mirror of
https://github.com/TheAlgorithms/Ruby
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76 lines
2 KiB
Ruby
76 lines
2 KiB
Ruby
# 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.
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# 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.
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# Example 1:
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# Input: grid = [
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# ["1","1","1","1","0"],
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# ["1","1","0","1","0"],
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# ["1","1","0","0","0"],
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# ["0","0","0","0","0"]
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# ]
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# Output: 1
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# Example 2:
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# Input: grid = [
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# ["1","1","0","0","0"],
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# ["1","1","0","0","0"],
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# ["0","0","1","0","0"],
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# ["0","0","0","1","1"]
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# ]
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# Output: 3
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# Constraints:
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# m == grid.length
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# n == grid[i].length
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# 1 <= m, n <= 300
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# grid[i][j] is '0' or '1'.
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# DFS, Recursive Bottom Up Approach - O(n*m) Time / O(1) Space
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# Init num_of_islands = 0, return if the grid is empty
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# Start a double loop with index to iterate through each plot (each value is a plot of either water or land in this case)
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# if the plot is land, dfs(grid, x, y)
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# num_of_islands += 1
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# Return num_of_islands
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# dfs(grid, x, y)
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# Return if x or y are out of bounds, or if the plot is water
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# Make the current plot water
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# Call dfs again for up, down, left, and right
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# @param {Character[][]} grid
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# @return {Integer}
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def num_islands(grid)
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return 0 if grid.empty?
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# init num of islands
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islands = 0
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# loop through each element (plot) in the 2d array
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grid.each_with_index do |row, x|
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row.each_with_index do |plot, y|
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# if the plot is water, start a dfs
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next unless plot == '1'
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dfs(grid, x, y)
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# add 1 to islands once all connected land plots are searched
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islands += 1
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end
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end
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# return ans
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islands
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end
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def dfs(grid, x, y)
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# don't search if out of bounds, or if it's already water
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return if x < 0 || x >= grid.length || y < 0 || y >= grid[0].length || grid[x][y] == '0'
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# set the plot to water
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grid[x][y] = '0'
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# search each adjacent plot
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dfs(grid, x - 1, y) # up
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dfs(grid, x + 1, y) # down
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dfs(grid, x, y - 1) # left
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dfs(grid, x, y + 1) # right
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end
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