Making A Large Island

Identifying Problem Isomorphism

“Making A Large Island” involves a 2D grid where you have to identify a 0 that, if changed to 1, would result in the largest possible contiguous island of 1s. This problem involves graph traversal and understanding of connected components.

A problem that follows a similar strategy is “Surrounded Regions”. In “Surrounded Regions”, we are also given a 2D grid and the task is to find ‘O’s that are not surrounded by ‘X’s. Both problems require us to understand how changing one element in the grid affects the surrounding area.

They are approximately isomorphic because they require different approaches to solve them. “Making A Large Island” is a more difficult problem because it requires not only graph traversal but also careful consideration of which ‘0’ to change to ‘1’ to form the largest island, whereas “Surrounded Regions” mainly requires identifying and marking the regions that are not completely surrounded by ‘X’s.

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from typing import List
from collections import defaultdict

class Solution:
    def largestIsland(self, grid: List[List[int]]) -> int:
        def dfs(r, c, index):
            if 0 <= r < len(grid) and 0 <= c < len(grid) and grid[r][c] == 1:
                grid[r][c] = index
                return 1 + dfs(r+1, c, index) + dfs(r-1, c, index) + dfs(r, c+1, index) + dfs(r, c-1, index)
            return 0

        areas = defaultdict(int)
        index = 2  # Start from 2 as grid already contains 0s and 1s
        for r in range(len(grid)):
            for c in range(len(grid)):
                if grid[r][c] == 1:
                    areas[index] = dfs(r, c, index)
                    index += 1

        res = max(areas.values() or [0])
        for r in range(len(grid)):
            for c in range(len(grid)):
                if grid[r][c] == 0:
                    possible = {grid[i][j] for i, j in ((r+1, c), (r-1, c), (r, c+1), (r, c-1)) if 0 <= i < len(grid) and 0 <= j < len(grid)}
                    res = max(res, 1 + sum(areas[i] for i in possible))

        return res

Problem Classification

This problem falls within the domain Graphs and connected components.

‘What’ Components:

  1. A binary 2D grid representing an island system, where 1s represent land and 0s represent water.
  2. An operation which allows changing at most one 0 (water cell) to 1 (land cell).
  3. The requirement to find the size of the largest island (maximum connected component of 1s) after potentially performing the operation.

The problem can be further classified as a Graph Traversal problem, with the twist of a single operation that can modify the graph. The problem can be solved by exploring all possible islands (connected components), measuring their sizes, and considering the potential size increase that could be achieved by converting a single water cell to land.

It falls under the categories of ‘Depth-First Search’ or ‘Breadth-First Search’ in Graph theory, as well as ‘Connected Components’ in Graph algorithms.

The problem requires knowledge of graph traversal algorithms to navigate through the 2D grid and identify connected components (islands). It requires the ability to identify adjacent cells in a 2D grid and understanding how changing a 0 to 1 might affect the connectedness of components in the grid. The complexity lies in efficiently exploring all possibilities and tracking the maximum possible island size. It’s also a dynamic problem in that the optimal solution may involve changing different 0s to 1s depending on the initial configuration of the grid.

Constraints

Given the problem statement and constraints, here are some characteristics or conditions that can be exploited:

  1. The Grid Size: The maximum size of the grid is 500x500. Although this is not small, it is still manageable for many computational tasks. Depth-first search (DFS) or breadth-first search (BFS) algorithms can be used to traverse the grid within acceptable time limits.

  2. Binary Grid: The grid only contains 0s and 1s. This simplicity can be used to our advantage. We can use DFS or BFS to identify islands (clusters of 1s) and also to identify 0s that are adjacent to these islands.

  3. Island Definition: The problem defines an island as a 4-directionally connected group of 1s. This significantly simplifies the problem compared to if the islands could also be diagonally connected. We only need to check up, down, left, and right from any given point to identify an island.

  4. Limited Changes: We can only change at most one 0 to a 1. This limits the number of potential changes we need to consider. If we first calculate the size of each island, we can then consider changing each 0 to a 1 and calculate the potential size of the new island that would form. We don’t need to consider multiple changes.

  5. The adjacency of 0s: A 0 that is surrounded by multiple distinct islands could potentially join those islands into a larger one. In contrast, a 0 that’s surrounded by a single island or by water will contribute less to the maximum possible island size. We can use this to guide our choice of which 0 to change to 1.

