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# Posts tagged as “union find”

In a N x N grid composed of 1 x 1 squares, each 1 x 1 square consists of a /\, or blank space.  These characters divide the square into contiguous regions.

(Note that backslash characters are escaped, so a \ is represented as "\\".)

Return the number of regions.

Example 1:

Input:[  " /",  "/ "]Output: 2Explanation: The 2x2 grid is as follows:

Example 2:

Input:[  " /",  "  "]Output: 1Explanation: The 2x2 grid is as follows:

Example 3:

Input:[  "\\/",  "/\\"]Output: 4Explanation: (Recall that because \ characters are escaped, "\\/" refers to \/, and "/\\" refers to /\.)The 2x2 grid is as follows:

Example 4:

Input:[  "/\\",  "\\/"]Output: 5Explanation: (Recall that because \ characters are escaped, "/\\" refers to /\, and "\\/" refers to \/.)The 2x2 grid is as follows:

Example 5:

Input:[  "//",  "/ "]Output: 3Explanation: The 2x2 grid is as follows:

Note:

1. 1 <= grid.length == grid[0].length <= 30
2. grid[i][j] is either '/''\', or ' '.

# Solution 1: Split grid into 4 triangles and Union Find Faces

Divide each grid into 4 triangles and union them if not split.
Time complexity: O(n^2*alphn(n^2))
Space complexity: O(n^2)

# Solution 3: Pixelation (Upscale 3 times)

Time complexity: O(n^2)
Space complexity: O(n^2)

# Problem

Given a non-empty array of unique positive integers A, consider the following graph:

• There are A.length nodes, labelled A[0] to A[A.length - 1];
• There is an edge between A[i] and A[j] if and only if A[i] and A[j] share a common factor greater than 1.

Return the size of the largest connected component in the graph.

Example 1:

Input: [4,6,15,35]
Output: 4



Example 2:

Input: [20,50,9,63]
Output: 2



Example 3:

Input: [2,3,6,7,4,12,21,39]
Output: 8



Note:

1. 1 <= A.length <= 20000
2. 1 <= A[i] <= 100000

# Solution: Union Find

For each number, union itself with all its factors.

E.g. 6, union(6,2), union(6,3)

Time complexity: $$O(\Sigma{sqrt(A[i])})$$

Space complexity: $$O(max(A))$$

# Problem

On a 2D plane, we place stones at some integer coordinate points.  Each coordinate point may have at most one stone.

Now, a move consists of removing a stone that shares a column or row with another stone on the grid.

What is the largest possible number of moves we can make?

Example 1:

Input: stones = [[0,0],[0,1],[1,0],[1,2],[2,1],[2,2]]
Output: 5


Example 2:

Input: stones = [[0,0],[0,2],[1,1],[2,0],[2,2]]
Output: 3


Example 3:

Input: stones = [[0,0]]
Output: 0


Note:

1. 1 <= stones.length <= 1000
2. 0 <= stones[i][j] < 10000

# Solution 2: Union Find

Find all connected components (islands)

Ans = # of stones – # of islands

# Problem

Two strings X and Y are similar if we can swap two letters (in different positions) of X, so that it equals Y.

For example, "tars" and "rats" are similar (swapping at positions 0 and 2), and "rats" and "arts" are similar, but "star" is not similar to "tars""rats", or "arts".

Together, these form two connected groups by similarity: {"tars", "rats", "arts"} and {"star"}.  Notice that "tars" and "arts" are in the same group even though they are not similar.  Formally, each group is such that a word is in the group if and only if it is similar to at least one other word in the group.

We are given a list A of strings.  Every string in A is an anagram of every other string in A.  How many groups are there?

Example 1:

Input: ["tars","rats","arts","star"]
Output: 2

Note:

1. A.length <= 2000
2. A[i].length <= 1000
3. A.length * A[i].length <= 20000
4. All words in A consist of lowercase letters only.
5. All words in A have the same length and are anagrams of each other.
6. The judging time limit has been increased for this question.

# Solution: Brute Force + Union Find

Time Complexity: O(n^2 * L)

Space Complexity: O(n)

C++

Problem:

Equations are given in the format A / B = k, where A and B are variables represented as strings, and k is a real number (floating point number). Given some queries, return the answers. If the answer does not exist, return -1.0.

Example:
Given a / b = 2.0, b / c = 3.0.
queries are: a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ? .
return [6.0, 0.5, -1.0, 1.0, -1.0 ].

The input is: vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries , where equations.size() == values.size(), and the values are positive. This represents the equations. Return vector<double>.

According to the example above:

The input is always valid. You may assume that evaluating the queries will result in no division by zero and there is no contradiction.

# Solution 2: Union Find

## Python3

Related Problems: