Number of Islands

Define Problem

1. What is the problem to be solved?

   • Find the number of islands in a given 2D binary grid.

2. What is the precondition? Where are you starting?

   • A 2D grid filled with ‘0’s (water) and ‘1’s (land).

3. What is the post condition? Where is your destination?

   • Count of distinct islands.

Define Step

• Traverse the grid cell by cell, marking visited land cells and expanding from them to find full islands.

Measure of Progress

• Number of cells visited or marked.

Define Loop Invariant

• At the start of each loop iteration, all visited land cells either form a complete island or are part of an island yet to be explored.

Main Steps

for each cell in grid:
  if cell is land and not visited:
    mark cell as visited
    initialize a new island
    explore this island (DFS/BFS)
    increment island count

Make Progress

• Each iteration either finds a new island or expands an existing one.

Maintain Loop Invariant

• The marking of cells ensures that each land cell is part of an identified island before the next loop iteration.

Establish the Loop Invariant

Initialize visited cells grid to all false
Initialize island count to 0

Exit Condition

• All cells in the grid have been visited.

Ending

• The exit condition ensures that all land has been explored, so the number of distinct islands is found.
• Return the island count.

return island_count

To find the number of islands in the given 2D grid, we can use Depth First Search (DFS). Here’s a step-by-step guide to implement the solution:

1. Loop through the Grid: Traverse each cell in the grid. If the cell has a value ‘1’, it’s part of an island, and we need to explore the island further.
2. Explore the Island (DFS): When we find a ‘1’, we’ll use DFS to explore all connected ‘1’s in the grid, marking them as visited along the way.
3. Count the Island: After exploring one island, increment the island count.
4. Continue Searching: Move to the next unvisited cell in the grid and repeat the process.

Here’s the code implementing this approach:

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class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        if not grid:
            return 0

        m, n = len(grid), len(grid[0])
        island_count = 0

        # DFS function to explore the island
        def dfs(i, j):
            if i < 0 or i >= m or j < 0 or j >= n or grid[i][j] == '0':
                return
            grid[i][j] = '0'  # Mark as visited
            dfs(i + 1, j)
            dfs(i - 1, j)
            dfs(i, j + 1)
            dfs(i, j - 1)

        # Loop through the grid
        for i in range(m):
            for j in range(n):
                if grid[i][j] == '1':
                    dfs(i, j)  # Explore the island
                    island_count += 1  # Increment the island count

        return island_count

Explanation:

• The dfs function explores the island by moving in all four cardinal directions and marks the visited land cells as ‘0’.
• The outer loops iterate through each cell in the grid, calling the dfs function when a ‘1’ is encountered.
• Each call to the dfs function explores an entire island, so the island count is incremented after each call.

The time complexity of this solution is (O(m \times n)), where (m) and (n) are the dimensions of the grid, and the space complexity is (O(m \times n)) in the worst case when the grid is filled with land cells.

Problem Classification

The problem falls under the domain of Graph Theory and 2D Array Manipulation. It involves traversing through a 2D grid and identifying connected components (islands).

1. 2D Binary Grid: The problem provides a 2D grid consisting of ‘1’s representing land and ‘0’s representing water.

2. Island Definition: An island is defined as adjacent ‘1’s (lands) that are connected either horizontally or vertically.

3. Grid Boundaries: All four edges of the grid are surrounded by water (‘0’).

4. Output: The goal is to find and return the total number of islands.

5. Constraints:

   • m = grid.length, n = grid[i].length
   • 1 <= m, n <= 300
   • grid[i][j] is either ‘0’ or ‘1’

This problem can be classified as a “Connected Components Identification” problem. We need to navigate through each cell in the grid to identify distinct islands, which are essentially connected components in a graph.

The traversal will need to happen both horizontally and vertically to meet the problem’s island definition, making it a graph-based search problem. Given the constraints, the search operation must also be efficient.

Clarification Questions

1. Edge Cases: Is it possible to have a grid with no islands at all (i.e., a grid filled only with ‘0’s)?

2. Diagonal Connections: Are diagonal connections between ‘1’s considered part of the same island, or only horizontal and vertical connections?

3. Grid Uniformity: Can we assume that all sub-arrays (rows) in the grid have the same length?

4. Input Format: Will the input grid always be a list of lists in Python, or could it be in some other format?

5. Output Format: Should the output be a simple integer representing the number of islands, or is any other format required?

6. Island Definition: To clarify, an island is any connected mass of ‘1’s that doesn’t have to take a particular shape, correct?

7. Memory Constraints: Are there any memory constraints we should be aware of when solving this problem?

8. Multiple Calls: Will the function be called multiple times with different grids, and if so, is there any state that needs to be maintained between calls?

9. Time Complexity: What is the maximum time complexity that is acceptable for solving this problem?

10. Empty Grid: What should be returned if the input grid is empty?

These questions help to clarify the scope and constraints of the problem, ensuring a more accurate and optimized solution.

10 Prerequisite LeetCode Problems

For the Number of Islands problem, the following is a good preparation:

1. Flood Fill - Helps you understand how to traverse and modify a 2D grid.
2. Surrounded Regions - Another grid traversal problem where you identify regions.
3. Max Area of Island - Direct extension of the current problem, focusing on area.
4. Walls and Gates - Teaches 2D grid traversal with BFS.
5. Word Search - Involves exploring paths within a 2D grid.
6. Pacific Atlantic Water Flow - Introduces the idea of water flow in a 2D grid.
7. Rotting Oranges - BFS on a grid with the added complexity of time.
8. Shortest Path in Binary Matrix - Adds path length as an additional criterion.
9. Minimum Path Sum - Traversing a 2D grid with attention to path cost.
10. Unique Paths - Counting paths from one corner to another, good for understanding traversal paths.

Each problem adds a layer of complexity or a new concept that will be useful when tackling Number of Islands.

Identifying Problem Isomorphism

“Number of Islands” can be approximately mapped to the “Connected Components in an Undirected Graph” problem.

Reasoning: Both problems involve finding connected components in a graph-like structure. In “Number of Islands,” the grid is implicitly treated as a graph where each cell is a node connected to its adjacent cells. In “Connected Components in an Undirected Graph,” you are given an explicit graph and need to find connected components, just like the islands.

Which is simpler: “Connected Components in an Undirected Graph” is generally simpler because the graph structure is given explicitly, making it more straightforward to apply graph traversal algorithms like DFS or BFS.

The mapping is approximate because the specific rules for connectivity differ between the two problems. In “Number of Islands,” connectivity is defined as horizontal or vertical adjacency in a 2D grid, while in the graph problem, connectivity is defined by edges in the graph.

Problem Analysis and Key Insights

Analyzing the Number of Islands problem, we can gather the following key insights:

1. Grid Representation: The problem uses a 2D grid where ‘1’s represent land and ‘0’s represent water.

2. Connectivity: Islands are formed by connecting adjacent lands horizontally or vertically, not diagonally.

3. Boundaries: The problem assumes that all four edges of the grid are surrounded by water, simplifying edge cases.

4. Size Constraints: The maximum dimensions of the grid are 300x300, indicating that a brute-force solution might not be efficient enough.

5. Output Type: The problem requires counting the number of islands, not detailing their properties or sizes.

6. Uniqueness of Islands: Islands are distinct entities separated by at least one ‘0’.

7. State Change: Once a land (‘1’) is counted as part of an island, it should not be recounted for another island.

8. Binary Values: The grid contains only binary values (‘0’ or ‘1’), making it simpler to process.

9. Search Space: The 2D grid search space is finite and well-defined.

These insights provide a strong basis for planning an algorithmic approach to solve the problem.

