LeetCode Link: 200. Number of Islands
Language: C#
Problem Statement
Given an m x n 2D binary grid grid which represents a map of ‘1’s (land) and ‘0’s (water), return the number of islands.
An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.
Examples
Example 1:
Input: grid = [
[“1”,”1”,”1”,”1”,”0”],
[“1”,”1”,”0”,”1”,”0”],
[“1”,”1”,”0”,”0”,”0”],
[“0”,”0”,”0”,”0”,”0”]
]
Output: 1
Example 2:
Input: grid = [
[“1”,”1”,”0”,”0”,”0”],
[“1”,”1”,”0”,”0”,”0”],
[“0”,”0”,”1”,”0”,”0”],
[“0”,”0”,”0”,”1”,”1”]
]
Output: 3
Constraints
- m == grid.length
- n == grid[i].length
- 1 <= m, n <= 300
- grid[i][j] is ‘0’ or ‘1’.
Solution
public class Solution
{
public int NumIslands(char[][] grid)
{
int count = 0;
for (int i=0; i<grid.Length; i++)
{
for (int j=0; j< grid[i].Length; j++)
{
if (grid[i][j] == '1')
{
count++;
}
DFS(grid, i, j);
}
}
return count;
}
private void DFS(char[][] grid, int i, int j)
{
if (i < 0 || j < 0 || i > grid.Length - 1 || j > grid[i].Length -1 || grid[i][j] == '0')
{
return;
}
if (grid[i][j] == '1')
{
grid[i][j] = '0';
DFS(grid, i-1, j); // UP
DFS(grid, i, j-1); // LEFT
DFS(grid, i, j+1); // RIGHT
DFS(grid, i+1, j); // DOWN
}
}
}
Complexity
Time Complexity: O(N)
Space Complexity: O(N)