These conditions can guide the formation of an efficient algorithm. For example, one approach could be to first use DFS or BFS to identify all islands and calculate their sizes. We could also store which island each cell belongs to. Then we can iterate over all the 0s, and for each 0, calculate the potential size of the island that would be formed by changing that 0 to a 1. This calculation can be done quickly by adding up the sizes of the distinct islands adjacent to the 0. The largest such potential size is the solution to the problem.

Thought Process

Let’s take a step-by-step approach to solving this problem.

Cues from the problem statement include:

  1. The problem requires us to maximize the size of an island by converting a ‘0’ cell to a ‘1’ in a binary grid, hinting at an optimization problem.

  2. The statement specifies that islands are 4-directionally connected groups of 1s. This suggests we’ll be working with grid traversal algorithms, likely DFS or BFS.

  3. The constraints on grid size and values (n x n, 1 <= n <= 500, grid[i][j] is either 0 or 1) further inform our approach.

Given these cues, a reasonable approach seems to be identifying the current islands, then looking for the ‘0’ cell that, if converted to ‘1’, would join the largest total area of distinct islands.

Now let’s write Python code for this approach:

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from collections import defaultdict

def largestIsland(grid):
    def dfs(r, c, index):
        if 0 <= r < len(grid) and 0 <= c < len(grid) and grid[r][c] == 1:
            grid[r][c] = index
            return 1 + dfs(r+1, c, index) + dfs(r-1, c, index) + dfs(r, c+1, index) + dfs(r, c-1, index)
        return 0

    areas = defaultdict(int)
    index = 2  # Start from 2 as grid already contains 0s and 1s
    for r in range(len(grid)):
        for c in range(len(grid)):
            if grid[r][c] == 1:
                areas[index] = dfs(r, c, index)
                index += 1

    res = max(areas.values() or [0])
    for r in range(len(grid)):
        for c in range(len(grid)):
            if grid[r][c] == 0:
                possible = {grid[i][j] for i, j in ((r+1, c), (r-1, c), (r, c+1), (r, c-1)) if 0 <= i < len(grid) and 0 <= j < len(grid)}
                res = max(res, 1 + sum(areas[i] for i in possible))

    return res

The dfs() function uses Depth-First Search to find and label each island with a unique index while counting the area of each island. It then finds and stores these areas in a dictionary.

The second part of the function iterates through the grid to find ‘0’ cells, calculates the potential total area if the ‘0’ cell is converted to ‘1’, and keeps track of the maximum such area.

The res = max(res, 1 + sum(areas[i] for i in possible)) line is key here. For each ‘0’ cell, it finds the unique indices of its neighboring cells (possible), sums their areas, adds 1 for the ‘0’ cell itself, and updates res if this value is larger than the current res.

Solution Approach and Analysis

Let’s break down the approach into smaller steps:

  1. Labeling the Islands and Calculating Their Sizes: This is the first phase where we need to label each island with a unique identifier and calculate the size of each island. This can be achieved using Depth-First Search (DFS) or Breadth-First Search (BFS). Start by iterating through each cell in the grid. If the cell is a ‘1’ and it hasn’t been visited yet, perform a DFS or BFS starting from this cell. As part of the search, mark each visited cell with a unique identifier for the island and increment a count for the size of the current island. Continue the search until all cells in the current island have been visited. Then, move to the next unvisited ‘1’ cell in the grid and repeat the process with a new unique identifier.

  2. Checking Each ‘0’ Cell: Once all the islands have been labeled and their sizes calculated, we will proceed to the next phase where we examine each ‘0’ cell in the grid. For each ‘0’ cell, look at its neighboring cells. If a neighboring cell is part of an island (i.e., not a ‘0’), note down the identifier of that island.

  3. Calculating Potential Island Size: After identifying all unique neighboring islands for a ‘0’ cell, calculate the total size of these islands if they were to be connected by converting the ‘0’ to a ‘1’. Add ‘1’ to this sum to account for the ‘0’ cell itself being turned into a ‘1’.

  4. Finding the Maximum Island Size: Keep track of the maximum size calculated in step 3 as you iterate through all the ‘0’ cells. This maximum value will be the largest possible island size that can be achieved by changing a single ‘0’ to a ‘1’.

Now, let’s discuss how specific changes in the problem’s parameters would affect this solution:

  • Grid Size: Increasing the grid size will increase the runtime of the algorithm since we need to potentially visit every cell multiple times. However, the algorithm should still be able to handle the maximum grid size specified in the problem (500x500) within a reasonable time frame.

