**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 :

` ````
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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