LeetCode-in-Ruby.github.io

200. Number of Islands

Medium

Given an m x n 2D binary grid grid which represents a map of '1's (land) and '0's (water), return the number of islands.

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.

Example 1:

Input:

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

Output: 1

Example 2:

Input:

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

Output: 3

Constraints:

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

Solution

# @param {Character[][]} grid
# @return {Integer}
def num_islands(grid)
  islands = 0
  if grid && !grid.empty? && !grid[0].empty?
    (0...grid.length).each do |i|
      (0...grid[0].length).each do |j|
        if grid[i][j] == '1'
          dfs_islands(grid, i, j)
          islands += 1
        end
      end
    end
  end
  islands
end

private

def dfs_islands(grid, x, y)
  return if x.negative? || grid.length <= x || y.negative? || grid[0].length <= y || grid[x][y] != '1'

  grid[x][y] = 'x'
  dfs_islands(grid, x + 1, y)
  dfs_islands(grid, x - 1, y)
  dfs_islands(grid, x, y + 1)
  dfs_islands(grid, x, y - 1)
end

This site is open source. Improve this page.