How do you solve Similar String Groups on LeetCode?

Similar String Groups (LeetCode #839) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Understanding the similarity condition is crucial for forming groups.

What is the brute force solution for Similar String Groups?

The brute force approach for Similar String Groups is Brute Force, with O(n²) time complexity and O(1) space complexity. We can compare each string with every other string to check if they are similar. This is straightforward but inefficient, as it involves checking all pairs of strings. The optimal Optimal Solution approach improves this to O(n²).

What are the interview tips for Similar String Groups?

Clearly explain your thought process and the reasoning behind your approach. Be prepared to discuss the trade-offs between different methods. Practice implementing union-find as it is a common pattern in graph-related problems.

What are common mistakes when solving Similar String Groups?

Not considering the case where strings are identical. Failing to optimize the union-find operations, leading to inefficient group counting.

#839

Similar String Groups

Hard

ArrayHash TableStringDepth-First SearchBreadth-First SearchUnion-FindHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n²)Space O(n)

Using a union-find (disjoint set) data structure allows us to group similar strings efficiently. We can union strings that are similar and count the distinct groups at the end.

⚙️

Algorithm

3 steps

1Step 1: Initialize a union-find structure to manage string groups.
2Step 2: For each pair of strings, if they are similar, union them in the structure.
3Step 3: Count the number of unique roots in the union-find structure to determine the number of groups.

solution.py41 lines

1# Full working Python code
2class UnionFind:
3    def __init__(self, size):
4        self.parent = list(range(size))
5
6    def find(self, x):
7        if self.parent[x] != x:
8            self.parent[x] = self.find(self.parent[x])
9        return self.parent[x]
10
11    def union(self, x, y):
12        rootX = self.find(x)
13        rootY = self.find(y)
14        if rootX != rootY:
15            self.parent[rootY] = rootX
16
17def are_similar(s1, s2):
18    if s1 == s2:
19        return True
20    diff = []
21    for a, b in zip(s1, s2):
22        if a != b:
23            diff.append((a, b))
24        if len(diff) > 2:
25            return False
26    return len(diff) == 2 and diff[0] == diff[1][::-1]
27
28def num_similar_groups(strs):
29    n = len(strs)
30    uf = UnionFind(n)
31
32    for i in range(n):
33        for j in range(i + 1, n):
34            if are_similar(strs[i], strs[j]):
35                uf.union(i, j)
36
37    groups = len(set(uf.find(i) for i in range(n)))
38    return groups
39
40# Example usage
41print(num_similar_groups(['tars', 'rats', 'arts', 'star']))  # Output: 2

ℹ

Complexity note: The time complexity remains O(n²) due to the pairwise comparison of strings. However, the space complexity is O(n) because we store the parent array for union-find operations.

1Understanding the similarity condition is crucial for forming groups.
2Union-Find is a powerful technique for grouping connected components.

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