Enclosed Islands - Google Top Interview Questions


Problem Statement :


You are given a two-dimensional integer matrix of 1s and 0s. 
A 1 represents land and 0 represents water. 
From any land cell you can move up, down, left or right to another land cell or go off the matrix.

Return the number of land cells from which we cannot go off the matrix.

Constraints

n, m ≤ 250 where n and m are the number of rows and columns in matrix

Example 1

Input

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

Output

4

Explanation

There's 4 land squares in the middle from which we cannot walk off the matrix.


Solution :



title-img 




                        Solution in C++ :

def find_answer():
    for cell in matrix:
        if cell is edge_cell:
            push cell in queue

    while queue is not empty:
        current_cell = 0 # marking as visited
        for neighbor of current_cell:
            if neighbor is land:
                push neighbor in queue

    return  number of lands in matrix
                    


                        Solution in Java :

import java.util.*;

class Solution {
    private int[][] matrix;
    public int solve(int[][] matrix) {
        this.matrix = matrix;
        // sink all islands touching an edge, then count land cells in interior.
        /// check for the vertical edges
        for (int i = 0; i < matrix.length; i++) {
            if (matrix[i][0] == 1) {
                floodfill(i, 0);
            }
            if (matrix[i][matrix[0].length - 1] == 1) {
                floodfill(i, matrix[0].length - 1);
            }
        }
        // check for the horizontal edges
        for (int j = 0; j < matrix[0].length; j++) {
            if (matrix[0][j] == 1) {
                floodfill(0, j);
            }
            if (matrix[matrix.length - 1][j] == 1) {
                floodfill(matrix.length - 1, j);
            }
        }
        int ret = 0;
        for (int i = 1; i < matrix.length - 1; i++) {
            for (int j = 1; j < matrix[0].length - 1; j++) {
                ret += matrix[i][j];
            }
        }
        return ret;
    }
    public void floodfill(int i, int j) {
        if (i == -1 || j == -1 || i == matrix.length || j == matrix[0].length
            || matrix[i][j] == 0) {
            return;
        }
        matrix[i][j] = 0;
        floodfill(i + 1, j);
        floodfill(i - 1, j);
        floodfill(i, j + 1);
        floodfill(i, j - 1);
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, matrix):
        q = [
            (i, j)
            for i in range(len(matrix))
            for j in range(len(matrix[i]))
            if matrix[i][j]
            and (i == 0 or i == len(matrix) - 1 or j == 0 or j == len(matrix[i]) - 1)
        ]
        idx = 0
        for x, y in q:
            matrix[x][y] = 0
        while idx < len(q):
            x, y = q[idx]
            for dx, dy in [(-1, 0), (0, -1), (0, 1), (1, 0)]:
                nx, ny = x + dx, y + dy
                if 0 <= nx < len(matrix) and 0 <= ny < len(matrix[nx]) and matrix[nx][ny]:
                    matrix[nx][ny] = 0
                    q.append((nx, ny))
            idx += 1
        return sum(sum(row) for row in matrix)


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