#2157

Hard

auto_awesomeString plus Bit Manipulation

LeetCode Problem Workspace

Groups of Strings

Group words into connected sets using operations on characters with string and bit manipulation techniques.

Problem Statement

You are given a list of strings where each string contains distinct lowercase English letters. Two strings are connected if one can be transformed into the other by adding, deleting, or modifying one letter. You need to determine how many distinct groups of connected strings exist in the list.

Each string is part of a group if it can be transformed into any other string in the same group. The goal is to compute the size of the largest group and the count of such groups. You can solve this by using graph-like methods with bit manipulation or union-find to track connections between strings.

Examples

Example 1

Input: words = ["a","b","ab","cde"]

Output: [2,3]

words[0] can be used to obtain words[1] (by replacing 'a' with 'b'), and words[2] (by adding 'b'). So words[0] is connected to words[1] and words[2].
words[1] can be used to obtain words[0] (by replacing 'b' with 'a'), and words[2] (by adding 'a'). So words[1] is connected to words[0] and words[2].
words[2] can be used to obtain words[0] (by deleting 'b'), and words[1] (by deleting 'a'). So words[2] is connected to words[0] and words[1].
words[3] is not connected to any string in words. Thus, words can be divided into 2 groups ["a","b","ab"] and ["cde"]. The size of the largest group is 3.

Example 2

Input: words = ["a","ab","abc"]

Output: [1,3]

words[0] is connected to words[1].
words[1] is connected to words[0] and words[2].
words[2] is connected to words[1]. Since all strings are connected to each other, they should be grouped together. Thus, the size of the largest group is 3.

Constraints

1 <= words.length <= 2 * 104
1 <= words[i].length <= 26
words[i] consists of lowercase English letters only.
No letter occurs more than once in words[i].

Solution Approach

Union-Find (Disjoint Set Union)

One of the most efficient ways to solve this problem is by using a union-find (disjoint-set) approach. Each word can be treated as a node in a graph, and an edge can be created between two nodes if the words are connected. Union-find allows us to group these nodes into sets based on their connectivity. This method efficiently handles the problem's constraints.

Bit Manipulation

Bit manipulation is particularly useful here as we can represent sets of characters using bits. Each word can be encoded as a bitmask where each bit represents the presence of a specific character. By comparing bitmasks, we can easily determine if two words are connected by checking if they differ by one bit (for a single character transformation).

Graph Representation

Another approach is to represent the problem as a graph where each string is a node and an edge exists if the two strings are connected. By building a graph and using BFS or DFS, we can traverse the connected components and group the words. This method is intuitive but may not be as time-efficient as union-find or bit manipulation.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the chosen approach. With union-find, the time complexity is nearly O(N * α(N)), where N is the number of words, and α is the inverse Ackermann function. Bit manipulation involves checking every pair of words, leading to a time complexity of O(N^2 * M), where M is the maximum length of a word. The space complexity depends on the method used: union-find uses O(N), while bit manipulation uses O(N * M) due to storing the bitmasks.

What Interviewers Usually Probe

Candidate's ability to apply graph theory or union-find for problem-solving.
Knowledge of bit manipulation and its application in optimizing string-based problems.
Understanding of time and space complexity trade-offs when choosing between union-find and bit manipulation.

Common Pitfalls or Variants

Common pitfalls

Not considering all possible connections between words, leading to incorrect groupings.
Overcomplicating the solution by trying to implement complex graph algorithms when a simpler union-find approach suffices.
Failure to properly handle edge cases, such as words with no connections or all words being connected.

Follow-up variants

Adding constraints where some strings may contain more than 26 characters.
Modifying the problem to account for case-insensitive letter transformations.
Allowing a larger set of operations, such as swapping letters, instead of just adding, deleting, or replacing.

FAQ

What is the optimal approach for solving the Groups of Strings problem?

The optimal approach combines Union-Find for grouping connected strings and bit manipulation for efficiently checking transformations between words.

Can bit manipulation help reduce time complexity for the Groups of Strings problem?

Yes, bit manipulation can significantly reduce time complexity by representing each word as a bitmask, allowing you to check for connections with a single bitwise operation.

What is the time complexity of the Groups of Strings problem?

The time complexity varies depending on the approach: Union-Find is near O(N * α(N)), while bit manipulation can be O(N^2 * M), where M is the word length.

How do you handle edge cases in the Groups of Strings problem?

Edge cases include words that aren't connected to any other words, or where all words are in a single connected group. These should be handled during the grouping process.

How does the GhostInterview solver help with the Groups of Strings problem?

GhostInterview assists by suggesting efficient algorithms like Union-Find and bit manipulation, offering clear solutions and highlighting potential pitfalls.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def groupStrings(self, words: List[str]) -> List[int]:
        def find(x):
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        def union(a, b):
            nonlocal mx, n
            if b not in p:
                return
            pa, pb = find(a), find(b)
            if pa == pb:
                return
            p[pa] = pb
            size[pb] += size[pa]
            mx = max(mx, size[pb])
            n -= 1

        p = {}
        size = Counter()
        n = len(words)
        mx = 0
        for word in words:
            x = 0
            for c in word:
                x |= 1 << (ord(c) - ord('a'))
            p[x] = x
            size[x] += 1
            mx = max(mx, size[x])
            if size[x] > 1:
                n -= 1
        for x in p.keys():
            for i in range(26):
                union(x, x ^ (1 << i))
                if (x >> i) & 1:
                    for j in range(26):
                        if ((x >> j) & 1) == 0:
                            union(x, x ^ (1 << i) | (1 << j))
        return [n, mx]

Continue Practicing

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