433. Minimum Genetic Mutation

• categories: Code, Interview Question, leetcode, Medium
• source: https://leetcode.com/problems/minimum-genetic-mutation
• topics: Graph, BFS

Description

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.

Idea

• BFS in graph

Code

class Solution:
    def minMutation(self, startGene: str, endGene: str, bank: List[str]) -> int:
        if startGene == endGene and startGene in bank:
            return 0
        if not bank or endGene not in bank:
            return -1
 
        adj = defaultdict(set)
 
        def distance(s1, s2):
            counter = 0
            for i in range(len(s1)):
                if s1[i] != s2[i]:
                    counter += 1
            return counter
 
        for s1 in bank:
            for s2 in bank:
                if s1 != s2 and distance(s1, s2) == 1:
                    adj[s1].add(s2)
                    adj[s2].add(s1)
 
        visited = set(s for s in bank if distance(s, startGene) == 1)
        queue = deque(visited)
        step = 0
        
        while queue:
            step += 1
            for _ in range(len(queue)):
                curr = queue.popleft()
                
                if curr == endGene:
                    return step
 
                for neighbour in adj[curr]:
                    if neighbour not in visited:
                        visited.add(neighbour)
                        queue.append(neighbour)
 
        return -1