Leetcode #2157: Groups of Strings
In this guide, we solve Leetcode #2157 Groups of Strings in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given a 0-indexed array of strings words. Each string consists of lowercase English letters only.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Bit Manipulation, Union Find, String
Intuition
We need to merge components and check connectivity efficiently.
Union-Find supports near-constant-time merges and finds.
Approach
Initialize each node as its own parent and union pairs as you scan.
Use path compression to keep operations fast.
Steps:
- Initialize parent arrays.
- Union related nodes.
- Use find to check connectivity.
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].
- 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.
Python Solution
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
The time complexity is Near O(n) (amortized). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.