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Groups of Strings

Difficulty: Hard

Problem Description

You are given a 0-indexed array of strings words. Each string consists of lowercase English letters only. No letter occurs more than once in any string of words. Two strings s1 and s2 are said to be connected if the set of letters of s2 can be obtained from the set of letters of s1 by any one of the following operations:

  • Adding exactly one letter to the set of the letters of s1.
  • Deleting exactly one letter from the set of the letters of s1.
  • Replacing exactly one letter from the set of the letters of s1 with any letter, including itself.

The array words can be divided into one or more non-intersecting groups. A string belongs to a group if any one of the following is true:

  • It is connected to at least one other string of the group.
  • It is the only string present in the group.

Return an array ans of size 2 where:

  • ans[0] is the maximum number of groups words can be divided into,
  • ans[1] is the size of the largest group.

Key Insights

  • Strings can be treated as nodes in a graph where edges represent the connection between them.
  • The problem can be solved using graph traversal techniques like Depth-First Search (DFS) or Union-Find to find connected components.
  • The key operations (add, delete, replace) can be abstracted into bit manipulation by representing each string as a bitmask of letters.

Space and Time Complexity

Time Complexity: O(n * m^2) where n is the number of strings and m is the maximum length of the strings, due to checking connections between strings. Space Complexity: O(n + m) for storing the graph and visited nodes.

Solution

To solve the problem, we can use a Union-Find data structure to dynamically group connected strings. Each string is represented by a bitmask where each bit corresponds to a letter. We check for connections by modifying the bitmask to see if we can add, delete, or replace a letter.

  1. Create a Union-Find structure to manage connected components.
  2. For each pair of strings, check if they can be connected using the defined operations.
  3. Merge groups in the Union-Find structure accordingly.
  4. Finally, calculate the number of unique groups and the size of the largest group.

Code Solutions

class UnionFind:
    def __init__(self, size):
        self.parent = list(range(size))
        self.rank = [1] * size

    def find(self, u):
        if self.parent[u] != u:
            self.parent[u] = self.find(self.parent[u])  # Path compression
        return self.parent[u]

    def union(self, u, v):
        rootU = self.find(u)
        rootV = self.find(v)
        if rootU != rootV:
            if self.rank[rootU] > self.rank[rootV]:
                self.parent[rootV] = rootU
            elif self.rank[rootU] < self.rank[rootV]:
                self.parent[rootU] = rootV
            else:
                self.parent[rootV] = rootU
                self.rank[rootU] += 1

def groupsOfStrings(words):
    n = len(words)
    uf = UnionFind(n)
    
    # Create a mapping of each word to its bitmask
    bitmasks = []
    for word in words:
        bitmask = 0
        for char in word:
            bitmask |= (1 << (ord(char) - ord('a')))
        bitmasks.append(bitmask)

    # Check connections
    for i in range(n):
        for j in range(i + 1, n):
            # Count the number of differing bits
            diff = bitmasks[i] ^ bitmasks[j]
            # Check if they can be connected
            if bin(diff).count('1') <= 1:  # One or zero differing bits
                uf.union(i, j)

    # Count groups and size of largest group
    group_count = {}
    max_group_size = 0
    for i in range(n):
        root = uf.find(i)
        if root not in group_count:
            group_count[root] = 0
        group_count[root] += 1
        max_group_size = max(max_group_size, group_count[root])

    return [len(group_count), max_group_size]
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