Problem Description

You are given a list of strings where every string is an anagram of every other string (they contain the same characters but possibly in different order).

Two strings X and Y are considered similar if:

• They are identical, OR
• We can make them equal by swapping at most two letters at distinct positions in string X

For example:

• "tars" and "rats" are similar (swap positions 0 and 2: t↔r and a↔t)
• "rats" and "arts" are similar (swap positions 0 and 2: r↔a and a↔r)
• "star" is NOT similar to "tars", "rats", or "arts"

Strings form groups based on similarity. A string belongs to a group if it is similar to at least one other string in that group. This creates a transitive relationship - if string A is similar to B, and B is similar to C, then A and C belong to the same group even if A and C are not directly similar.

From the example above:

• {"tars", "rats", "arts"} form one group (even though "tars" and "arts" are not directly similar, they are connected through "rats")
• {"star"} forms another group by itself

Your task is to determine how many such groups exist in the given list of anagram strings.

The solution uses a Union-Find data structure to track connected components. For each pair of strings, if they differ in at most 2 positions, they are similar and should be in the same group. The union operation merges their groups, and the final answer is the number of remaining distinct groups (connected components).

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: This problem involves strings that form connections based on similarity. Each string can be viewed as a node, and two nodes are connected if the strings are similar (differ by at most 2 character positions). We need to find connected components in this graph.

Is it a tree?

• No: The graph is not necessarily a tree. Multiple strings can be similar to each other forming cycles, and there can be multiple disconnected components.

Is the problem related to directed acyclic graphs?

• No: The similarity relationship is symmetric (if A is similar to B, then B is similar to A), making this an undirected graph problem.

Is the problem related to shortest paths?

• No: We're not finding shortest paths between nodes. We need to count the number of connected components (groups).

Does the problem involve connectivity?

• Yes: The core of the problem is finding how many connected groups exist. Strings in the same group are connected through similarity relationships, either directly or transitively.

Disjoint Set Union

• Yes: DSU (Union-Find) is the recommended approach for connectivity problems, especially when we need to count connected components.

Conclusion: The flowchart leads us to use Disjoint Set Union (Union-Find) for this connectivity problem. While DFS could also solve this by exploring each connected component, Union-Find is more efficient for counting groups as we can merge similar strings and track the number of distinct components directly.

Note: Although the solution uses Union-Find, DFS would also work by building an adjacency list of similar strings and performing DFS to mark all strings in each connected component, counting how many times we initiate a new DFS traversal.

Intuition

The key insight is recognizing that this is fundamentally a graph connectivity problem in disguise. Each string is like a person in a social network, and two people are "friends" if their names (strings) are similar enough (differ by at most 2 characters). We want to find how many friend groups exist where everyone in a group is connected through a chain of friendships.

Consider the example with "tars", "rats", "arts", and "star". Even though "tars" and "arts" aren't directly similar, they're in the same group because both are similar to "rats". This transitivity property is exactly what connected components in a graph represent.

Since all strings are anagrams, comparing any two strings for similarity is straightforward - we just count how many positions have different characters. If this count is 0 (identical) or 2 (one swap away), they're similar.

The challenge is efficiently grouping these strings. We could use DFS to explore each connected component, but Union-Find is particularly elegant here because:

1. We're iterating through pairs of strings anyway to check similarity
2. When we find two similar strings, we simply union them into the same group
3. The number of successful unions tells us how many times we merged groups
4. Starting with n individual groups (one per string), each successful union reduces the group count by 1

The beauty of this approach is that Union-Find handles the transitivity automatically. When we union "tars" with "rats", and later union "rats" with "arts", the data structure ensures all three end up in the same group without us explicitly checking if "tars" and "arts" should be together.

The final answer is simply the number of remaining distinct groups after all possible unions have been performed.

Learn more about Depth-First Search, Breadth-First Search and Union Find patterns.

Solution Approach

The solution implements Union-Find (Disjoint Set Union) with path compression and union by size optimization to efficiently manage the grouping of similar strings.