Problem Boundary

The scope of the Number of Islands problem is fairly contained. It involves the following:

1. Input Scope: You are given a 2D grid of dimensions m x n containing binary values (‘0’ or ‘1’).

2. Operational Scope: You need to scan the grid to identify and count islands. An island is formed by ‘1’s connected horizontally or vertically.

3. Constraints: The dimensions of the grid are bound between 1 and 300 for both m and n.

4. Output Scope: The output is a single integer representing the number of distinct islands in the grid.

5. Assumptions: The grid is surrounded by water on all sides, meaning you don’t have to handle edge cases where the island might extend beyond the grid.

6. Computational Scope: While the grid can be as large as 300x300, the problem statement suggests that a brute-force approach may not be efficient enough, hinting that the solution should ideally be more optimized.

7. Uniqueness: Each land (‘1’) cell can be part of only one island. Once accounted for, it should not be re-counted.

The scope is primarily computational and algorithmic, focused on efficient data traversal and state management within a 2D array.

To establish the boundary of the Number of Islands problem, consider the following aspects:

1. Input Boundary:

   • Minimum: A 1x1 grid
   • Maximum: A 300x300 grid
   • Values: Only ‘0’ or ‘1’

2. Output Boundary:

   • Minimum: 0 (if all cells are ‘0’)
   • Maximum: All cells are islands if they are ‘1’ and not connected (but this is an unlikely scenario based on the problem)

3. Operational Boundary:

   • You can only move horizontally or vertically to connect land (‘1’) cells.
   • The grid is surrounded by water (‘0’) on all sides.

4. Algorithmic Boundary:

   • You must traverse each cell at least once.
   • The algorithm must halt and produce an output.

5. Time Complexity: Ideally less than O(m * n), where m and n are the dimensions of the grid. A more optimized solution is desirable if possible.

6. Functional Boundary:

   • The function takes a 2D grid as input and returns an integer.
   • No external data storage or API calls.

7. Constraints:

   • The 2D grid is a fixed size defined at the time of input.
   • No mutations to the grid dimensions during the operation.

By defining these boundaries, you clarify the inputs, outputs, and operational limits, making it easier to focus on finding an optimized solution.

Distilling the Problem to Its Core Elements

Fundamental Concept

The fundamental concept here is Graph Traversal, particularly Depth-First Search (DFS) or Breadth-First Search (BFS). The grid can be considered as a 2D graph, where each cell is a node connected to its neighbors.

Simplest Description

Imagine a grid filled with water and islands. An island is made up of connected lands in a horizontal or vertical line. Your job is to count how many separate islands are there in the grid.

Core Problem

The core problem is to identify and count distinct groups of connected ‘1’s in a 2D grid, considering only horizontal and vertical connections.

Key Components

1. Grid Representation: A 2D array consisting of ‘1’s (land) and ‘0’s (water).
2. Island Definition: Groups of ‘1’s connected either horizontally or vertically.
3. Traversal: Visiting each cell in the grid at least once to identify islands.
4. Count: A mechanism to count each separate island only once.

Minimal Set of Operations

1. Initialize a counter for the number of islands.
2. Loop through each cell in the grid.
3. If a cell contains ‘1’, start a DFS or BFS to traverse the entire island and mark it as visited.
4. Increment the island counter.
5. Continue until all cells have been visited.

These operations encapsulate the essential steps needed to solve the problem.

Visual Model of the Problem

To visualize this problem, think of the 2D grid as a map where each cell is a piece of land (‘1’) or water (‘0’). You could literally draw this map on paper, where squares represent cells and different shades or markings differentiate between land and water.

Ways to Visualize:

1. Grid as Map: Draw a grid on a piece of paper. Fill in squares according to the ‘1’s and ‘0’s. Use one color for land and another for water.

2. Island Highlighting: Once the map is drawn, use a different color or pattern to shade in the groups of connected ‘1’s, each representing a distinct island.

3. Traversal Paths: If you want to go a step further, you can use arrows to indicate the order in which you’ll traverse the grid, either row-by-row or column-by-column.

4. Graph Representation: Convert the grid into a graph where each cell is a node. Add edges between adjacent ‘1’s to represent the “connections” that form islands. This could be a mental conversion rather than a physical drawing, but it can help to conceptualize the problem.

5. Counter Indicators: Place a number next to each identified island, effectively counting them as you go along.

6. Animation: If you’re digitally inclined, you could animate the DFS or BFS process, showing the traversal and discovery of new islands in real-time.

Visualizing the problem this way makes it easier to understand how to approach solving it, and what “finding an island” really means in terms of the 2D grid.

Problem Restatement

You’re given a 2D grid, like a map, filled with ‘1’s and ‘0’s. A ‘1’ stands for land, and a ‘0’ stands for water. Your task is to count the number of distinct islands on this map. An island is a group of adjacent ‘1’s, connected either horizontally or vertically. You can assume that the entire border of the grid is water.

Requirements:

• Count and return the number of separate islands.
• Two lands (‘1’s) are part of the same island if they are adjacent either horizontally or vertically, not diagonally.

Constraints:

• The grid has dimensions m x n.
• m and n are at least 1 and at most 300.
• Each cell contains only ‘0’ or ‘1’.

This exercise boils down to identifying and counting clusters of ‘1’s in a 2D array, where each cluster represents an island.

Abstract Representation of the Problem

In abstract terms, this problem can be described as finding and counting connected components in a 2D grid. The grid is made up of two types of elements: ‘A’ and ‘B’. A connected component consists of adjacent ‘A’ elements that are connected horizontally or vertically. The challenge is to identify and count these connected components.

Abstract Requirements:

• Identify and count all connected components made up of ‘A’ elements.
• Adjacency is defined as horizontal or vertical connectivity between elements.

Abstract Constraints:

• The grid has dimensions m x n.
• m and n are within a defined limit.
• Each cell in the grid contains only ‘A’ or ‘B’.

This abstract representation focuses on the underlying mathematical or computational concept of “connected components” within a grid, rather than the real-world notion of “islands” and “water”.

Terminology

List any specialized terms, jargon, or technical concepts that are crucial to understanding this problem or solution. Define them and explain their role within the context of this problem.

Problem Simplification and Explanation

Key Concepts:

1. 2D Grid: Think of this like a chessboard where each square is either land (‘1’) or water (‘0’).

2. Island: A collection of adjacent land squares (horizontally or vertically). An island is isolated when surrounded by water.

3. Adjacent: Two squares are adjacent if they share a side.

Interaction of Concepts:

You’re essentially navigating through the grid, moving from square to square, to find all the separate islands. Every time you land on a ‘1’, you mark it and check its neighbors to see if they are also ‘1’. If they are, they belong to the same island. You repeat this process until you’ve explored the whole grid.

Analogy:

Imagine the grid as a classroom with desks and chairs. Each desk represents a ‘1’ (land), and each empty chair represents a ‘0’ (water). An island would be a group of desks close enough that you can move from one to another without stepping on a chair. Your task is to walk around the classroom and count how many such groups (islands) of desks exist.

By breaking down the problem into these core concepts and understanding their interactions, you can approach the problem in a more structured manner.

Constraints

Let’s examine the characteristics and constraints of the problem that can be leveraged for an efficient solution:

1. Grid Size: The grid is limited to a 300x300 size, making it possible to exploit iterative or recursive techniques without overly exceeding computational limits.

