Number of Islands

Clarify the problem:

• The problem requires us to count the number of islands in a given 2D grid.
• An island is represented by a group of connected '1's (representing land) surrounded by '0's (representing water).
• We need to implement a function that takes in the grid as input and returns the number of islands.

Analyze the problem:

• Input: 2D grid of characters.
• Output: Number of islands (integer).
• Constraints:
  • The grid can have '0's and '1's only.
  • An island is formed by connecting horizontally or vertically adjacent '1' cells.
  • Diagonal connections are not considered.
  • The grid size is between 1 and 10^4.

Design an algorithm:

• We can solve this problem using a depth-first search (DFS) approach.
• The idea is to traverse the grid and explore each cell to find the connected island cells.
• We can define a recursive function to perform the DFS.
• The function will take the current cell coordinates as input.
• If the current cell is valid (within the grid bounds) and represents land ('1'), we mark it as visited and recursively call the function on its neighboring cells.
• We mark the visited cells to avoid counting them again.
• Each time we encounter a new unvisited island cell ('1'), we increment the island count.
• Finally, we iterate over the entire grid, calling the DFS function on each unvisited island cell, and return the total island count.

Explain your approach:

• We will implement a depth-first search (DFS) algorithm to count the number of islands.
• First, we define a recursive function dfs that takes the current cell coordinates as input.
• Inside the function, we check if the current cell is valid and represents land ('1').
• If it is, we mark it as visited by updating its value to '0'.
• Then, we recursively call the dfs function on the neighboring cells (up, down, left, and right).
• After defining the dfs function, we initialize a variable island_count to 0.
• Next, we iterate over the grid and call the dfs function on each unvisited island cell ('1').
• Each time we encounter a new unvisited island cell, we increment the island_count.
• Finally, we return the island_count.

Write clean and readable code:

python

class Solution:
    def numIslands(self, grid):
        def dfs(i, j):
            if 0 <= i < rows and 0 <= j < cols and grid[i][j] == '1':
                grid[i][j] = '0'
                dfs(i + 1, j)
                dfs(i - 1, j)
                dfs(i, j + 1)
                dfs(i, j - 1)

        rows = len(grid)
        cols = len(grid[0])
        island_count = 0

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

        return island_count

solution = Solution()
grid = [['1', '1', '1', '1', '0'],
        ['1', '1', '0', '1', '0'],
        ['1', '1', '0', '0', '0'],
        ['0', '0', '0', '0', '0']]
num_islands = solution.numIslands(grid)
print(num_islands)  # Output: 1

Test your code:

• We test the code with a grid containing a single island.
• We chose this test case to ensure that the code can correctly identify and count a single island.
• We expect the output to be 1, as there is only one island in the given grid.

Optimize if necessary:

• The provided solution already uses the most efficient approach to solve the problem, which is DFS.
• There are no further optimizations needed for this solution.

Handle error cases:

• The code assumes that the input grid is a valid 2D list containing only '0's and '1's.
• If the input is empty or the grid size is outside the constraints, the behavior of the code may be undefined.

Discuss complexity analysis:

• Let M be the number of rows and N be the number of columns in the grid.
• The time complexity of the solution is O(M*N) as we visit each cell of the grid exactly once.
• The space complexity is O(M*N) in the worst case, considering the recursive stack space used by the DFS algorithm.
• The solution is efficient and performs well for grids with a large number of cells.