How do you solve Groups of Strings on LeetCode?

Groups of Strings (LeetCode #2157) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m²) time complexity and O(n) space complexity. Key insight: Understanding how to determine connectivity between strings is crucial.

What is the brute force solution for Groups of Strings?

The brute force approach for Groups of Strings is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check every pair of strings to see if they are connected. This involves comparing each string with every other string in the list, which can be slow but is straightforward. The optimal Optimal Solution approach improves this to O(n * m²).

What algorithm or data structure is used to solve Groups of Strings?

Groups of Strings uses the following concepts: Array, Hash Table, String, Bit Manipulation, Union-Find. The recommended patterns to study are: Union-Find, Graph Traversal.

What are the interview tips for Groups of Strings?

Explain your thought process clearly while coding. Consider edge cases and test your code with them. Be prepared to discuss the trade-offs of different approaches.

What are common mistakes when solving Groups of Strings?

Not properly implementing the union-find structure, leading to incorrect group counts. Failing to consider edge cases, such as strings with lengths differing by more than one.

#2157

Groups of Strings

Hard

ArrayHash TableStringBit ManipulationUnion-FindUnion-FindGraph Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * m²)
Space	O(1)	O(n)

💡

Intuition

Time O(n * m²)Space O(n)

We can use a Union-Find (Disjoint Set Union) structure to efficiently group connected strings. This allows us to quickly find and union groups without checking every pair explicitly.

⚙️

Algorithm

3 steps

1Step 1: Initialize a Union-Find structure to manage groups.
2Step 2: For each string, compare it with all others to determine if they are connected, and union them if they are.
3Step 3: Count the number of unique groups and find the size of the largest group.

solution.py44 lines

1# Full working Python code
2class UnionFind:
3    def __init__(self, size):
4        self.parent = list(range(size))
5        self.rank = [1] * size
6
7    def find(self, x):
8        if self.parent[x] != x:
9            self.parent[x] = self.find(self.parent[x])
10        return self.parent[x]
11
12    def union(self, x, y):
13        rootX = self.find(x)
14        rootY = self.find(y)
15        if rootX != rootY:
16            if self.rank[rootX] > self.rank[rootY]:
17                self.parent[rootY] = rootX
18            elif self.rank[rootX] < self.rank[rootY]:
19                self.parent[rootX] = rootY
20            else:
21                self.parent[rootY] = rootX
22                self.rank[rootX] += 1
23
24def groups_of_strings(words):
25    n = len(words)
26    uf = UnionFind(n)
27
28    for i in range(n):
29        for j in range(i + 1, n):
30            if are_connected(words[i], words[j]):
31                uf.union(i, j)
32
33    group_count = len(set(uf.find(i) for i in range(n)))
34    group_sizes = [0] * n
35    for i in range(n):
36        group_sizes[uf.find(i)] += 1
37    max_group_size = max(group_sizes)
38    return [group_count, max_group_size]
39
40def are_connected(s1, s2):
41    set1, set2 = set(s1), set(s2)
42    if abs(len(set1) - len(set2)) > 1:
43        return False
44    return len(set1.symmetric_difference(set2)) <= 1

ℹ

Complexity note: This complexity arises because each string can be compared with every other string, and the union-find operations are nearly constant time due to path compression.

1Understanding how to determine connectivity between strings is crucial.
2Using Union-Find can significantly optimize the grouping process.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.