2. Binary Grid: The grid contains only ‘0’s and ‘1’s. This simplifies the problem as we only have two states to consider: land or water.

3. Adjacent Land: Islands are formed by adjacent lands either horizontally or vertically but not diagonally. This restricts the directions we need to search in, reducing the complexity of the search.

4. Edges Are Water: The four edges of the grid are surrounded by water. This can eliminate the need for edge-case checking while traversing the grid, making our traversal more straightforward.

5. Constraints on m and n: Since the minimum value of m and n is 1, we don’t have to worry about empty grids.

6. Islands Are Isolated: Once you find a land cell (‘1’), you can traverse all its connected lands and mark them as visited. This helps reduce repeated work and makes it easier to count distinct islands.

7. Grid Mutability: The problem doesn’t state that we must keep the grid unchanged, so marking visited cells directly in the grid could be an option, thus saving extra memory for tracking visited cells.

Identifying these specific characteristics can guide us in crafting a more efficient algorithm to solve the problem.

Analyzing the constraints provides the following key insights:

1. Limited Grid Size: With a maximum grid size of 300x300, the problem is solvable using techniques that might have quadratic time complexity.

2. Binary Nature: The grid contains only ‘1’s and ‘0’s, making it straightforward to identify land and water.

3. Well-Defined Adjacency: Since adjacency is limited to horizontal and vertical directions, we don’t need to check diagonal neighbors, simplifying traversal.

4. Guaranteed Water Border: The edges being surrounded by water can simplify the traversal algorithm, eliminating the need for edge-case checking.

5. Non-Empty Grid: Minimum size constraints mean the grid won’t be empty, which simplifies initial checks.

6. In-place Operation: The absence of a constraint on grid mutability suggests we can change the grid in-place to mark visited cells, saving memory.

Understanding these constraints helps narrow down the types of algorithms and data structures that are likely to provide an efficient solution.

Case Analysis

Let’s explore additional examples to get a better understanding of the problem space.

Small Grids

1. Single Cell, Land

   • Input: ["1"]
   • Output: 1
   • Reasoning: A single land cell is an island by itself.

2. Single Cell, Water

   • Input: ["0"]
   • Output: 0
   • Reasoning: A single water cell doesn’t form an island.

Rectangular Grids

3. All Water

   • Input:

     ["0", "0", "0"],
     ["0", "0", "0"]
     

   • Output: 0
   • Reasoning: No land to form an island.

4. Single Row of Land

   • Input: ["1", "1", "1", "1"]
   • Output: 1
   • Reasoning: All adjacent land cells form a single island.

Disjoint Islands

5. Multiple Isolated Lands

   • Input:

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

   • Output: 4
   • Reasoning: Each ‘1’ is an isolated island.

6. Adjacent Islands

   • Input:

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

   • Output: 4
   • Reasoning: Top left and top right are separate islands, bottom left and bottom right are separate islands.

Complex Islands

7. L-Shaped Island

   • Input:

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

   • Output: 1
   • Reasoning: All ‘1’s are connected either horizontally or vertically.

8. Donut-Shaped Island

   • Input:

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

   • Output: 1
   • Reasoning: The hole in the middle doesn’t break the island into two.

Edge Cases

9. Maximum Size Grid, All Land

   • Input: 300x300 grid of ‘1’
   • Output: 1
   • Reasoning: All land is connected, forming a single island.

10. Maximum Size Grid, All Water

    • Input: 300x300 grid of ‘0’
    • Output: 0
    • Reasoning: No land to form an island.

By examining these cases, you should have a good understanding of how islands can be formed and what to watch for when solving the problem.

Visualizing these cases can be achieved by thinking of each grid as a map, where ‘1’ represents land and ‘0’ represents water. Imagine each cell as a square in a matrix, and adjacent cells are directly touching each other on their sides, not corners. Here’s how to visualize each of the cases:

Small Grids

1. Single Cell, Land

   • A single dot on a blank canvas, representing a tiny island.

2. Single Cell, Water

   • A blank canvas, no islands.

Rectangular Grids

3. All Water

   • Imagine a blank canvas, representing a sea with no land.

4. Single Row of Land

   • Think of it as a straight line, representing a long and narrow island.

Disjoint Islands

5. Multiple Isolated Lands

   • Picture four dots placed far apart from each other on a canvas, each representing a separate island.

6. Adjacent Islands

   • Imagine two groups of dots, each group closely packed, but the two groups are far apart. Each group represents an island.

Complex Islands

7. L-Shaped Island

   • Visualize an ‘L’ shape made out of dots, representing an L-shaped island.

8. Donut-Shaped Island

   • Imagine a ring or a donut shape, representing an island with a lake in the middle.

Edge Cases

9. Maximum Size Grid, All Land

   • Picture a large square canvas filled completely, representing a large island.

10. Maximum Size Grid, All Water

    • Again, a blank canvas indicating an ocean with no islands.

Each visualization should give you a mental image of what the “map” looks like and how the islands are formed based on the adjacency of land cells.

Analyzing different cases offers the following key insights:

1. Small Grids Matter: Single-cell grids can serve as base cases for a recursive or iterative algorithm.

2. Shape Irrelevant: The shape of the landmass doesn’t matter; it can be a line, L-shape, or more complex. It’s the connectivity that defines an island.

3. Isolation is Key: Isolated groups of ‘1’s form distinct islands. Connectivity can be horizontal or vertical but not diagonal.

4. Water as Boundaries: A single contiguous landmass surrounded entirely by water forms one island. Water acts as a natural boundary.

5. Empty Space: A grid with all ‘0’s has no islands. This could be another base case or an early exit condition for the algorithm.

6. Edge Cases: Handling maximum size grids efficiently is important for performance. You’ll need an algorithm that scales well.

7. Complex Shapes: Donut-shaped or other complex-shaped islands could potentially trip up an incorrectly designed algorithm. They emphasize the need for a thorough traversal of each island.

8. Adjacent Islands: Cases where islands are adjacent but not connected highlight the need for accurate tracking of visited land cells to avoid double-counting.

Each of these insights informs the design of an efficient algorithm by highlighting what to look for and what to avoid.

Identification of Applicable Theoretical Concepts

1. Graph Theory: The problem essentially asks us to find connected components in a graph. Each land cell (‘1’) is a node, and adjacent land cells are connected.

2. Depth-First Search (DFS): DFS is a well-suited algorithmic concept for traversing connected components in a graph.

3. Breadth-First Search (BFS): An alternative to DFS, BFS can also be applied to explore each island.

4. Dynamic Programming: The notion of memoization can be used to store already-visited cells to reduce redundant work.

5. Union-Find: This data structure can be used to keep track of connected components efficiently.

6. Matrix Operations: The grid can be considered as a binary matrix, which opens the door to matrix manipulation techniques.

7. Recursion: Recursive algorithms are particularly useful in situations where a problem can be broken down into similar sub-problems.

8. Spatial Complexity: Understanding 2D arrays is crucial as it’s the data structure you’ll be interacting with.

9. Boolean Logic: The grid cells can be toggled between ‘1’ and ‘0’, which can be exploited to mark visited cells.

10. Time Complexity Analysis: Given the problem’s constraints, understanding the time complexity will be critical for developing an efficient solution.

11. Queue Data Structure: Useful for BFS-based solutions.

Each of these concepts has methodologies or algorithms that can be directly applied to make solving the problem more efficient.

Simple Explanation

Imagine you have a map of a water body with islands. Each island is represented by sections marked as ‘1’, and water is marked as ‘0’. Your job is to figure out how many separate islands are on this map.