  • Number of Islands: Increasing the number of islands (i.e., more dispersed ‘1’s in the grid) can potentially increase the runtime slightly, as we’ll need to perform more DFS/BFS searches in the first step. However, it could also decrease the runtime of the second phase of the algorithm, as there will be fewer ‘0’ cells to consider changing to ‘1’.

  • Island Size: Larger islands will lead to fewer DFS/BFS searches in the first step, potentially speeding up this part of the algorithm. However, they could also lead to more ‘0’ cells to consider in the second phase, potentially slowing this part down.

Let’s take the first example from the problem statement to illustrate this process:

grid = [[1, 0], 
        [0, 1]]

In the first phase, we’ll label the islands and calculate their sizes. We’ll end up with:

grid = [[2, 0], 
        [0, 3]]

Here, ‘2’ and ‘3’ are just unique identifiers for the islands, and the sizes of the islands are {2: 1, 3: 1}.

In the second phase, we’ll look at each ‘0’ cell. Both ‘0’ cells have a single neighboring island, so turning either of them into a ‘1’ would result in an island of size 2. Thus, the maximum island size is 2.

Now, if we change the input grid to:

grid

= [[1, 0, 1], [0, 1, 0], [1, 0, 1]]

After the first phase, we’ll have:

grid = [[2, 0, 4], 
        [0, 3, 0],
        [5, 0, 6]]

with island sizes {2: 1, 3: 1, 4: 1, 5: 1, 6: 1}.

In the second phase, the ‘0’ cell at grid[1][1] has four neighboring islands. If we change this ‘0’ to a ‘1’, we’d get an island of size 1 + 1 + 1 + 1 + 1 = 5. Thus, the maximum island size is 5.

Language Agnostic Coding Drills

Here are the distinct coding concepts or drills present in the solution:

  1. Two-Dimensional Array Traversal: This is the most basic concept utilized in the code. It involves looping over all elements in a two-dimensional grid. This is generally the first step taken to analyze or manipulate any kind of 2D data structure.

  2. Depth-First Search (DFS): This is an algorithm for traversing or searching tree or graph data structures. It is used here to explore each island in the grid and calculate its area. This is more complex than simple array traversal, as it involves recursive calls and backtracking.

  3. Use of Dictionary for Mapping and Lookups: Dictionaries (or equivalent data structures like hash maps) are used to store and retrieve values based on unique keys. This is slightly more complex than simple array operations, as it involves understanding hashing and key-value mapping. Here, it is used to store the areas of each distinct island, mapped by their unique indices.

  4. Set Operations: In this problem, we’re using sets to gather unique indices of neighboring islands and to avoid summing the area of the same island more than once. This concept involves understanding the properties of set data structures, including automatic removal of duplicates and fast membership tests.

  5. Conditional Checks and Edge Cases Handling: This involves adding conditionals to the code to ensure we’re not exceeding the grid boundaries or accessing invalid elements. Understanding this requires a good grasp of if-else conditional statements and logical operators, along with understanding the problem constraints thoroughly.

  6. Using Custom Comparison for Priority Queue: Though not included in the given solution, some approaches to this problem might involve using a priority queue with custom comparison logic to sort elements based on a specific criterion. This is more advanced as it involves understanding sorting, comparison functions, and data structure customization.

Here is how these concepts contribute to the overall solution:

  • The problem starts with a 2D array traversal. We iterate over each cell in the grid to identify and process all “1s” (which represent land in the problem context).
  • For each land cell, we perform a DFS to find the entire island that this cell belongs to. In the process, we mark all cells of this island with a unique index and calculate the total area of the island.
  • We store the area of each distinct island in a dictionary, using the island index as the key. This will allow us to quickly look up the area of an island later on.
  • Once we’ve processed all the land cells, we do another 2D array traversal. This time, we’re looking for “0s” (which represent water).
  • For each water cell, we look at its neighboring cells. If changing this water cell to land would connect two or more islands, we calculate the total area of the new island that would form. We use a set to store the indices of the neighboring islands to avoid double-counting any island.
  • Finally, we return the maximum area we can achieve by changing at most one water cell to land. This involves keeping a running maximum of all the areas we’ve calculated.

