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
function numDistinctIslands(grid: number[][]): number {
    const m = grid.length;
    const n = grid[0].length;
    const paths: Set<string> = new Set();
    const path: number[] = [];
    const dirs: number[] = [-1, 0, 1, 0, -1];
    const dfs = (i: number, j: number, k: number) => {
        grid[i][j] = 0;
        path.push(k);
        for (let h = 1; h < 5; ++h) {
            const [x, y] = [i + dirs[h - 1], j + dirs[h]];
            if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y]) {
                dfs(x, y, h);
            }
        }
        path.push(k);
    };
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            if (grid[i][j]) {
                dfs(i, j, 0);
                paths.add(path.join(','));
                path.length = 0;
            }
        }
    }
    return paths.size;
}