Think of it like a puzzle. If you can hop from one piece of land to an adjacent piece of land (left, right, up, or down), it’s the same island. Otherwise, it’s a different island.

Metaphor: Think about a classroom floor made of tiles. Some tiles are colored, and some are not. You start on a colored tile. You can move to the next tile if it’s also colored and is directly next to, or above or below the one you’re currently on. Count how many separate colored tile areas (islands) you can find by moving this way.

That’s the essence of the problem: counting these distinct ‘islands’ in the given map.

Problem Breakdown and Solution Methodology

Approach to Solving the Problem

1. Initialize a Counter: Start with a counter set to zero. This will keep track of the number of islands.

2. Scan the Grid: Traverse each cell in the grid one by one.

3. Land Check: When you hit a cell marked ‘1’, it indicates a piece of land that is part of an island.

   • Metaphor: It’s like stepping onto a colored tile on a classroom floor.

4. Start Exploration: If a ‘1’ is encountered, start an exploration process to find all the connected ‘1’s.

   • Metaphor: Imagine you’re at a node in a spider web. You’re trying to explore all paths from that node without lifting your foot off the web.

5. Mark as Visited: As you explore, change the value of the visited ‘1’ cells to another value (like ‘2’), so you don’t revisit them.

6. Island Counter: Each time you start a new exploration (Step 4), increment your island counter by 1.

7. Continue Scanning: Move on to the next cell in the grid and go back to Step 3.

8. Stop: Once you’ve looked at every cell, your counter will tell you the total number of distinct islands.

Parameters’ Effect

• Size of Grid: The larger the grid, the more computational time.

• Density of ‘1’s: The more ‘1’s there are, the longer it will take because of more exploration.

Example

Let’s consider the grid:

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

1. The first cell is ‘1’. Start exploration. The connected cells are all the ‘1’s in the top-left corner. Mark them as visited.

   • Counter = 1

2. Continue scanning. You hit another ‘1’ at (2,2). Another exploration starts.

   • Counter = 2

3. The last ‘1’ you hit is a single isolated cell.

   • Counter = 3

So, you have 3 islands in total.

That’s the full process, broken down.

Inference of Problem-Solving Approach from the Problem Statement

Key Terms and Concepts

1. 2D Binary Grid:

   • This informs us that we’ll be working with a two-dimensional array containing only ‘1’s and ‘0’s. A typical approach to navigating 2D grids involves nested loops.

2. m x n:

   • Defines the dimensions of the grid, guiding us on how to set up the loops for traversal.

3. ‘1’s (Land) and ‘0’s (Water):

   • These values help in identifying what to look for when scanning through the grid. Land (‘1’) is our point of interest.

4. Islands:

   • The core concept to understand. An island is defined as adjacent ‘1’s either horizontally or vertically. This guides us towards using depth/breadth-first search algorithms for connected components.

5. Return the number of islands:

   • Indicates that our final output is a single integer. We’d typically use a counter to keep track of this.

6. Adjacent Lands:

   • Implies we’ll be examining top, bottom, left, and right cells from any given ’land’ cell to explore the full extent of an island.

7. Surrounded by water:

   • Suggests that islands are fully enclosed by ‘0’s or the grid boundary, simplifying the island identification process.

8. Constraints:

   • Bounds the problem’s input sizes, informing us that an O(m*n) solution would be appropriate, as m, n <= 300.

Each of these terms helps in choosing algorithms, data structures, and also in optimizing the solution. For instance, the notion of “Islands” and “Adjacent Lands” directly points towards a graph traversal problem, which can be solved by DFS or BFS. The constraints tell us that our solution’s time complexity needs to be manageable within the given input size.

Recognizing Properties through Tables and Diagrams

1. 2D Binary Grid:

   • A simple table with rows and columns can be drawn. Fill it with ‘1’s and ‘0’s to represent the grid.

2. m x n:

   • In your table, specify the dimensions (m rows and n columns) to make sure the drawing fits the problem constraints.

3. ‘1’s (Land) and ‘0’s (Water):

   • Use different colors or shading to distinguish between ‘1’s and ‘0’s in the table.

4. Islands:

   • Circle or highlight groups of adjacent ‘1’s in the table to identify islands.

5. Return the number of islands:

   • You could place a unique label or number next to each island in your diagram for easy counting.

6. Adjacent Lands:

   • Draw arrows or lines connecting adjacent ‘1’s to show how you would move to explore an island.

7. Surrounded by Water:

   • Use another color or symbol to emphasize the ‘0’s that surround each group of ‘1’s, confirming them as islands.

8. Constraints:

   • You can note down the max dimensions (300 x 300) to remind yourself of the constraints.

A table filled with ‘1’s and ‘0’s can serve as your primary diagram. Additional markers, colors, or notations can be added to the table to highlight each key term or concept, aiding in visual recognition and problem-solving.

How did you infer from the problem statement that this problem can be solved using ?

Simple Explanation of the Proof

I’m having trouble understanding the proof of this algorithm. Could you explain it in a way that’s easy to understand?

Stepwise Refinement

1. High-Level Approach

   • Explore the 2D grid cell-by-cell.
   • When you find an unvisited land (‘1’), start a traversal to mark all the connecting lands.
   • Count the number of unique traversals. This count is the number of islands.

2. Granular, Actionable Steps

   1. Initialize Count and Visited Matrix

      • Initialize an island counter to zero.
      • Create a visited matrix with the same dimensions as the grid.

   2. Iterate Through Grid

      • Go through each cell in the grid.

   3. Check for Unvisited Land

      • If the cell is ‘1’ and has not been visited, increment the island counter.

   4. Land Traversal

      • When you find a ‘1’, traverse all its horizontally and vertically adjacent ‘1’s and mark them as visited.

   5. Update Visited Matrix

      • Mark all the visited cells in the visited matrix.

   6. Return Count

      • The island counter gives the number of islands.

3. Independent Parts

   • The initialization of the visited matrix and the island counter can happen independently.
   • Traversing each land (‘1’) to mark its adjacent lands is independent of other lands.

4. Repeatable Patterns

   • The pattern of checking a cell’s value and its visited status repeats for every cell.
   • The pattern of traversing adjacent lands is also repeatable. Once you find a land (‘1’), the logic to mark all its connecting lands is the same, regardless of its position.

Breaking down the problem in this manner helps you understand the individual components and how they interact, enabling a focused and efficient problem-solving approach.

Solution Approach and Analysis

Think of the grid as a map where you’re an explorer. Your goal is to find all the islands (‘1’s) and count them. An island is a landmass made up of adjacent lands, either horizontally or vertically. Let’s start exploring!

Smaller Steps

1. Prepare Your Tools

   • Initialize a counter (island_count) to zero.
   • Create a 2D matrix (visited) with the same dimensions as the grid to keep track of visited cells.

2. Start the Exploration

   • Loop through each cell in the grid.

3. Land Ahoy!

   • If you encounter an unvisited ‘1’, you’ve found a new island.
   • Increment island_count.

4. Explore the Island

   • Mark the current land as visited in the visited matrix.
   • Search horizontally and vertically to mark all connected lands as visited.

5. Continue Exploring the Map

   • Move to the next cell in the grid and repeat steps 3-4 until you’ve explored the entire map.

6. Count the Islands

   • Return the final value of island_count.

Effects of Changing Parameters

• Grid Size: As the grid grows, the time complexity will increase. The exploration will take longer.
• Island Shape and Size: Different shapes or isolated lands will require more checks but won’t affect the overall methodology.

