#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'. #DFS, Recursive Bottom Up Approach - O(n*m) Time / O(1) Space #Init num_of_islands = 0, return if the grid is empty #Start a double loop with index to iterate through each plot (each value is a plot of either water or land in this case) #if the plot is land, dfs(grid, x, y) #num_of_islands += 1 #Return num_of_islands #dfs(grid, x, y) #Return if x or y are out of bounds, or if the plot is water #Make the current plot water #Call dfs again for up, down, left, and right # @param {Character[][]} grid # @return {Integer} def num_islands(grid) return 0 if grid.empty? #init num of islands islands = 0 #loop through each element (plot) in the 2d array grid.each_with_index do |row, x| row.each_with_index do |plot, y| #if the plot is water, start a dfs if plot == "1" dfs(grid, x, y) #add 1 to islands once all connected land plots are searched islands += 1 end end end #return ans islands end def dfs(grid, x, y) #don't search if out of bounds, or if it's already water return if x < 0 || x >= grid.length || y < 0 || y >= grid[0].length || grid[x][y] == "0" #set the plot to water grid[x][y] = "0" #search each adjacent plot dfs(grid, x - 1, y) #up dfs(grid, x + 1, y) #down dfs(grid, x, y - 1) #left dfs(grid, x, y + 1) #right end