Union-Find Data Structure Setup:

The UnionFind class maintains two arrays:

• p: Parent array where p[x] stores the parent of node x. Initially, each node is its own parent (p[i] = i)
• size: Size array tracking the size of each component for union by size optimization

The find method implements path compression - when finding the root of a node, it updates all nodes along the path to point directly to the root, flattening the tree structure for faster future lookups.

The union method merges two components by:

1. Finding the roots of both nodes using find
2. If they're already in the same component, return False
3. Otherwise, attach the smaller component to the larger one (union by size) to keep the tree balanced
4. Return True to indicate a successful merge

Main Algorithm:

1. Initialize variables:

   • n = number of strings (also serves as our group counter)
   • m = length of each string
   • Create a Union-Find structure with n nodes (one for each string)

2. For each pair of strings (i, s) and (j, t) where j < i:

   • Count the number of differing positions: sum(s[k] != t[k] for k in range(m))
   • If the count is ≤ 2, the strings are similar
   • Attempt to union indices i and j
   • If union is successful (they weren't already in the same group), decrement n by 1

3. Return n as the final number of groups

Why this works:

Starting with n groups (each string in its own group), every successful union operation merges two groups into one, reducing the total count by 1. The condition sum(s[k] != t[k] for k in range(m)) <= 2 efficiently checks if two anagram strings are similar:

• 0 differences: strings are identical
• 2 differences: exactly one swap can make them equal
• More than 2: not similar

The nested loop structure for i, s in enumerate(strs): for j, t in enumerate(strs[:i]) ensures we check each pair exactly once, avoiding redundant comparisons.

Example Walkthrough

Let's walk through the algorithm with a small example: strs = ["tars", "rats", "arts", "star"]

Initial Setup:

• n = 4 (number of groups, initially each string is its own group)
• m = 4 (length of each string)
• Union-Find structure initialized with 4 nodes:
  • Parent array: [0, 1, 2, 3] (each node is its own parent)
  • Size array: [1, 1, 1, 1] (each group has size 1)

Step-by-Step Comparison:

1. i=1, s="rats": Compare with all previous strings

   • Compare with j=0, t="tars":
     • Differences at positions: 0 (r≠t), 2 (t≠r) → 2 differences
     • Similar! Union(1, 0) succeeds
     • Parent array becomes: [0, 0, 2, 3] (node 1 now points to 0)
     • n = 3 (one group merged)

2. i=2, s="arts": Compare with all previous strings

   • Compare with j=0, t="tars":
     • Differences at positions: 0 (a≠t), 1 (r≠a) → 2 differences
     • Similar! Union(2, 0) succeeds
     • Parent array becomes: [0, 0, 0, 3] (node 2 joins group with root 0)
     • n = 2
   • Compare with j=1, t="rats":
     • Differences at positions: 0 (a≠r), 2 (t≠a) → 2 differences
     • Similar! Union(2, 1) attempted
     • Both already have root 0, so no merge needed
     • n remains 2

3. i=3, s="star": Compare with all previous strings

   • Compare with j=0, t="tars":
     • Differences at positions: 0 (s≠t), 2 (a≠r), 3 (r≠s) → 3 differences
     • Not similar (>2 differences)
   • Compare with j=1, t="rats":
     • Differences at positions: 0 (s≠r), 1 (t≠a), 2 (a≠t), 3 (r≠s) → 4 differences
     • Not similar
   • Compare with j=2, t="arts":
     • Differences at positions: 0 (s≠a), 1 (t≠r), 3 (r≠s) → 3 differences
     • Not similar

Final Result:

• Parent array: [0, 0, 0, 3]
• Groups: {0, 1, 2} and {3}
• Return n = 2

The algorithm correctly identifies 2 groups:

• Group 1: ["tars", "rats", "arts"] - all connected through similarity
• Group 2: ["star"] - isolated, not similar to any other string

Solution Implementation

Time and Space Complexity

Time Complexity: O(n² × (m + α(n)))

The time complexity breaks down as follows:

• The outer loop iterates through all n strings: O(n)
• The inner loop iterates through up to i strings for each i, giving us O(n²) total iterations across both loops
• For each pair of strings being compared:
  • String comparison takes O(m) time to check character differences
  • The union operation calls find twice, each taking O(α(n)) amortized time due to path compression
  • Total per pair: O(m + α(n))
• Overall: O(n² × (m + α(n)))

Where n is the number of strings, m is the length of each string, and α(n) is the inverse Ackermann function (effectively a very small constant in practice).