Example Cases

1. Case 1

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

   • Start at (0,0): Found Island 1. Explore it fully.
   • Move to (0,2): It’s water. Skip.
   • Continue this way.
   • End at (2,2): Found Island 2.
   • Final Count: 2 islands

2. Case 2

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

   • Each ‘1’ is a separate island as they are not connected.
   • Final Count: 3 islands

By breaking down the problem into these actionable steps, you can implement a solution that covers all scenarios and edge cases.

Identify Invariant

The loop invariant in this problem could be: “At the start of each iteration of the outer loop, all ‘1’s in the visited rows and columns have been explored and counted, and island_count contains the accurate number of islands found so far.”

This invariant helps us ensure two things:

1. We don’t double-count any part of an island.
2. We only update the island count when we find the start of a new island.

Before the loop begins, the invariant is true because we have visited zero rows and zero columns, and island_count is zero.

During each iteration, if we find an unvisited ‘1’, we fully explore the new island and mark all its land cells as visited, maintaining the loop invariant.

After the loop, the invariant ensures that all ‘1’s have been explored and correctly counted, allowing us to return island_count as the final answer.

Identify Loop Invariant

The loop invariant in this problem is: “At the start of each iteration of the main loop, all ‘1’s in the grid up to the current row and column have been identified and marked as visited, and island_count accurately represents the number of distinct islands found up to that point.”

This loop invariant serves the purpose of:

1. Ensuring that no part of an island is double-counted.
2. Confirming that island_count is up to date at each iteration.

The loop invariant is established before the loop starts, as we have visited zero cells, and island_count is initialized to zero.

During each iteration, if we encounter a ‘1’ that has not been visited, we traverse and mark the entire island, ensuring the invariant holds for the next iteration.

After the loop concludes, the invariant guarantees that all islands have been accounted for, allowing us to confidently return island_count as the answer.

In the context of this problem, the term “invariant” could refer to properties that must hold true throughout the entire execution of the algorithm, while “loop invariant” specifically targets the properties that should hold true before, during, and after the execution of the main loop.

For this problem, the loop invariant is specifically focused on the behavior and state within the main loop that iterates through the grid. It ensures that the algorithm behaves as expected during each iteration of the loop, helping us make claims about the correctness of the algorithm.

So, while all loop invariants are invariants, not all invariants are loop invariants. However, in this particular problem, the main focus is on the loop invariant, as the loop is the primary mechanism driving the solution. Therefore, the loop invariant could be considered the critical invariant for this problem.

No, the invariant and the loop invariant are not the same for this problem.

• Invariant: The invariant here is that a ‘1’ represents land and ‘0’ represents water in the 2D grid. This understanding remains constant throughout the problem-solving process and is crucial for the DFS function to work as expected.

• Loop Invariant: Before and after each iteration of the for-loop that goes through each cell, the variable count correctly represents the number of islands found so far. This ensures that when you find a new ‘1’, you can confidently increment count by 1.

The invariant is a broader property that the problem relies on, while the loop invariant is specific to the correct functioning of the loop in our algorithm. They are related but serve different roles in understanding and solving the problem.

For the problem of counting the number of islands in a 2D grid, you can use Depth-First Search (DFS) to traverse through all the ‘1’s (lands) in a connected component and mark them as visited. Let’s say your for-loop starts from lines x to y to iterate over each cell in the grid. The loop invariant here is: before and after each iteration, the variable count correctly represents the number of islands found up to that point.

Initialization:

Prior to the first iteration of the loop, we have count initialized to zero. This is accurate because we haven’t started looking for islands yet.

Maintenance:

To see that each iteration maintains the loop invariant, lines a - b perform a DFS from the current ‘1’ cell to explore all its connected ‘1’s, and then increment the count by 1. The loop invariant is maintained here because the DFS accurately identifies a new island and updates the count. After the DFS, all cells belonging to that island are marked as visited (‘0’ or any other symbol to indicate they are no longer ‘1’) so they won’t be counted again in future iterations.

Termination:

At termination, the count will represent the total number of islands in the 2D grid. The loop invariant guarantees this, as it ensures that every found ‘1’ leads to a complete traversal of an island, incrementing the count by exactly 1 for each unique island.

So the loop invariant holds true at all stages of the loop, and it provides a useful property to show the correctness of the algorithm when the loop terminates.

Thought Process

1. Understand the Problem: The task is to find the number of islands in a 2D grid. An island is defined as a collection of ‘1’s connected horizontally or vertically.

2. Constraints and Cues:

   • The grid is a 2D array with m x n dimensions.
   • ‘1’ represents land and ‘0’ represents water.
   • The size of the grid is limited, so we don’t have to worry too much about time complexity.

3. Approach: Depth-First Search (DFS) seems suitable for exploring each land (‘1’) and all of its neighbors to form an island, marking them as visited along the way.

4. Data Structures:

   • A 2D array (the grid itself) to represent the map.
   • A stack or recursion for DFS.

5. Pseudocode:

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initialize count = 0
for each cell (i, j) in grid:
    if cell (i, j) is '1':
        count += 1
        DFS to mark all connected '1's to '0'
return count

6. Loop Invariant: Before each iteration of the for-loop, the number of islands found should be stored in count.

7. Main Steps:

   • Run a nested loop to go through each cell in the grid.
   • If we find land (‘1’), we perform a DFS to find the entire island and mark it as visited (‘0’).

Code

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def numIslands(grid):
    if not grid:
        return 0

    count = 0
    rows, cols = len(grid), len(grid[0])

    # DFS function to explore an island
    def dfs(r, c):
        if r < 0 or c < 0 or r >= rows or c >= cols or grid[r][c] == '0':
            return
        grid[r][c] = '0'  # Mark as visited
        dfs(r-1, c)
        dfs(r+1, c)
        dfs(r, c-1)
        dfs(r, c+1)

    for i in range(rows):
        for j in range(cols):
            if grid[i][j] == '1':
                count += 1
                dfs(i, j)

    return count

PENDING TEST

Insights

• The choice of DFS is crucial; it helps to simplify the problem into manageable chunks.
• The loop invariant helps us confidently increment the count variable, ensuring its accuracy at each stage of the iteration.
• The constraints and data structure choice enable an efficient solution.

Establishing Preconditions and Postconditions

For the example problem of counting the number of islands in a 2D grid using Depth-First Search (DFS), let’s go through each point:

Parameters:

1. The inputs to the method are a 2D grid represented as a list of lists. Each inner list represents a row.
2. The types of these parameters are lists of lists containing strings or integers (‘0’ for water and ‘1’ for land).
3. These parameters represent the 2D world where islands and water are located.

Preconditions:

1. The 2D grid must not be empty, and each row should have the same number of columns.
2. Each element in the grid must either be ‘0’ or ‘1’.
3. No specific state of the program is required before the method is called.

Method Functionality:

1. The method is expected to count the total number of islands in the 2D grid.
2. It takes the grid as input and mutates its state to mark visited cells, using DFS to explore each island fully.

Postconditions:

1. After the method returns, each ‘1’ that has been visited in the grid may be mutated to indicate it’s visited.
2. The return value represents the total number of islands found in the grid.
3. Side effects include changes to the input grid to mark visited cells.

Error Handling:

1. If the grid is empty or not properly formatted, the method should respond accordingly.
2. Depending on your design, it could throw an exception, return a special value like -1, or print an error message.

Here, the parameters, preconditions, and postconditions form the contract that the method should fulfill, providing a clear idea of its functionality and limitations.

