mark connected land

&#x20;                       queue\_for\_marking.append((new\_x, new\_y))

&#x20;                   elif grid\[new\_x]\[new\_y] == 0:

&#x20;                       \# Add boundary water cells to queue for BFS expansion

&#x20;                       queue.append((x, y))

&#x20;   def expand(self, grid: List\[List\[int]], queue: deque) -> int:

&#x20;       directions = \[(1, 0), (-1, 0), (0, 1), (0, -1)]

&#x20;       steps = 0

&#x20;       \# Start BFS to expand towards the second island

&#x20;       while queue:

&#x20;           for \_ in range(len(queue)):

&#x20;               x, y = queue.popleft()

&#x20;               for direction in directions:

&#x20;                   new\_x, new\_y = x + direction\[0], y + direction\[1]

&#x20;                   if 0 <= new\_x < len(grid) and 0 <= new\_y < len(grid\[0]):

&#x20;                       if grid\[new\_x]\[new\_y] == 1:  # Found the second island

&#x20;                           return steps

&#x20;                       if grid\[new\_x]\[new\_y] == 0:  # Expand to water, turn it into land

&#x20;                           grid\[new\_x]\[new\_y] = 2

&#x20;                           queue.append((new\_x, new\_y))

&#x20;           steps += 1  # Increase the step count as we expand one layer outwards

&#x20;       return -1  # Should never reach here if there are two islands

Playground: Shortest Bridge Between Islands

Introducción

Grafos y Árboles: Estructuras de Datos Avanzadas

Estructuras de Datos: Introducción a Árboles y Sus Propiedades

Recursión: Concepto y Aplicaciones Prácticas con Ejemplos

Aplicaciones Prácticas de Grafos en Tecnología e Industria

Representación de Grafos: Matriz y Lista de Adyacencia

DFS

Búsqueda en Profundidad (DFS) en Árboles y Grafos

Implementación de DFS recursivo para búsqueda en árboles

Búsqueda en Profundidad (DFS) para Grafos: Enfoque Iterativo y Recursivo

Recorridos y Profundidad en Árboles Binarios y Enearios

Suma de Caminos en Árboles Binarios

Suma de Números de Raíz a Hoja en Árboles

Playground: Sum Root to Leaf Numbers

Implementación de Algoritmo DFS en Árboles Binarios con Golang

Resolución del Problema de Número de Islas con DFS

Conteo de Islas en Matrices con DFS

Playground: Number of Islands

Implementación de "Número de Islas" con Recursión en Python

Ejercicios Prácticos de Búsqueda en Profundidad (DFS)

Algoritmos de Búsqueda en Profundidad (DFS) en Problemas Comunes

BFS

Algoritmo BFS: Recorrido en Anchura de Grafos y Árboles

Implementación de BFS en Árboles usando Python

Movimiento mínimo de caballo en ajedrez infinito

Resolviendo el Problema Mínimo de Movimiento del Caballo en Ajedrez

Playground: Minimum Knights Moves

Resolución de Problemas de Caballos de Ajedrez con BFS en Python

Propagación de Plagas en Cultivos: Cálculo de Días para Contagio Total

Resolución de Rotting Oranges usando BFS

Playground: Rotting Oranges

Propagación de Plagas en Matrices usando BFS en Java

Construcción de Puentes Cortos entre Islas en Matrices Binarias

Resolución del Problema Shortest Bridge con DFS y BFS

Playground: Shortest Bridge Between Islands

Búsqueda del camino más corto entre islas usando BFS en Python

Búsqueda en anchura: Ejercicios prácticos y aplicaciones

Ejercicios avanzados de búsqueda en anchura (BFS) en programación

Backtrack

Algoritmo Backtracking: Solución de Problemas Complejos

Combinaciones de Letras en Números Telefónicos

Combinaciones de Letras a partir de un Número de Teléfono

Generación de combinaciones de letras con teclados numéricos en C++

Playground: Letter Combinations of a Phone Number

Generación de Direcciones IP Válidas a partir de Cadenas Numéricas

Generación de IPs válidas con backtracking en C++

Playground: Restore IP Addresses

Búsqueda de Palabras en Matrices: Solución y Complejidad

Búsqueda de Palabras en Matrices usando Backtracking y DFS

Playgrund: Word Search

Implementación de búsqueda de palabras en matrices con DFS en JavaScript

Resolución del problema de las n reinas en ajedrez

Ejercicios de Backtracking: Combinaciones y Permutaciones

Combinaciones y Permutaciones con Backtracking

Próximos pasos

Algoritmos de Grafos: MIN/MAX-HIP, TRI, Topological Sort y Dijkstra

Algoritmos y Estructuras de Datos en la Ingeniería