How hard is LeetCode 2157. Groups of Strings?

LeetCode 2157. Groups of Strings is rated Hard on LeetCode.

What is the time complexity of LeetCode 2157. Groups of Strings?

The Python solution on this page runs in Near O(n) (amortized).

What is the space complexity of LeetCode 2157. Groups of Strings?

The Python solution on this page uses O(n) auxiliary space.

What topics does LeetCode 2157. Groups of Strings cover?

LeetCode 2157. Groups of Strings is tagged Bit Manipulation, Union Find and String on LeetCode.

Groups of Strings — LeetCode 2157 Python Solution

HardBit ManipulationUnion FindString

Problem: #2157
Pattern: Union-Find
Reading time: 7 min
Source: leetcode.com

The problem

You are given a 0-indexed array of strings words. Each string consists of lowercase English letters only.

Example

Input: words = ["a","b","ab","cde"]
Output: [2,3]
Explanation: - 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].

Python solution

Python

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]

Complexity

Measure	Complexity
Time	Near O(n) (amortized)
Space	O(n) auxiliary

Pattern: Union-Find

Merge groups and ask whether two things are connected, both in near-constant time. LeetCode 2157. Groups of Strings is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Union Find.

The union-find guide has the Python template for the pattern and the 83 LeetCode problems that use it.

Frequently asked questions

How hard is LeetCode 2157. Groups of Strings?: LeetCode 2157. Groups of Strings is rated Hard on LeetCode.
What is the time complexity of LeetCode 2157. Groups of Strings?: The Python solution on this page runs in Near O(n) (amortized).
What is the space complexity of LeetCode 2157. Groups of Strings?: The Python solution on this page uses O(n) auxiliary space.
What topics does LeetCode 2157. Groups of Strings cover?: LeetCode 2157. Groups of Strings is tagged Bit Manipulation, Union Find and String on LeetCode.