Problem Decomposition

The problem involves counting the number of islands in a 2D grid. An island is a collection of ‘1’s (land) connected either horizontally or vertically. The grid contains ‘0’s (water) and ‘1’s (land). The goal is to return the number of distinct islands.

Initial Breakdown:

1. Loop through each cell in the grid.
2. If a cell contains ‘1’, perform Depth-First Search (DFS) to explore the island.
3. Mark cells as visited.
4. Count the island and continue.

Subproblem Refinement:

1. Loop through each cell: Implement a nested loop to traverse rows and columns.
2. Perform DFS: Create a helper function to perform DFS.
   • Mark the starting cell as visited.
   • Visit neighboring land cells recursively.
3. Mark cells: Change the cell value to mark it as visited.
4. Count: Maintain a counter to keep track of visited islands.

Task Identification:

1. Traversing the grid is a repeated task.
2. Performing DFS on a cell is a repeated task.
3. Marking cells as visited is a repeated task.

Task Abstraction:

1. traverseGrid: Loops through each cell in the grid.
2. performDFS: Takes a starting cell and explores the island using DFS.
3. markVisited: Marks a cell as visited.

Method Naming:

1. traverseGrid: For looping through the grid.
2. performDFS: For DFS traversal starting from an unvisited ‘1’.
3. markVisited: For marking a cell as visited.

Subproblem Interactions:

1. traverseGrid is the main driver and calls other tasks as needed.
2. performDFS is dependent on markVisited to avoid revisiting cells.
3. performDFS updates the island counter when it completes.
4. markVisited is an independent operation but is called within performDFS.

Order of Execution:

1. Start with traverseGrid.
2. For each ‘1’ cell, call performDFS.
3. Inside performDFS, use markVisited.

Dependencies:

• performDFS depends on markVisited.
• Both performDFS and markVisited are called within the scope of traverseGrid.

From Brute Force to Optimal Solution

Brute Force Solution

A brute-force way to solve the problem is to loop through each cell in the grid. When we encounter a ‘1’, we can recursively mark all connected ‘1’s as visited by turning them into ‘0’s and increase our island count. This is essentially Depth-First Search (DFS).

Brute Force Code in Python3

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def numIslands(grid):
    if not grid:
        return 0
    
    rows, cols = len(grid), len(grid[0])
    island_count = 0
    
    def dfs(x, y):
        if x < 0 or y < 0 or x >= rows or y >= cols or grid[x][y] == '0':
            return
        grid[x][y] = '0'
        
        dfs(x+1, y)
        dfs(x-1, y)
        dfs(x, y+1)
        dfs(x, y-1)
        
    for i in range(rows):
        for j in range(cols):
            if grid[i][j] == '1':
                island_count += 1
                dfs(i, j)
                
    return island_count

Inefficiencies

• Time Complexity: The time complexity of the brute-force solution is (O(R \times C)), where (R) is the number of rows and (C) is the number of columns.
• Space Complexity: The space complexity is (O(R \times C)) due to the recursive stack in the worst case.

Optimizing the Solution

The brute-force solution is already quite efficient for this problem. The DFS approach inherently requires visiting all cells, so the (O(R \times C)) time complexity is arguably the best we can do for a general solution. However, the space complexity can be reduced.

Space Complexity

The space complexity primarily comes from the recursive DFS function. To avoid this, we can use an iterative DFS with an explicit stack. This would replace the recursive stack, although the complexity would remain the same.

Optimized Code in Python3

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def numIslands(grid):
    if not grid:
        return 0
    
    rows, cols = len(grid), len(grid[0])
    island_count = 0
    
    for i in range(rows):
        for j in range(cols):
            if grid[i][j] == '1':
                island_count += 1
                stack = [(i, j)]
                grid[i][j] = '0'
                
                while stack:
                    x, y = stack.pop()
                    
                    for dx, dy in [(1, 0), (-1, 0), (0, 1), (0, -1)]:
                        newX, newY = x + dx, y + dy
                        if 0 <= newX < rows and 0 <= newY < cols and grid[newX][newY] == '1':
                            stack.append((newX, newY))
                            grid[newX][newY] = '0'
                
    return island_count

Time and Space Complexity After Optimization

• Time Complexity: (O(R \times C)) - unchanged, still efficient.
• Space Complexity: (O(R \times C)) - remains the same but the explicit stack is generally smaller than the recursive stack.

In summary, the initial brute-force solution was already efficient in terms of time complexity. The optimization focuses on making the space usage a bit more manageable by using an explicit stack instead of recursion, although the space complexity remains (O(R \times C)).

Code Explanation and Design Decisions

1. Initial Parameters

• grid: A 2D list containing ‘1’s representing land and ‘0’s representing water. The grid represents the geographical layout. The dimensions of the grid are given by rows and cols.

2. Primary Loop

The primary loop iterates over each cell in the grid. Each iteration checks if the current cell is a ‘1’ (land) that hasn’t been visited yet.

• Significance: Each unvisited ‘1’ signifies the starting point of a new island.
• Contribution: Once we find a new starting point, we explore the entire island to mark it as visited and increase the island count.

3. Conditions within Loop

Inside the primary loop, there’s a condition if grid[i][j] == '1':.

• Significance: This condition checks if we’ve hit a piece of land that belongs to an unexplored island.

4. Updates to Parameters

• island_count += 1: We increment island_count when we find a new island’s starting point.
• grid[i][j] = '0': We mark cells as visited (water) to avoid recounting.

5. Invariant

• Invariant: At any point, island_count represents the number of completely explored islands.
• Significance: Maintaining this invariant ensures that we count each island only once and helps validate the final island count.

6. Final Output

• Significance: The final output, island_count, represents the total number of distinct islands in the given grid.
• Satisfaction: This output satisfies the problem’s requirement to count the number of distinct islands.

By understanding these components, we get a clear roadmap of how the solution works in the context of the problem domain. This helps not only in grasping the current solution but also in identifying potential areas for further improvement or modification.

Coding Constructs

1. High-Level Strategies

The code employs Depth-First Search (DFS) for traversal. It uses two nested loops to go through each cell in the grid and, upon finding a land cell, traverses all its connected land cells to mark the whole island as visited.

2. Explaining to a Non-Programmer

The code is like a drone flying over a map. When the drone sees a piece of land, it explores the whole island to see how big it is. Then it moves on. In the end, it tells you how many separate islands it found.

3. Logical Elements

• Iteration: Loop through each grid cell.
• Conditionals: Check if a cell is land (‘1’) or water (‘0’).
• Recursion: Visiting all cells of a found island.
• Variables: Storing the count of islands.

4. Algorithmic Approach in Plain English

• Start at the top-left corner of the map.
• Move cell by cell.
• If you find land, explore the whole island.
• Mark the island as visited.
• Continue this until you’ve looked at every cell.
• The number of islands you found is your answer.

5. Key Steps or Operations

• Loop Through Grid: To examine each cell.
• Identify Land Cell: To locate possible new islands.
• DFS Traversal: To explore an entire island once a land cell is found.
• Mark Visited: To avoid re-exploring cells.
• Count Islands: Increment the counter each time a new island is found.

6. Algorithmic Patterns

• Graph Traversal: Specifically, Depth-First Search.
• Nested Iteration: For 2D grid traversal.
• State Marking: To keep track of visited cells.

By understanding these components, you can easily grasp how the algorithm solves the problem. This understanding is key for future improvements or variations.

