Count Number of Surrounded Islands in a Matrix using Python

Suppose we have a binary matrix. Where 1 represents land and 0 represents water. As we know an island is a group of 1s that are grouped together whose perimeter is surrounded by water. We have to find the number of completely surrounded islands.

So, if the input is like

then the output will be 2, as there are three islands, but two of them are completely surrounded.

To solve this, we will follow these steps −

• Define a function dfs() . This will take i, j
• if i and j are not in range of matrix, then
  • return False
• if matrix[i, j] is same as 0, then
  • return True
• matrix[i, j] := 0
• a := dfs(i + 1, j)
• b := dfs(i - 1, j)
• c := dfs(i, j + 1)
• d := dfs(i, j - 1)
• return a and b and c and d
• From the main method do the following:
• R := row count of matrix, C := column count of matrix
• ans := 0
• for i in range 0 to R, do
  • for j in range 0 to C, do
    • if matrix[i, j] is same as 1, then
      • if dfs(i, j) is true, then
        • ans := ans + 1
• return ans

Let us see the following implementation to get better understanding −

Example 

Live Demo

class Solution:
   def solve(self, matrix):
      def dfs(i, j):
         if i < 0 or j < 0 or i >= R or j >= C:
            return False
         if matrix[i][j] == 0:
            return True
         matrix[i][j] = 0
         a = dfs(i + 1, j)
         b = dfs(i - 1, j)
         c = dfs(i, j + 1)
         d = dfs(i, j - 1)
         return a and b and c and d

      R, C = len(matrix), len(matrix[0])
      ans = 0
      for i in range(R):
         for j in range(C):
            if matrix[i][j] == 1:
               if dfs(i, j):
                  ans += 1
      return ans

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

Input

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

Output

2