Number of Islands - Amazon Top Interview Questions


Problem Statement :


Given a two-dimensional integer matrix of 1s and 0s, return the number of "islands" in the matrix. A 1 represents land and 0 represents water, so an island is a group of 1s that are neighboring whose perimeter is surrounded by water.

Note: Neighbors can only be directly horizontal or vertical, not diagonal.

Constraints

n, m ≤ 100 where n and m are the number of rows and columns in matrix.

Example 1

Input

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

Output

1

Example 2

Input

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

Output

4

Example 3

Input

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

Output

2


Solution :



title-img 




                        Solution in C++ :

void dfs(vector<vector<int>>& matrix, int i, int j) {
    if (i < 0 or i >= matrix.size() or j < 0 or j >= matrix[i].size() or !matrix[i][j]) return;
    matrix[i][j] = 0;
    dfs(matrix, i - 1, j);
    dfs(matrix, i + 1, j);
    dfs(matrix, i, j - 1);
    dfs(matrix, i, j + 1);
}

int solve(vector<vector<int>>& matrix) {
    int ans = 0;
    for (int i = 0; i < matrix.size(); i++) {
        for (int j = 0; j < matrix[i].size(); j++) {
            if (matrix[i][j] == 1) {
                ans++;
                dfs(matrix, i, j);
            }
        }
    }
    return ans;
}
                    




                        Solution in Python : 
                            
class Solution:
    def solve(self, matrix):
        n, m = len(matrix), len(matrix[0])

        # append dummy values to the bottom row and right column
        matrix.append([0] * m)
        for i in range(n + 1):
            matrix[i].append(0)

        # flood fill
        def ff(i, j):
            if matrix[i][j]:  # no need to check for bounds!
                matrix[i][j] = 0
                ff(i + 1, j)
                ff(i - 1, j)
                ff(i, j + 1)
                ff(i, j - 1)

        # count number of fills needed
        c = 0
        for i in range(n):
            for j in range(m):
                if matrix[i][j]:
                    c += 1
                    ff(i, j)
        return c


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