Language Agnostic Coding Drills

1. Distinct Coding Concepts

1. Variable Declaration: Basic understanding of variables to store data.
2. Arrays and Matrices: Understanding of 1D and 2D arrays to represent data.
3. Loops: Basic loops to iterate over arrays.
4. Conditional Statements: Use of if, else to make decisions.
5. Functions: Basic function declarations and calls.
6. Recursion: Functions calling themselves for DFS.
7. State Management: Changing and checking the state of elements.
8. Complexity Analysis: Understanding time and space complexity.

2. Concepts in Order of Increasing Difficulty

1. Variable Declaration: Easiest, foundation of all programming.
2. Arrays and Matrices: Basic but slightly more complex data structures.
3. Loops: Repeated actions, the cornerstone of many algorithms.
4. Conditional Statements: Introduces logic into the code.
5. Functions: Encapsulation of logic, a step towards modular programming.
6. State Management: Involves more abstract thinking, tied into the logic.
7. Recursion: Conceptually challenging for many, involves stack memory.
8. Complexity Analysis: More theoretical, involves mathematical intuition.

3. Problem-Solving Approach

1. Variable Declaration: Declare a variable to keep track of the number of islands.
2. Arrays and Matrices: Understand the grid representation of the map, where ‘1’ signifies land and ‘0’ signifies water.
3. Loops: Loop through each cell in the 2D grid to check if it’s land or water.
4. Conditional Statements: If a cell is water, skip it. If it’s land, you have a new island.
5. Functions: Implement a separate function to perform Depth-First Search to explore the entire island starting from a given land cell.
6. State Management: As you go through each land cell in an island, mark it as visited so you don’t recount the same island.
7. Recursion: Within the Depth-First Search, make recursive function calls to explore all directions from the current land cell.
8. Complexity Analysis: Finally, analyze the time and space complexity of your solution to ensure it meets the problem’s constraints.

Each of these drills contributes to the overall solution. You start by setting up your variables and data structures, then move into the core logic of the program, which is encapsulated within loops, conditionals, and function calls. As you progress, the concepts become more complex, ultimately allowing you to solve the problem efficiently.

Targeted Drills in Python

1. Python-Based Coding Drills for General Concepts

Drill 1: Variable Declaration

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# Declaring an integer variable
count = 0

Drill 2: Arrays and Matrices

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# Declaring a 1D array
one_d_array = [1, 2, 3]

# Declaring a 2D array
two_d_array = [[0, 1], [1, 0], [1, 1]]

Drill 3: Loops

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# Iterating through a 1D array
for element in one_d_array:
    print(element)

Drill 4: Conditional Statements

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# If-else statement
if count == 0:
    print("Count is zero.")
else:
    print("Count is not zero.")

Drill 5: Functions

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# Defining a simple function
def print_hello():
    print("Hello")

Drill 6: Recursion

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# Factorial function using recursion
def factorial(n):
    if n == 0:
        return 1
    return n * factorial(n-1)

Drill 7: State Management

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# Toggling a boolean state
state = False
if not state:
    state = True

Drill 8: Complexity Analysis

No code for this drill, it’s a conceptual understanding.

2. Problem-Specific Coding Drills

Drill 9: Depth-First Search (DFS)

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# Implementing DFS to explore a 2D grid
def dfs(grid, i, j):
    if i < 0 or i >= len(grid) or j < 0 or j >= len(grid[0]):
        return
    if grid[i][j] == "0":
        return
    grid[i][j] = "0"  # Mark as visited

DFS is essential for our problem to navigate through each “island” efficiently, marking parts as visited.

3. Integrating Drills into the Final Solution

1. Variable Declaration: Start by declaring the count variable to keep track of islands.
2. Arrays and Matrices: Initialize your 2D grid based on the problem’s input.
3. Loops: Use nested loops to traverse each cell in the grid.
4. Conditional Statements: Inside the loop, check if a cell is “1” (land) or “0” (water).
5. Functions and Recursion: When a land cell is found, call the dfs function recursively to explore the entire island.
6. State Management: Mark visited land cells as “0” in the 2D grid.
7. Depth-First Search (DFS): This is the core of the solution, as it navigates through each island.
8. Complexity Analysis: Finally, understand that the solution is O(rows * cols) in time complexity and O(rows * cols) in space complexity if we count the grid’s space.

Integrating these drills in the given order forms a complete solution for counting the number of islands in a 2D grid.

Here are 10 distinct problems that use similar underlying concepts:

1. 200. Number of Islands

   • Relation: While this might seem like the problem we discussed, it actually serves as a base for understanding similar problems. It involves traversing a 2D grid and uses DFS.

2. 695. Max Area of Island

   • Relation: This problem extends the concept of counting islands to finding the maximum area of an island. It uses similar grid traversal and DFS techniques.

3. 130. Surrounded Regions

   • Relation: This problem asks you to flip surrounded regions. The concept of visiting each cell in a grid and making decisions based on its neighbors is similar.

4. 79. Word Search

   • Relation: This problem involves traversing a 2D grid of characters to form a given word. DFS is used to explore possible paths.

5. 417. Pacific Atlantic Water Flow

   • Relation: In this problem, you need to find the cells where water can flow to both the Pacific and Atlantic oceans, which involves grid traversal and state management.

6. 286. Walls and Gates

   • Relation: You need to fill each empty room with the distance to its nearest gate. This is another problem that involves grid traversal but uses BFS instead of DFS.

7. 994. Rotting Oranges

   • Relation: This problem involves a 2D grid where you have to determine the minimum time to rot all oranges. It involves BFS for grid traversal.

8. 547. Number of Provinces

   • Relation: Though not a 2D grid, this problem involves counting connected components in a graph, similar to counting islands. DFS or Union-Find can be applied.

9. 1091. Shortest Path in Binary Matrix

   • Relation: Involves finding the shortest path in a grid, using BFS or A* algorithms. It extends the grid traversal concept to include path length.

10. 127. Word Ladder

    • Relation: While this doesn’t involve a grid, it involves transforming one word into another through intermediate words. The transformation can be thought of as a graph traversal problem, often solved using BFS.

Each of these problems involves some form of graph traversal, state management, or pathfinding, which are the core concepts of the problem we’ve been discussing.

Define the Terms What is an island? The 1 must be surrounded by water which is 0

Implication: Derived from the problem statement. Why do we need to consider something that is outside of the given mxn matrix?

We must be surrounded by water top, bottom, left and right.

Define the Interface Input: array of arrays consisting of characters 1, 0 Output: integer (number of islands)

Can we mutate the input? Is it mutable?

We can mutate the array.

Analyze the Input and Output

Identify the Invariants

Identify the constraints

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 300
• grid[i][j] is ‘0’ or ‘1’.

Search the Problem Space

Classify the Problem Grid Traversal

Analyze the Examples

Solve the Problem by Hand We have to traverse the grid and count the number of islands Return the value

We need to keep track of which cells I have already visited.

Traversal - What type of traversal can we do here?

DFS

• row, column
• grid
• visited / keep track of it in the grid
• ‘#’ means it has been visited

We need to check for bounds to make sure we don’t go outside of the grid. If I see a cell is ‘0’, return from the recursive call If I see a cell is ‘#’, return from the recursive call After visiting all ‘1’ from a given starting point of ‘1’ we will increment the counter by one.

We will initialize the counter outside of the dfs method in num_islands method.

What many dfs calls do we need to make? - Top - Down - Left - Right

Where to increment the counter? How to keep track of the counter

Return count

I have to traverse the entire grid to check.

