Number of Distinct Islands
<p>You are given an <code>m x n</code> binary matrix <code>grid</code>. An island is a group of <code>1</code>&#39;s (representing land) connected <strong>4-directionally</strong> (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.</p>

<p>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.</p>

<p>Return <em>the number of <b>distinct</b> islands</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0600-0699/0694.Number%20of%20Distinct%20Islands/images/distinctisland1-1-grid.jpg" style="width: 413px; height: 334px;" />
<pre>
<strong>Input:</strong> grid = [[1,1,0,0,0],[1,1,0,0,0],[0,0,0,1,1],[0,0,0,1,1]]
<strong>Output:</strong> 1
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0600-0699/0694.Number%20of%20Distinct%20Islands/images/distinctisland1-2-grid.jpg" style="width: 413px; height: 334px;" />
<pre>
<strong>Input:</strong> grid = [[1,1,0,1,1],[1,0,0,0,0],[0,0,0,0,1],[1,1,0,1,1]]
<strong>Output:</strong> 3
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 50</code></li>
	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>
Depth-First Search; Breadth-First Search; Union Find; Hash Table; Hash Function
func numDistinctIslands(grid [][]int) int {
	m, n := len(grid), len(grid[0])
	paths := map[string]bool{}
	path := []byte{}
	dirs := [5]int{-1, 0, 1, 0, -1}
	var dfs func(i, j, k int)
	dfs = func(i, j, k int) {
		grid[i][j] = 0
		path = append(path, byte(k))
		for h := 1; h < 5; h++ {
			x, y := i+dirs[h-1], j+dirs[h]
			if x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 1 {
				dfs(x, y, h)
			}
		}
		path = append(path, byte(k))
	}
	for i, row := range grid {
		for j, x := range row {
			if x == 1 {
				dfs(i, j, 0)
				paths[string(path)] = true
				path = path[:0]
			}
		}
	}
	return len(paths)
}