How do you solve Minimum Genetic Mutation on LeetCode?

Minimum Genetic Mutation (LeetCode #433) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m) time complexity and O(n) space complexity. Key insight: BFS is effective for shortest path problems in unweighted graphs.

What is the brute force solution for Minimum Genetic Mutation?

The brute force approach for Minimum Genetic Mutation is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries every possible mutation of the gene string until it finds the end gene or exhausts all options. This method is simple but inefficient, as it can lead to a large number of unnecessary checks. The optimal Optimal Solution approach improves this to O(n * m).

What algorithm or data structure is used to solve Minimum Genetic Mutation?

Minimum Genetic Mutation uses the following concepts: Hash Table, String, Breadth-First Search. The recommended patterns to study are: Breadth-First Search, Graph Traversal.

What are the interview tips for Minimum Genetic Mutation?

Explain your thought process clearly, especially when generating mutations. Consider edge cases, such as when the bank is empty or when startGene equals endGene. Practice BFS problems to become familiar with the pattern.

What are common mistakes when solving Minimum Genetic Mutation?

Not checking if endGene is in the bank before starting the BFS. Failing to mark genes as visited, leading to infinite loops.

#433

Minimum Genetic Mutation

Medium

Hash TableStringBreadth-First SearchBreadth-First SearchGraph Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * m)
Space	O(1)	O(n)

💡

Intuition

Time O(n * m)Space O(n)

The optimal approach uses Breadth-First Search (BFS) to explore the shortest path from the startGene to the endGene. This ensures that we find the minimum number of mutations efficiently by exploring all possibilities level by level.

⚙️

Algorithm

5 steps

1Step 1: Initialize a queue and add the startGene along with a mutation count of 0.
2Step 2: Use a set to keep track of visited genes to avoid cycles.
3Step 3: While the queue is not empty, dequeue the current gene and check if it matches the endGene.
4Step 4: If it matches, return the mutation count; otherwise, generate all valid mutations and enqueue them if they haven't been visited.
5Step 5: If the queue is exhausted without finding the endGene, return -1.

solution.py21 lines

1# Full working Python code
2from collections import deque
3
4def minMutation(startGene, endGene, bank):
5    if endGene not in bank:
6        return -1
7    bank_set = set(bank)
8    queue = deque([(startGene, 0)])
9    visited = set([startGene])
10    while queue:
11        current, mutations = queue.popleft()
12        if current == endGene:
13            return mutations
14        for i in range(len(current)):
15            for char in 'ACGT':
16                if current[i] != char:
17                    mutation = current[:i] + char + current[i+1:]
18                    if mutation in bank_set and mutation not in visited:
19                        visited.add(mutation)
20                        queue.append((mutation, mutations + 1))
21    return -1

ℹ

Complexity note: The time complexity is O(n * m) where n is the number of genes in the bank and m is the length of the gene strings (which is constant at 8). Each gene can generate up to 32 mutations (4 choices for each of the 8 positions).

1BFS is effective for shortest path problems in unweighted graphs.
2Using a set for the bank allows O(1) lookups.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.