Shortest Bridge

The problem statement requires us to connect two islands with the minimum number of flips from 0 to 1. We can achieve this by performing a Breadth-First Search (BFS) to explore the grid and find the shortest path to connect the two islands.

Here’s the step-by-step process:

Find the First Island: Use Depth-First Search (DFS) to find the first island and mark all its cells with a different value (e.g., 2) to distinguish it from the second island.
Initialize BFS Queue: Add all the cells of the first island to the BFS queue.
BFS Traversal: Start a BFS from the cells of the first island, exploring the neighboring cells in 4 directions (up, down, left, right).
Connecting Islands: Continue BFS until you reach a cell that is part of the second island (value 1). Keep track of the number of steps taken (i.e., the number of 0’s flipped).
Return the Result: The number of steps taken to reach the second island will be the answer.

Here’s the code:

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from collections import deque

class Solution:
    def shortestBridge(self, grid: List[List[int]]) -> int:
        n = len(grid)
        directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]

        # DFS function to find the first island
        def dfs(x, y):
            grid[x][y] = 2
            queue.append((x, y))
            for dx, dy in directions:
                nx, ny = x + dx, y + dy
                if 0 <= nx < n and 0 <= ny < n and grid[nx][ny] == 1:
                    dfs(nx, ny)

        # Finding the first island
        queue = deque()
        found = False
        for i in range(n):
            for j in range(n):
                if grid[i][j] == 1:
                    dfs(i, j)
                    found = True
                    break
            if found:
                break

        # BFS to connect the islands
        steps = 0
        while queue:
            size = len(queue)
            for _ in range(size):
                x, y = queue.popleft()
                for dx, dy in directions:
                    nx, ny = x + dx, y + dy
                    if 0 <= nx < n and 0 <= ny < n:
                        if grid[nx][ny] == 1:
                            return steps
                        elif grid[nx][ny] == 0:
                            grid[nx][ny] = 2
                            queue.append((nx, ny))
            steps += 1

        return -1  # This line is not reachable as there are exactly two islands

Example: Input: grid = [[0,1,0],[0,0,0],[0,0,1]] Output: 2 Explanation: Flipping two 0’s will connect the islands.

Identifying Problem Isomorphism

“Shortest Bridge” can be approximately mapped to “Rotting Oranges”.

Reasoning:

Both problems require the application of a breadth-first search (BFS) algorithm to find the shortest path or time in a grid.

“Shortest Bridge” involves finding the shortest bridge (i.e., minimum number of adjacent steps) from one island to another in a 2D grid.
“Rotting Oranges” involves figuring out the minimum amount of time it takes for every orange in a 2D grid to become rotten, given that a rotten orange at a certain cell will rot adjacent (up, down, left, right) fresh orange cells in each time frame.

While the contexts and specific conditions of these two problems differ, they share a similar structure in solving the problem, which is to find the shortest path or time from certain cells (a.k.a., sources) to the rest of the cells.

Both problems are at a similar level of complexity as they both require breadth-first search traversal on the grid.

10 Prerequisite LeetCode Problems

“Shortest Bridge” involves concepts related to depth-first search (DFS) or breadth-first search (BFS), and understanding how to traverse through a grid. Here are 10 LeetCode problems of lesser complexity that can help you prepare for this problem:

Number of Islands (LeetCode 200): This problem can help you get familiar with the idea of traversing a grid and counting connected components using DFS or BFS.
Flood Fill (LeetCode 733): This problem introduces you to the concept of changing the color of an image, starting from a pixel, which can help you understand how to traverse a grid.
Surrounded Regions (LeetCode 130): This problem can help you practice DFS in a matrix, which is helpful in understanding how to traverse the grid in “Shortest Bridge”.
Word Search (LeetCode 79): This problem can help you understand how to search for a sequence in a 2D grid, a similar concept to searching for the bridge in “Shortest Bridge”.
Max Area of Island (LeetCode 695): This problem requires you to find the maximum area of an island in a 2D array, a useful concept for understanding “Shortest Bridge”.
Walls and Gates (LeetCode 286): This problem requires you to fill each empty room with the distance to its nearest gate, which involves BFS in a grid.
Rotting Oranges (LeetCode 994): This problem requires understanding of BFS in a grid where the state of cells can change over time.
Island Perimeter (LeetCode 463): This problem helps you understand how to calculate the perimeter of an island in a grid, which involves traversing the grid and checking the neighbors of each cell.
01 Matrix (LeetCode 542): This problem can help you understand how to update a grid based on certain conditions using BFS or DFS.
Friend Circles (LeetCode 547): This problem introduces you to the concept of finding connected components in a graph, which is relevant to the “Shortest Bridge” problem.