Space Complexity: O(n)

The space complexity consists of:

• UnionFind data structure stores two arrays:
  • Parent array p: O(n)
  • Size array size: O(n)
• No additional significant space is used in the main algorithm
• Total: O(n)

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Similarity Check for Anagrams

A critical pitfall is assuming that two anagram strings differing in exactly 2 positions can always be made equal with one swap. This is only true if those 2 positions contain swapped characters.

Incorrect approach:

# Simply counting differences is NOT sufficient!
diff_count = sum(s[k] != t[k] for k in range(len(s)))
if diff_count <= 2:  # This can give false positives
    # Mark as similar

The problem: Consider "abcd" and "abdc". They differ at positions 2 and 3 (c≠d and d≠c). These ARE similar because swapping positions 2 and 3 in the first string gives the second. However, "abcd" and "efcd" also differ in exactly 2 positions (0 and 1), but they are NOT similar because no single swap can transform one into the other.

Correct approach:

def are_similar(s: str, t: str) -> bool:
    if s == t:
        return True
  
    diff_positions = []
    for i in range(len(s)):
        if s[i] != t[i]:
            diff_positions.append(i)
  
    # Must have exactly 2 differences for a swap to work
    if len(diff_positions) != 2:
        return False
  
    # Check if swapping would actually make them equal
    i, j = diff_positions
    return s[i] == t[j] and s[j] == t[i]

2. Inefficient String Comparison for Large Datasets

When the number of strings is small but string length is large, comparing all pairs becomes expensive. The current solution has O(n² × m) complexity where n is the number of strings and m is string length.

Optimization for when n >> m: Instead of comparing all pairs, group strings by their sorted character representation first, then only compare within groups:

def numSimilarGroups(self, strs: List[str]) -> int:
    from collections import defaultdict
  
    # Group anagrams together first
    anagram_groups = defaultdict(list)
    for i, s in enumerate(strs):
        key = ''.join(sorted(s))
        anagram_groups[key].append(i)
  
    n = len(strs)
    uf = UnionFind(n)
    num_groups = n
  
    # Only compare strings within the same anagram group
    for indices in anagram_groups.values():
        for i in range(len(indices)):
            for j in range(i + 1, len(indices)):
                idx1, idx2 = indices[i], indices[j]
                if are_similar(strs[idx1], strs[idx2]):
                    if uf.union(idx1, idx2):
                        num_groups -= 1
  
    return num_groups

3. Memory Issues with Large String Lists

The Union-Find structure uses O(n) space, but the nested loop approach can cause memory issues when creating intermediate data structures.

Solution: Use generators and iterate efficiently:

# Instead of creating a slice with strs[:i]
for j in range(i):  # More memory efficient
    if are_similar(strs[i], strs[j]):
        # Process similarity

4. Not Handling Edge Cases

Common missed cases:

• Empty list of strings
• Single string in the list
• All identical strings
• Strings of different lengths (though the problem states they're anagrams)

Robust implementation:

def numSimilarGroups(self, strs: List[str]) -> int:
    if not strs:
        return 0
    if len(strs) == 1:
        return 1
  
    # Validate that all strings are anagrams (same length, same character frequency)
    if len(set(len(s) for s in strs)) > 1:
        raise ValueError("All strings must have the same length")
  
    # Continue with main algorithm...

The most critical pitfall is #1 - the similarity check must verify that the differing positions actually contain swapped characters, not just count the differences. This is essential for correctness when working with anagram strings.