Introduction

In this blog post, we will explore how to solve the LeetCode problem "839" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Two strings, X and Y, are considered similar if either they are identical or we can make them equivalent by swapping at most two letters (in distinct positions) within the string X. For example, "tars" and "rats" are similar (swapping at positions 0 and 2), and "rats" and "arts" are similar, but "star" is not similar to "tars", "rats", or "arts". Together, these form two connected groups by similarity: {"tars", "rats", "arts"} and {"star"}. Notice that "tars" and "arts" are in the same group even though they are not similar. Formally, each group is such that a word is in the group if and only if it is similar to at least one other word in the group. We are given a list strs of strings where every string in strs is an anagram of every other string in strs. How many groups are there? Example 1: Input: strs = ["tars","rats","arts","star"] Output: 2 Example 2: Input: strs = ["omv","ovm"] Output: 1 Constraints: 1 <= strs.length <= 300 1 <= strs[i].length <= 300 strs[i] consists of lowercase letters only. All words in strs have the same length and are anagrams of each other.

Explanation

Here's the breakdown of the solution:

• Connected Components: The problem can be modeled as finding the number of connected components in a graph where nodes are the strings in strs, and an edge exists between two strings if they are similar.
• Union-Find (Disjoint Set Union): An efficient way to determine the number of connected components is using the Union-Find data structure. We iterate through all pairs of strings. If two strings are similar, we union their corresponding sets.

• Similarity Check: A helper function efficiently checks if two strings are similar as defined by the problem (identical or differ by at most two swaps).

• Runtime Complexity: O(n² m), where n is the number of strings and m is the length of each string. *Storage Complexity: O(n).

Code

    class UnionFind:
    def __init__(self, n):
        self.parent = list(range(n))
        self.rank = [0] * n
        self.count = n

    def find(self, x):
        if self.parent[x] != x:
            self.parent[x] = self.find(self.parent[x])
        return self.parent[x]

    def union(self, x, y):
        root_x = self.find(x)
        root_y = self.find(y)
        if root_x != root_y:
            if self.rank[root_x] < self.rank[root_y]:
                self.parent[root_x] = root_y
            elif self.rank[root_x] > self.rank[root_y]:
                self.parent[root_y] = root_x
            else:
                self.parent[root_y] = root_x
                self.rank[root_x] += 1
            self.count -= 1

def are_similar(s1, s2):
    diff_indices = []
    for i in range(len(s1)):
        if s1[i] != s2[i]:
            diff_indices.append(i)

    if not diff_indices:
        return True
    if len(diff_indices) == 2:
        i, j = diff_indices
        return s1[i] == s2[j] and s1[j] == s2[i]
    return False

def numSimilarGroups(strs):
    n = len(strs)
    uf = UnionFind(n)

    for i in range(n):
        for j in range(i + 1, n):
            if are_similar(strs[i], strs[j]):
                uf.union(i, j)

    return uf.count