Targeted Drills in Python

Let’s break this down:

  1. Two-Dimensional Array Traversal

This is a basic exercise of looping through all elements of a two-dimensional array (grid).

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def traverse_2d_array(grid):
    for i in range(len(grid)):
        for j in range(len(grid[0])):
            print(grid[i][j])

grid = [[1, 0], [0, 1]]
traverse_2d_array(grid)
  1. Depth-First Search (DFS)

DFS is a recursive algorithm for searching all the vertices of a graph or tree. It’s used here to explore each “island” in the grid.

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def dfs(grid, i, j):
    if i<0 or j<0 or i>=len(grid) or j>=len(grid[0]) or grid[i][j] != 1:
        return 0
    grid[i][j] = 0
    return 1 + dfs(grid, i+1, j) + dfs(grid, i-1, j) + dfs(grid, i, j+1) + dfs(grid, i, j-1)

grid = [[1, 1, 0], [0, 1, 0], [0, 0, 1]]
print(dfs(grid, 0, 0))  # Output: 3
  1. Use of Dictionary for Mapping and Lookups

Dictionaries in Python are used to store key-value pairs. Here it is used to map each distinct island to its area.

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def dict_usage():
    d = {'island1': 3, 'island2': 2, 'island3': 1}
    print(d['island1'])  # Output: 3

dict_usage()
  1. Set Operations

Sets are used to store multiple items in a single variable. Sets are unordered and unindexed, and they do not allow duplicate values.

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def set_usage():
    s = set([1, 2, 2, 3, 3, 3])
    print(s)  # Output: {1, 2, 3}

set_usage()
  1. Conditional Checks and Edge Cases Handling

This involves adding conditionals to the code to ensure we’re not exceeding the grid boundaries or accessing invalid elements.

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def check_boundaries(i, j, n):
    if i >= 0 and i < n and j >= 0 and j < n:
        return True
    return False

print(check_boundaries(1, 2, 2))  # Output: False

Integrating these coding drills into one comprehensive solution, the process would look like this:

  • Start by traversing the 2D grid to locate each island (group of connected 1s). For each island, run a DFS to explore its entire area and store the result in a dictionary.
  • Next, traverse the grid again, this time looking for water cells (0s). For each 0, gather the indices of neighboring land cells using a set to avoid duplicates.
  • Using the dictionary, calculate the potential total area if the water cell were turned into land. Keep track of the maximum area seen.
  • Finally, return the maximum area. If no 0s were found, return the area of the grid as all cells are land.

By practicing these separate drills and then integrating them together, one can build a deeper understanding of the coding concepts involved and how they contribute to the solution of the problem.

Q&A

I think the official implementation behind python3 is wrong. For this case:

[[1,0,1,0],[0,0,0,1],[1,0,1,0]]

The largest island should be 4. But when I run the code with custom test cases, the expected result is 3, which is wrong.

It seems there might be a misunderstanding about the problem statement. The problem allows you to change at most one ‘0’ to a ‘1’. It doesn’t necessarily guarantee that the change will connect all disjoint islands.

In your example:

[[1,0,1,0],
 [0,0,0,1],
 [1,0,1,0]]

If we change the ‘0’ at the location (1,1) to ‘1’, it does not connect all islands into a single one. It only creates a larger island with a size of 3 (there are 3 ‘1’s connected now). If we change the ‘0’ at (0,1), we get a single island with 4 ‘1’s, which is the correct solution.

Therefore, for each ‘0’ in the grid, we need to change it to a ‘1’ and then calculate the size of the island formed, choosing the maximum one as our answer.

If you are using a solution based on this approach and it’s returning 3 for this case, then it’s likely there’s a mistake in the solution. It could be incorrectly handling cases where changing a ‘0’ to a ‘1’ results in multiple islands becoming connected.

Grid should be n*n! according to question!

Yes, you’re correct. The problem statement specifies that the grid is an n x n binary matrix. This means the number of rows equals the number of columns. It’s a square grid where each cell contains either a 0 (representing water) or a 1 (representing land).

For example, a 3x3 grid might look like this:

[[1, 0, 1],
 [0, 1, 0],
 [1, 0, 1]]

Here, n is 3 because there are 3 rows and 3 columns.

In the context of the problem, when you’re looping through the grid to count islands or change a ‘0’ to ‘1’, you should keep in mind that the grid dimensions are n x n, and any indices used to access the grid’s cells should be in the range [0, n-1].