1020. Number of Enclaves

Source code notebook Author Update time

Given a 2D array A, each cell is 0 (representing sea) or 1 (representing land)

A move consists of walking from one land square 4-directionally to another land square, or off the boundary of the grid.

Return the number of land squares in the grid for which we cannot walk off the boundary of the grid in any number of moves.

Example 1:

Input: [[0,0,0,0],[1,0,1,0],[0,1,1,0],[0,0,0,0]]
Output: 3
Explanation:
There are three 1s that are enclosed by 0s, and one 1 that isn't enclosed because its on the boundary.

Example 2:

Input: [[0,1,1,0],[0,0,1,0],[0,0,1,0],[0,0,0,0]]
Output: 0
Explanation:
All 1s are either on the boundary or can reach the boundary.

Note:

  1. 1 <= A.length <= 500
  2. 1 <= A[i].length <= 500
  3. 0 <= A[i][j] <= 1
  4. All rows have the same size.
# @lc code=start
using LeetCode

# BFS algorithm
function num_of_enclaves_bfs(grid::Vector{Vector{Int}})::Int
    m, n, queue = length(grid), length(grid[1]), Tuple{Int,Int}[]
    (m <= 2 || n <= 2) && return 0
    for i in 1:m ## left and right border
        grid[i][1] == 1 && push!(queue, (i, 1))
        grid[i][n] == 1 && push!(queue, (i, n))
    end
    for i in 2:(n - 1)
        grid[1][i] == 1 && push!(queue, (1, i))
        grid[m][i] == 1 && push!(queue, (m, i))
    end
    walkoff = copy(queue)
    while length(queue) != 0
        i, j = popfirst!(queue)
        for (a, b) in [(i - 1, j), (i + 1, j), (i, j - 1), (i, j + 1)]
            if a >= 1 &&
                b >= 1 &&
                a <= m &&
                b <= n &&
                grid[a][b] != 0 &&
                !((a, b) in walkoff)
                push!(walkoff, (a, b))
                push!(queue, (a, b))
            end
        end
    end
    return sum(sum(line) for line in grid) - length(walkoff)
end
# @lc code=end
num_of_enclaves_bfs (generic function with 1 method)

This page was generated using DemoCards.jl and Literate.jl.