Minimum Genetic Mutation

Published: Nov 2, 2022

Medium Breadth-First Search String

Introduction

When words are given, and each character in the word should be replaced and checked one by one, the breadth-first search is a good algorithm to find the answer. Use queue to save the path (word mutation) along with the steps. Also use a set to avoid repetition. When the popped out word from queue is the end word, return the steps.

Problem 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 start to a gene string end 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 start and end and the gene bank bank, return the minimum number of mutations needed to mutate from start to end. 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.

Constraints:

  • start.length == 8
  • end.length == 8
  • 0 <= bank.length <= 10
  • bank[i].length == 8
  • start, end, and bank[i] consist of only the characters ['A', 'C', 'G', 'T'].

https://leetcode.com/problems/minimum-genetic-mutation/

Examples

Example 1
Input: start = "AACCGGTT", end = "AACCGGTA", bank = ["AACCGGTA"]
Output: 1

Example 2
Input: start = "AACCGGTT", end = "AAACGGTA", bank = ["AACCGGTA","AACCGCTA","AAACGGTA"]
Output: 2

Example 3
Input: start = "AAAAACCC", end = "AACCCCCC", bank = ["AAAACCCC","AAACCCCC","AACCCCCC"]
Output: 3

Analysis

Use a queue data structure to save paths, in another word, mutations. Also, save steps along with the mutated word. While going over each word, change each character to one of ACGT. If one-character mutated word is in the bank, add it to the queue along with current step plus one. When the pooped out word from the queue is the end, the answer is found.

Solution

class MinimumGeneticMutation:
    def minMutation(self, start: str, end: str, bank: List[str]) -> int:
        queue = [(start, 0)]  # word, steps
        seen = {start}
        while queue:
            w, st = queue.pop(0)
            if w == end:
                return st
            for i in range(len(w)):
                neighbors = [w[:i] + c + w[i + 1:] for c in 'ACGT']
                for neigh in neighbors:
                    if neigh not in seen and neigh in bank:
                        queue.append([neigh, st + 1])
                        seen.add(neigh)
        return -1

Complexities

  • Time: O(n) – n: length of bank
  • Space: O(1) – The word length is always 8. Both queue and seen have constant space.