Introduction

In this blog post, we will explore how to solve the LeetCode problem "433" 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

A gene string can be represented by an 8-character long string, with choices from 'A', 'C', 'G', and 'T'. Suppose we need to investigate a mutation from a gene string startGene to a gene string endGene where one mutation is defined as one single character changed in the gene string. For example, "AACCGGTT" --> "AACCGGTA" is one mutation. There is also a gene bank bank that records all the valid gene mutations. A gene must be in bank to make it a valid gene string. Given the two gene strings startGene and endGene and the gene bank bank, return the minimum number of mutations needed to mutate from startGene to endGene. If there is no such a mutation, return -1. Note that the starting point is assumed to be valid, so it might not be included in the bank. Example 1: Input: startGene = "AACCGGTT", endGene = "AACCGGTA", bank = ["AACCGGTA"] Output: 1 Example 2: Input: startGene = "AACCGGTT", endGene = "AAACGGTA", bank = ["AACCGGTA","AACCGCTA","AAACGGTA"] Output: 2 Constraints: 0 <= bank.length <= 10 startGene.length == endGene.length == bank[i].length == 8 startGene, endGene, and bank[i] consist of only the characters ['A', 'C', 'G', 'T'].

Explanation

Here's a solution that uses Breadth-First Search (BFS) to find the shortest mutation path.

• BFS Approach: Treat the gene strings as nodes in a graph. Two nodes are connected if they differ by only one character (a single mutation). Use BFS to find the shortest path from the startGene to the endGene.
• Gene Bank as Valid Nodes: The bank array represents the set of valid gene strings. Only these strings can be part of the mutation path.

• Visited Set: Keep track of visited gene strings to avoid cycles and redundant exploration.

• Runtime Complexity: O(B L), where B is the number of genes in the bank and L is the length of each gene (fixed at 8). *Storage Complexity: O(B) in the worst case, to store the queue and visited set.

Code

    from collections import deque

def minMutation(startGene: str, endGene: str, bank: list[str]) -> int:
    """
    Finds the minimum number of mutations needed to mutate from startGene to endGene,
    given a bank of valid gene mutations.
    """

    bank_set = set(bank)
    if endGene not in bank_set:
        return -1

    queue = deque([(startGene, 0)])  # (gene, mutations)
    visited = {startGene}
    nucleotides = ['A', 'C', 'G', 'T']

    while queue:
        gene, mutations = queue.popleft()

        if gene == endGene:
            return mutations

        for i in range(len(gene)):
            for nucleotide in nucleotides:
                mutated_gene = gene[:i] + nucleotide + gene[i+1:]

                if mutated_gene in bank_set and mutated_gene not in visited:
                    queue.append((mutated_gene, mutations + 1))
                    visited.add(mutated_gene)

    return -1