# Posts tagged as “Graph”

There are `N` rooms and you start in room `0`.  Each room has a distinct number in `0, 1, 2, ..., N-1`, and each room may have some keys to access the next room.

Formally, each room `i` has a list of keys `rooms[i]`, and each key `rooms[i][j]` is an integer in `[0, 1, ..., N-1]` where `N = rooms.length`.  A key `rooms[i][j] = v` opens the room with number `v`.

Initially, all the rooms start locked (except for room `0`).

You can walk back and forth between rooms freely.

Return `true` if and only if you can enter every room.

Example 1:

```Input: [[1],[2],[3],[]]
Output: true
Explanation:
We start in room 0, and pick up key 1.
We then go to room 1, and pick up key 2.
We then go to room 2, and pick up key 3.
We then go to room 3.  Since we were able to go to every room, we return true.
```

Example 2:

```Input: [[1,3],[3,0,1],[2],[0]]
Output: false
Explanation: We can't enter the room with number 2.
```

Note:

1. `1 <= rooms.length <= 1000`
2. `0 <= rooms[i].length <= 1000`
3. The number of keys in all rooms combined is at most `3000`.

Idea:

DFS

Solution:

``````/**
* @param {number[][]} rooms
* @return {boolean}
*/
var canVisitAllRooms = function(rooms) {
let visited = [];
dfs(rooms, visited, 0);
return visited.length === rooms.length;
};

function dfs(rooms, visited, roomNum){
// exit recursion:
if (visited.includes(roomNum)) return;

visited.push(roomNum);
// possible solution
for (let key of rooms[roomNum])
dfs(rooms, visited, key);
}``````

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.

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'`.

Idea:

Traverse connected components: Use DFS to traverse 4 directions and mark them “0” if found “1”

Solution:

``````/**
* @param {character[][]} grid
* @return {number}
*/
var numIslands = function(grid) {
let row = grid.length;
if (row === 0) return 0;
let col = grid[0].length;
let res = 0;
for (let x = 0; x < col ; x++) {
for (let y = 0; y < row; y++) {
if (grid[y][x] === "1") {
res++;
dfs(grid, x, y, col, row);
}
}
}
return res;
};

var dfs = function(grid, x, y, n, m) {
// over boundry cases and not "1" => return
if (x < 0 || y < 0 || x >= n || y >= m || grid[y][x] === "0") return;

// already visited mark it "0"
grid[y][x] = "0";
// traverse 4 directions if any "1" => mark them "0"
dfs(grid, x + 1, y, n, m);
dfs(grid, x - 1, y, n, m);
dfs(grid, x, y + 1, n, m);
dfs(grid, x, y - 1, n, m);
}``````

Given a reference of a node in a connected undirected graph.

Return a deep copy (clone) of the graph.

Each node in the graph contains a val (`int`) and a list (`List[Node]`) of its neighbors.

```class Node {
public int val;
public List<Node> neighbors;
}
```

Test case format:

For simplicity sake, each node’s value is the same as the node’s index (1-indexed). For example, the first node with `val = 1`, the second node with `val = 2`, and so on. The graph is represented in the test case using an adjacency list.

Adjacency list is a collection of unordered lists used to represent a finite graph. Each list describes the set of neighbors of a node in the graph.

The given node will always be the first node with `val = 1`. You must return the copy of the given node as a reference to the cloned graph.

Example 1:

```Input: adjList = [[2,4],[1,3],[2,4],[1,3]]
Output: [[2,4],[1,3],[2,4],[1,3]]
Explanation: There are 4 nodes in the graph.
1st node (val = 1)'s neighbors are 2nd node (val = 2) and 4th node (val = 4).
2nd node (val = 2)'s neighbors are 1st node (val = 1) and 3rd node (val = 3).
3rd node (val = 3)'s neighbors are 2nd node (val = 2) and 4th node (val = 4).
4th node (val = 4)'s neighbors are 1st node (val = 1) and 3rd node (val = 3).
```

Example 2:

```Input: adjList = [[]]
Output: [[]]
Explanation: Note that the input contains one empty list. The graph consists of only one node with val = 1 and it does not have any neighbors.
```

Example 3:

```Input: adjList = []
Output: []
Explanation: This an empty graph, it does not have any nodes.
```

Example 4:

```Input: adjList = [[2],[1]]
Output: [[2],[1]]
```

Constraints:

• `1 <= Node.val <= 100`
• `Node.val` is unique for each node.
• Number of Nodes will not exceed 100.
• There is no repeated edges and no self-loops in the graph.
• The Graph is connected and all nodes can be visited starting from the given node.

Idea:

BFS Queue + HashMap +

• Time complexity: O(V+E) 点+边
• Space complexity: O(V+E)

Solution 1:

``````/**
* // Definition for a Node.
* function Node(val, neighbors) {
*    this.val = val === undefined ? 0 : val;
*    this.neighbors = neighbors === undefined ? [] : neighbors;
* };
*/
/**
* @param {Node} node
* @return {Node}
*/
var cloneGraph = function(node) {
if (node === null) return null;
let map = new Map();
map.set(node, new Node(node.val));
let q = []; // queue
q.push(node);
while (q.length !== 0) {
let cur = q.shift();
for (let n of cur.neighbors) {
if (!map.has(n)) {
map.set(n, new Node(n.val));
q.push(n);
}
map.get(cur).neighbors.push(map.get(n));
}
}
return map.get(node);
};``````

Solution 2:

``````var cloneGraph = function(node) {
if (node === null) return null;
let visited = new Map();
let map = new Map();
let q = []; // queue
q.push(node);
while (q.length !== 0) {
let cur = q.shift();
if (visited.has(cur)) continue;
visited.set(cur, true);
if (!map.has(cur)) map.set(cur, new Node(cur.val));
let temp = map.get(cur);
for (let n of cur.neighbors) {
if (!map.has(n)) map.set(n, new Node(n.val));
q.push(n);
temp.neighbors.push(map.get(n));
}
}
return map.get(node);
};``````