Describe the Pattern

Distill the Observations to Create Insights

Brute Force Approach

Analyze Time and Space Complexity

Time: O( ) Space: O( )

Outline the Solution

Main Method

1. counter = 0
2. Traverse from (0,0) —-> (m-1, n-1)
3. Check if the value at (r, c) is ‘1’ change the value for (r, c) to ‘#’
4. DFS traversal for this (r, c)
5. counter must be incremented

DFS

• Bounds check
• Check if the r,c is 0 or # (base case)
• DFS calls for left, right, up, down

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def dfs(grid, row, col)
   rows = grid.size
   cols = grid[0].size
    
   if row < 0 || col < 0 || row >= rows || col >= cols || grid[row][col] == '#' || grid[row][col] == '0'
      return 0
   end
    
    grid[row][col] = '#'
   dfs(grid, row-1, col)
   dfs(grid, row+1, col)
   dfs(grid, row, col-1)
   dfs(grid, row, col+1)
    
   return 1
end

# @param {Character[][]} grid
# @return {Integer}
def num_islands(grid)
  rows = grid.size
  cols = grid[0].size
  
  islands = 0

  for row in (0..rows-1)
     for col in (0..cols-1)
        if grid[row][col] == '1'
            islands += dfs(grid, row, col) 
        end
     end
  end
  
  return islands  
end

Identifying Problem Isomorphism

“Number of Islands” is isomorphic to the problem “Surrounded Regions”.

In “Number of Islands”, you are given a 2D grid map of ‘1’s (land) and ‘0’s (water). An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water. The problem is to find the number of islands.

In “Surrounded Regions”, you are given a 2D board containing ‘X’ and ‘O’ (the letter O). Capture all regions surrounded by ‘X’. A region is captured by flipping all ‘O’s into ‘X’s in that surrounded region.

The reason these problems are isomorphic is that both involve exploring a 2D grid using depth-first search (DFS) or breadth-first search (BFS). In both cases, we’re interested in contiguous regions of the grid: islands of ‘1’s in the “Number of Islands” problem, and regions of ‘O’s surrounded by ‘X’s in the “Surrounded Regions” problem. The tasks are different, but the methods to solve them are the same: start a DFS or BFS from each cell that contains ‘1’ or ‘O’ respectively, and traverse all the cells connected to it.

However, the mapping is approximate because the problems are asking for different things (counting islands versus capturing regions), and they treat the edge of the grid differently (“Number of Islands” considers the edge to be water, while “Surrounded Regions” doesn’t capture regions that reach the edge). Therefore, while the problems are structurally similar and use the same traversal algorithm, they aren’t exactly the same.

The problem “Number of Islands” can be approximately mapped to the “Connected Components in an Undirected Graph” problem.

Reasoning: Both problems involve finding connected components in a graph-like structure. In “Number of Islands,” the grid is implicitly treated as a graph where each cell is a node connected to its adjacent cells. In “Connected Components in an Undirected Graph,” you are given an explicit graph and need to find connected components, just like the islands.

“Connected Components in an Undirected Graph” is simpler because the graph structure is given explicitly, making it more straightforward to apply graph traversal algorithms like DFS or BFS.

The mapping is approximate because the specific rules for connectivity differ between the two problems. In “Number of Islands,” connectivity is defined as horizontal or vertical adjacency in a 2D grid, while in the graph problem, connectivity is defined by edges in the graph.

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class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        count = 0
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j] == '1':
                    print(i,j)
                    self.dfs(grid,i,j)
                    count  += 1
        #print(grid)
        return count
    # use a helper function to flip connected '1's to 0
    def dfs(self,grid,i,j):
        grid[i][j] = 0
        for dr,dc in (1,0), (-1,0), (0,-1), (0,1):
            r = i + dr
            c = j + dc
            if 0 <= r < len(grid) and 0 <= c < len(grid[0]) and grid[r][c]=='1':
                self.dfs(grid,r,c)

Problem Classification

Grid Traversal

Language Agnostic Coding Drills

The code given is solving the problem of counting the number of islands in a 2D grid map of ‘1’s (land) and ‘0’s (water). The islands are defined as a group of connected ‘1’s and the connections can only be made horizontally or vertically. Here are the smallest units of learning in increasing level of difficulty:

1. Understanding the concept of a 2D grid: Understanding how 2D lists (also known as matrices) work and how to navigate them.

2. Familiarity with nested for-loops: Ability to iterate over all elements in a 2D grid.

3. Basic conditional statements: Understanding of how to use if statements to perform actions based on certain conditions.

4. Recursion: Understanding how recursive calls work, particularly in the context of exploring a data structure.

5. Depth-First Search (DFS) algorithm: The ability to implement and understand the DFS algorithm, which is a common way to traverse trees or graphs.

6. In-place modifications: In this case, the original grid is being modified to ‘mark’ visited locations.

7. Boundary checks in a grid: When exploring a grid, it’s crucial to ensure that the algorithm does not go out of bounds. Understanding how to check these boundaries is a key part of many grid-based problems.

8. Counting patterns in a grid: Finally, the main goal of this code is to count occurrences of a certain pattern in a grid - a very common task in many grid-based problems. In this case, the pattern is a group of ‘1’s that are all connected. This requires all of the above concepts, making it the most complex unit.

Targeted Drills in Python

1. Understanding the concept of a 2D grid:

Drill: Create a 2D list (5x5) and print it out.

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grid = [[0]*5 for _ in range(5)]
print(grid)

2. Familiarity with nested for-loops:

Drill: Print each element in the 2D grid created above.

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for i in range(5):
    for j in range(5):
        print(grid[i][j])

3. Basic conditional statements:

Drill: In the 2D grid, if an element is zero, print “Water”, else print “Land”.

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for i in range(5):
    for j in range(5):
        if grid[i][j] == 0:
            print("Water")
        else:
            print("Land")

4. Recursion:

Drill: Write a function to calculate the factorial of a number using recursion.

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def factorial(n):
    if n == 0 or n == 1:
        return 1
    else:
        return n * factorial(n-1)

5. Depth-First Search (DFS) algorithm:

Drill: Implement a simple depth-first search on a tree data structure.

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class Node:
    def __init__(self, val, children=[]):
        self.val = val
        self.children = children

def DFS(node):
    print(node.val)
    for child in node.children:
        DFS(child)

root = Node(1, [Node(2), Node(3, [Node(4)])])
DFS(root)

6. In-place modifications:

Drill: Create a list of numbers and then double each number in the list without creating a new list.

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numbers = [1, 2, 3, 4, 5]
for i in range(len(numbers)):
    numbers[i] *= 2

7. Boundary checks in a grid:

Drill: In a 2D grid, write a function that takes coordinates as input and returns True if they are in bounds, False otherwise.

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def in_bounds(grid, coordinates):
    x, y = coordinates
    return 0 <= x < len(grid) and 0 <= y < len(grid[0])

print(in_bounds(grid, (1, 1)))
print(in_bounds(grid, (5, 5)))

8. Counting patterns in a grid:

Drill: Count the number of ‘1’s in a 2D grid.

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grid = [[0, 1, 0, 1], [1, 0, 1, 0], [0, 1, 0, 1], [1, 0, 1, 0]]
count = 0
for i in range(len(grid)):
    for j in range(len(grid[0])):
        if grid[i][j] == 1:
            count += 1
print(count)

These drills cover the fundamental building blocks that are combined in the complete solution code you posted. By mastering each of these skills, one will be well-equipped to understand and implement such a solution.