Regions Cut By Slashes
An n x n grid is composed of 1 x 1 squares where each 1 x 1 square consists of a '/', '\', or blank space ' '. These characters divide the square into contiguous regions.
Given the grid grid represented as a string array, return the number of regions.
Note that backslash characters are escaped, so a '\' is represented as '\\'.
Example 1:
Input: grid = [" /","/ "]
Output: 2
Example 2:
Input: grid = [" /"," "]
Output: 1
Example 3:
Input: grid = ["/\\\\","\\\\/"]
Output: 5
Explanation: Recall that because \ characters are escaped, "\\/" refers to \/, and "/\\" refers to /\.
Constraints:
n == grid.length == grid[i].length1 <= n <= 30grid[i][j]is either'/','\', or' '.
Solution
With the cutted sections, we will join the sections depending on the slashes. For example, '/' means that north joins west and south joins east.
Note that we must also join the north region of a square to the south region of the square to the top of it. Similar for the left square's east region and current square's west region.
This is best approached by DSU, in the very end after joining all sections, we just have to find the number of components in the map regions.
Implementation
1def regionsBySlashes(grid: List[str]) -> int:
2 regions = {}
3 def find(x):
4 y = regions.get(x, x)
5 if y != x:
6 regions[x] = y = find(y)
7 return y
8 def union(x, y):
9 regions[find(x)] = find(y)
10
11 size = len(grid)
12 for i in range(size):
13 for j in range(size):
14 if i > 0: # connect square with top
15 union((i - 1, j, 'S'), (i, j, 'N'))
16 if j > 0: # connect square with left
17 union((i, j - 1, 'E'), (i, j, 'W'))
18 if grid[i][j] != "/": # ' ' or '\'
19 union((i, j, 'N'), (i, j, 'E')) # connect NE
20 union((i, j, 'S'), (i, j, 'W')) # connect SW
21 if grid[i][j] != "\\": # ' ' or '/'
22 union((i, j, 'N'), (i, j, 'W')) # connect NW
23 union((i, j, 'S'), (i, j, 'E')) # connect SE
24 return len(set(map(find, regions)))
Alternatively, we can find the number of connected components as in Number of Coneected Components, but bare in mind that we should only decrease the component count when we know that two elements aren't from the same set.
What's the output of running the following function using input [30, 20, 10, 100, 33, 12]?
1def fun(arr: List[int]) -> List[int]:
2 import heapq
3 heapq.heapify(arr)
4 res = []
5 for i in range(3):
6 res.append(heapq.heappop(arr))
7 return res
81public static int[] fun(int[] arr) {
2 int[] res = new int[3];
3 PriorityQueue<Integer> heap = new PriorityQueue<>();
4 for (int i = 0; i < arr.length; i++) {
5 heap.add(arr[i]);
6 }
7 for (int i = 0; i < 3; i++) {
8 res[i] = heap.poll();
9 }
10 return res;
11}
121class HeapItem {
2 constructor(item, priority = item) {
3 this.item = item;
4 this.priority = priority;
5 }
6}
7
8class MinHeap {
9 constructor() {
10 this.heap = [];
11 }
12
13 push(node) {
14 // insert the new node at the end of the heap array
15 this.heap.push(node);
16 // find the correct position for the new node
17 this.bubble_up();
18 }
19
20 bubble_up() {
21 let index = this.heap.length - 1;
22
23 while (index > 0) {
24 const element = this.heap[index];
25 const parentIndex = Math.floor((index - 1) / 2);
26 const parent = this.heap[parentIndex];
27
28 if (parent.priority <= element.priority) break;
29 // if the parent is bigger than the child then swap the parent and child
30 this.heap[index] = parent;
31 this.heap[parentIndex] = element;
32 index = parentIndex;
33 }
34 }
35
36 pop() {
37 const min = this.heap[0];
38 this.heap[0] = this.heap[this.size() - 1];
39 this.heap.pop();
40 this.bubble_down();
41 return min;
42 }
43
44 bubble_down() {
45 let index = 0;
46 let min = index;
47 const n = this.heap.length;
48
49 while (index < n) {
50 const left = 2 * index + 1;
51 const right = left + 1;
52
53 if (left < n && this.heap[left].priority < this.heap[min].priority) {
54 min = left;
55 }
56 if (right < n && this.heap[right].priority < this.heap[min].priority) {
57 min = right;
58 }
59 if (min === index) break;
60 [this.heap[min], this.heap[index]] = [this.heap[index], this.heap[min]];
61 index = min;
62 }
63 }
64
65 peek() {
66 return this.heap[0];
67 }
68
69 size() {
70 return this.heap.length;
71 }
72}
73
74function fun(arr) {
75 const heap = new MinHeap();
76 for (const x of arr) {
77 heap.push(new HeapItem(x));
78 }
79 const res = [];
80 for (let i = 0; i < 3; i++) {
81 res.push(heap.pop().item);
82 }
83 return res;
84}
85Which type of traversal does breadth first search do?
Solution Implementation
Depth first search can be used to find whether two components in a graph are connected.
The three-steps of Depth First Search are:
- Identify states;
- Draw the state-space tree;
- DFS on the state-space tree.
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