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class Solution:
    def shortestBridge(self, A: List[List[int]]) -> int:
        m, n = len(A), len(A[0])
        i, j = next((i, j) for i in range(m) for j in range(n) if A[i][j])
        
        # dfs 
        stack = [(i, j)]
        seen = set(stack)
        while stack: 
            i, j = stack.pop()
            seen.add((i, j)) # mark as visited 
            for ii, jj in (i-1, j), (i, j-1), (i, j+1), (i+1, j): 
                if 0 <= ii < m and 0 <= jj < n and A[ii][jj] and (ii, jj) not in seen: 
                    stack.append((ii, jj))
                    seen.add((ii, jj))
        
        # bfs 
        ans = 0
        queue = list(seen)
        while queue:
            newq = []
            for i, j in queue: 
                for ii, jj in (i-1, j), (i, j-1), (i, j+1), (i+1, j): 
                    if 0 <= ii < m and 0 <= jj < n and (ii, jj) not in seen: 
                        if A[ii][jj] == 1: return ans 
                        newq.append((ii, jj))
                        seen.add((ii, jj))
            queue = newq
            ans += 1

Problem Classification

You are given an n x n binary matrix grid where 1 represents land and 0 represents water. An island is a 4-directionally connected group of 1’s not connected to any other 1’s. There are exactly two islands in grid. You may change 0’s to 1’s to connect the two islands to form one island. Return the smallest number of 0’s you must flip to connect the two islands.

Example 1:

Input: grid = [[0,1],[1,0]] Output: 1

Example 2:

Input: grid = [[0,1,0],[0,0,0],[0,0,1]] Output: 2

Example 3:

Input: grid = [[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]] Output: 1

Constraints:

n == grid.length == grid[i].length 2 <= n <= 100 grid[i][j] is either 0 or 1. There are exactly two islands in grid.

Language Agnostic Coding Drills

This code uses a combination of concepts, including list comprehension, nested loops, conditionals, tuples, breadth-first search (BFS), depth-first search (DFS), and operations on sets. Here is how you can break it down into smaller units:

Variable Initialization: Understanding how to declare variables and assign values to them.
Data Types (List, Tuples, and Sets): Learn about lists, tuples, and sets, and their operations such as append, pop, etc.
List Comprehension: Learn how to create lists in a compact way.
Loops: Understand how to use nested for-loops and while-loops to iterate over data.
Conditional Statements: Understand the usage of if-else conditions.
Tuples and Indexing: Learn how tuples work and how to access elements through indexing.
DFS and Recursion: Understand the concept of depth-first search and how recursion is used to implement it.
BFS and Queues: Understand the concept of breadth-first search and how queues are used to implement it.
Sets and Set Operations: Learn about sets and basic operations like add, pop, and the “in” keyword for checking membership.
Problem Solving and Algorithm Design: Learn how to design an algorithm that combines all these concepts to solve a complex problem. In this case, a combination of DFS and BFS is used to find the shortest bridge between islands represented in a 2D grid.

Each concept provides a foundation for the next, culminating in the ability to understand and implement a complex algorithm such as the one in the provided code.

Targeted Drills in Python

Variable Initialization:
1 2 3
a = 10 b = 20 print(a, b)

Data Types (List, Tuples, and Sets):

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my_list = [1, 2, 3, 4]
my_tuple = (5, 6, 7, 8)
my_set = {9, 10, 11, 12}

print(my_list, my_tuple, my_set)

List Comprehension:

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squares = [x**2 for x in range(10)]
print(squares)

Loops:
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for i in range(5): print(i)

Conditional Statements:

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x = 10
if x > 5:
    print('x is greater than 5')
else:
    print('x is less than or equal to 5')

Tuples and Indexing:

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my_tuple = (1, 2, 3, 4, 5)
print(my_tuple[2])  # prints 3

DFS and Recursion:

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def dfs(node, visited):
    if node not in visited:
        print(node)
        visited.add(node)
        for neighbor in node.neighbors:
            dfs(neighbor, visited)

BFS and Queues:

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from collections import deque

def bfs(start, visited):
    queue = deque([start])
    while queue:
        node = queue.popleft()
        if node not in visited:
            print(node)
            visited.add(node)
            queue.extend(node.neighbors for node in nodes if node not in visited)

Sets and Set Operations:

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s = set()
s.add(1)
s.add(2)
print(1 in s)  # prints True
s.remove(1)
print(1 in s)  # prints False

Problem Solving and Algorithm Design:

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# A simplified version of the final problem
def bfs(grid, start, goal):
    visited = set()
    queue = deque([start])
    while queue:
        x, y = queue.popleft()
        if (x, y) == goal:
            return True
        for dx, dy in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
            nx, ny = x + dx, y + dy
            if 0 <= nx < len(grid) and 0 <= ny < len(grid[0]) and grid[nx][ny] == 1 and (nx, ny) not in visited:
                visited.add((nx, ny))
                queue.append((nx, ny))
    return False