# Program to find the perimeter of an island shape in Python

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Suppose we have a binary matrix where 0 shows empty cell and 1 shows a block that forms a shape, now we have to find the perimeter of the shape. The shape will not hold any hole inside it.

So, if the input is like

 0 0 0 0 0 0 0 1 1 1 0 0 1 1 0 0 1 1 1 0 0 0 0 0 0

then the output will be 14.

To solve this, we will follow these steps −

• d := 0

• perimeter := 0

• height := row count of matrix

• length := column count of matrix

• for each row in matrix, do

• c := 0

• for each val in row, do

• if val is same as 1, then

• surround := 4

• if c is not same as length - 1, then

• if matrix[d, c + 1] is same as 1, then

• surround := surround - 1

• if c is not same as 0, then

• if matrix[d, c - 1] is same as 1, then

• surround := surround - 1

• if d is not same as height - 1, then

• if matrix[d + 1, c] is same as 1, then

• surround := surround - 1

• if d is not same as 0, then

• if matrix[d - 1, c] is same as 1, then

• surround := surround - 1

• perimeter := perimeter + surround

• c := c + 1

• d := d + 1

• return perimeter

Let us see the following implementation to get better understanding −

## Example

Live Demo

class Solution:
def solve(self, matrix):
d = 0
perimeter = 0
height = len(matrix)
length = len(matrix[0])
for line in matrix:
c = 0

for val in line:
if val == 1:
surround = 4
if c != length - 1:
if matrix[d][c + 1] == 1:
surround -= 1
if c != 0:
if matrix[d][c - 1] == 1:
surround -= 1
if d != height - 1:
if matrix[d + 1][c] == 1:
surround -= 1
if d != 0:
if matrix[d - 1][c] == 1:
surround -= 1
perimeter += surround
c += 1
d += 1
return perimeter

ob = Solution()
matrix = [
[0,0,0,0,0],
[0,0,1,1,1],
[0,0,1,1,0],
[0,1,1,1,0],
[0,0,0,0,0]
]
print(ob.solve(matrix))

## Input

matrix = [
[0,0,0,0,0],
[0,0,1,1,1],
[0,0,1,1,0],
[0,1,1,1,0],
[0,0,0,0,0]]

## Output

14
Updated on 09-Oct-2020 14:13:04