Problem
The grid holds 1 for land and 0 for water. Land cells that touch up, down, left or right belong to the same island. Count the islands; a recursive flood fill is a natural approach.
Implement count_islands(grid).
Input
grid: A rectangular grid of 0s and 1s.
Output
Return the number of islands.
Constraints
0 ≤ rows, columns ≤ 100
Examples
Input:
grid = [
[1, 1, 0, 0],
[0, 1, 0, 1],
[0, 0, 0, 1]
]
Expected output:
expected = 2Input:
std::vector<std::vector<long long>> grid = {
{1, 1, 0, 0},
{0, 1, 0, 1},
{0, 0, 0, 1}
};
Expected output:
long long expected = 2;Input:
long[][] grid = new long[][] {
new long[] {1, 1, 0, 0},
new long[] {0, 1, 0, 1},
new long[] {0, 0, 0, 1}
};
Expected output:
long expected = 2;Explanation
The top-left group of 3 cells is one island; the two 1s on the right form a second: 2.
Input:
grid = [
[1, 0, 1],
[0, 1, 0]
]
Expected output:
expected = 3Input:
std::vector<std::vector<long long>> grid = {
{1, 0, 1},
{0, 1, 0}
};
Expected output:
long long expected = 3;Input:
long[][] grid = new long[][] {
new long[] {1, 0, 1},
new long[] {0, 1, 0}
};
Expected output:
long expected = 3;Explanation
Diagonal neighbors don't connect, so each 1 is its own island: 3.
How to approach it
Try on your own first, then open one hint at a time.
Hint 1
Each land cell belongs to one island if it connects vertically or horizontally.
Hint 2
When you find new land, flood-fill its whole region so you do not count it again.
Hint 3
Watch the grid boundaries carefully during the recursive exploration.
Expected complexity
O(rows × columns) time, O(rows × columns) extra space
References
This challenge is adapted for CodeUtility. Follow the statement and constraints above.