LeetCode: Island Perimeter

Posted on February 27, 2018July 26, 2020 by braindenny

Island Perimeter

Similar Problems:

You are given a map in form of a two-dimensional integer grid where 1 represents land and 0 represents water. Grid cells are connected horizontally/vertically (not diagonally). The grid is completely surrounded by water, and there is exactly one island (i.e., one or more connected land cells). The island doesn’t have “lakes” (water inside that isn’t connected to the water around the island). One cell is a square with side length 1. The grid is rectangular, width and height don’t exceed 100. Determine the perimeter of the island.

Example:

[[0,1,0,0],
 [1,1,1,0],
 [0,1,0,0],
 [1,1,0,0]]

Answer: 16

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

Solution: array

## https://code.dennyzhang.com/island-perimeter
## Basic Ideas: array
##
## Add then substract
##
## Complexity: Time O(n*m), Space O(1)
class Solution:
    def islandPerimeter(self, grid: List[List[int]]) -> int:
        cnt = 0
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j] == 1:
                    cnt += 4
                    ## check below
                    if i<len(grid)-1 and grid[i+1][j] == 1:
                        cnt -= 2
                    ## check right
                    if j<len(grid[0])-1 and grid[i][j+1] == 1:
                        cnt -= 2
        return cnt

Solution: array

## https://code.dennyzhang.com/island-perimeter
## Basic Ideas: array
##
## Check each cell
##
## Complexity: Time O(n*m), Space O(1)
class Solution:
    def islandPerimeter(self, grid: List[List[int]]) -> int:
        def check(i, j):
            res = 0
            for i2, j2 in [(i + 1, j), (i - 1, j), (i, j + 1), (i, j - 1)]:
                if i2<0 or i2>=len(grid) or j2<0 or j2>=len(grid[0]) or grid[i2][j2] == 0:
                    res += 1
            return res

        cnt = 0
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j] == 1:
                    cnt += check(i, j)
        return cnt

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