# LeetCode – Number of Islands (Java)

Given a 2-d grid map of '1's (land) and '0's (water), count 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.

Example 1:

```11110
11010
11000
00000
```

Java Solution 1 - DFS

The basic idea of the following solution is merging adjacent lands, and the merging should be done recursively.

Each element is visited once only. So time is O(m*n).

```public int numIslands(char[][] grid) { if(grid==null || grid.length==0||grid[0].length==0) return 0;   int m = grid.length; int n = grid[0].length;   int count=0; for(int i=0; i<m; i++){ for(int j=0; j<n; j++){ if(grid[i][j]=='1'){ count++; merge(grid, i, j); } } }   return count; }   public void merge(char[][] grid, int i, int j){ int m=grid.length; int n=grid[0].length;   if(i<0||i>=m||j<0||j>=n||grid[i][j]!='1') return;   grid[i][j]='X';   merge(grid, i-1, j); merge(grid, i+1, j); merge(grid, i, j-1); merge(grid, i, j+1); }```

Java Solution 2 - Union-Find

Time is O(m*n*log(k)).

```public int numIslands(char[][] grid) { if(grid==null || grid.length==0 || grid[0].length==0) return 0;   int m = grid.length; int n = grid[0].length;   int[] dx={-1, 1, 0, 0}; int[] dy={0, 0, -1, 1};   int[] root = new int[m*n];   int count=0; for(int i=0; i<m; i++){ for(int j=0; j<n; j++){ if(grid[i][j]=='1'){ root[i*n+j] = i*n+j; count++; } } }   for(int i=0; i<m; i++){ for(int j=0; j<n; j++){ if(grid[i][j]=='1'){ for(int k=0; k<4; k++){ int x = i+dx[k]; int y = j+dy[k];   if(x>=0&&x<m&&y>=0&&y<n&&grid[x][y]=='1'){ int cRoot = getRoot(root, i*n+j); int nRoot = getRoot(root, x*n+y); if(nRoot!=cRoot){ root[cRoot]=nRoot; //update previous node's root to be current count--; }   } } } } }   return count; }   public int getRoot(int[] arr , int i){ while(arr[i]!=i){ i = arr[arr[i]]; }   return i; }```

Check out Number of Island II.

Category >> Algorithms >> Interview
If you want someone to read your code, please put the code inside <pre><code> and </code></pre> tags. For example:
```<pre><code>
String foo = "bar";
</code></pre>
```
• Sathish Kumar

int[] dx={-1, 1, 0, 0};
int[] dy={0, 0, -1, 1};

why are we having a hard coded array? can someone please explain why this is needed ?

• Piyush Dubey

What will be the space complexity of first solution (using DFS) ?

• Misha Arora

In the second method, complexity is m*n*log(k). What is k??

• Sudhir

isn’t the time complexity of DFS O(m)+O(n) where m is number of vertices and n is number of edges.??

• foo

def num_islands():

visited = set()

islands = 0

for i in range(4):

for j in range(5):

if m[i][j] == 1:

if i == 0 and j == 0:

islands += 1

elif i == 0:

if m[i][j-1] == 1:

continue # we have visited this island

else:

islands += 1

else:

if (i-1,j-1) in visited or

(i-1,j+1) in visited or

(i-1,j) in visited or

(i,j-1) in visited:

continue

else:

islands += 1

print visited

return islands

• DachiCoding

It’s necessary to call merge(gird,i-1,j) and merge(grid,i,j-1).

e.g
11111
00100
00101 <- this guy cannot be achieved by merge(grid,i+1,j) or merge(grid,i,j+1)
00111

• Guus

You don’t need to call `merge(grid, i - 1, j)` and `merge(grid, i, j - 1)`.