# Number of Distinct Islands

MEDIUM

Given a non-empty 2D array `grid` of 0’s and 1’s, an island is a group of `1`‘s (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.

Count the number of distinct islands. An island is considered to be the same as another if and only if one island can be translated (and not rotated or reflected) to equal the other.

Example 1:

```11000
11000
00011
00011
```

Given the above grid map, return `1`.

Example 2:

```11011
10000
00001
11011```

Given the above grid map, return `3`.

Notice that:

```11
1
```

and

``` 1
11
```

are considered different island shapes, because we do not consider reflection / rotation.

Note: The length of each dimension in the given `grid` does not exceed 50.

We can use direction string the calculate the uniqueness of each island.

```class Solution {
public int numDistinctIslands(int[][] grid) {

if(grid == null || grid.length == 0 || grid.length == 0){
return 0;
}
int m = grid.length;
int n = grid.length;
Set<String> distinctIsland = new HashSet<>();
for(int i = 0 ; i < m ; i++){
for(int j = 0 ; j < n ; j++){
if(grid[i][j] == 1){
StringBuilder encodeIsland = new StringBuilder();
DFS(grid, m, n, i, j, encodeIsland, "o"); // origin
String island = encodeIsland.toString();
if(!distinctIsland.contains(island)){
}
}
}
}
return distinctIsland.size();
}

private void DFS(int[][] grid, int m, int n, int i, int j,
StringBuilder encodeIsland, String direction) {
if (i < 0 || i == m || j < 0 || j == n || grid[i][j] == 0) {
return;
}
grid[i][j] = 0;
encodeIsland.append("1").append(direction);
DFS(grid, m, n, i + 1, j, encodeIsland, "b");  // bottom
DFS(grid, m, n, i - 1, j, encodeIsland, "t");  // top
DFS(grid, m, n, i, j + 1, encodeIsland, "r");  // right
DFS(grid, m, n, i, j - 1, encodeIsland, "l");  // left
encodeIsland.append("e"); // end
}
}
```

Complexity Analysis

• Time Complexity: `O(R∗C)`, where R is the number of rows in the given `grid`, and C is the number of columns. We visit every square once.
• Space Complexity: `O(R∗C)`, the space used by `seen` to keep track of visited squares, and `shapes`.

Categories: DFS

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