What is the time complexity of Minimum Genetic Mutation?

The optimal solution has a time complexity of O(B*L).

What data structures are used to solve Minimum Genetic Mutation?

This problem uses Hash Table, String, Breadth-First Search. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Genetic Mutation?

Insight 1: Breadth-First Search (BFS) is ideal for finding the shortest path in an unweighted graph. We can represent each gene as a node, and mutations between genes as edges. Insight 2: The gene bank acts as a constraint; we can only explore nodes (genes) that are present in the bank.

Which companies ask LeetCode 433. Minimum Genetic Mutation in interviews?

This problem is frequently asked at X.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Minimum Genetic Mutation - LeetCode 433 Solution

Minimum Genetic Mutation - Complete Solution Guide

Minimum Genetic Mutation is LeetCode problem 433, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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

Detailed Explanation

The problem asks us to find the minimum number of single-character mutations required to transform a starting gene string (`startGene`) into a target gene string (`endGene`). The mutations must be valid, meaning each intermediate gene string must exist in a given gene bank (`bank`). A gene string consists of 8 characters, each being one of 'A', 'C', 'G', or 'T'. The starting gene is considered valid, even if it is not in the bank. If a path to the end gene doesn't exist using valid mutations, return -1.

Solution Approach

The solution uses Breadth-First Search (BFS) to explore possible mutations. The `startGene` is the starting node. In each step, we generate all possible one-character mutations of the current gene. If a mutated gene is present in the `bank` and hasn't been visited, we add it to the queue for further exploration and mark it as visited. The BFS continues until we reach the `endGene` or the queue is empty (meaning no path exists).

Step-by-Step Algorithm

Step 1: Check if `startGene` is the same as `endGene`. If so, return 0 (no mutations needed).
Step 2: Convert the `bank` list to a set (`bank_set`) for faster lookups.
Step 3: Check if `endGene` is present in the `bank_set`. If not, return -1 because the target is unreachable.
Step 4: Initialize a queue with `startGene` and a mutation count of 0. Also, initialize a `visited` set to keep track of visited genes.
Step 5: While the queue is not empty, dequeue a gene and its mutation count.
Step 6: If the dequeued gene is the `endGene`, return the mutation count.
Step 7: For each character in the current gene (8 characters):
Step 8: Iterate through all possible characters ('A', 'C', 'G', 'T').
Step 9: If the current character is the same as the original character at that position, skip it.
Step 10: Create a new gene string by replacing the character at the current position with the new character.
Step 11: Check if the new gene is in `bank_set` and not in `visited`. If both are true:
Step 12: Add the new gene to `visited`. Enqueue the new gene and increment the mutation count.
Step 13: If the queue becomes empty without finding the `endGene`, return -1 (no path exists).

Key Insights

Insight 1: Breadth-First Search (BFS) is ideal for finding the shortest path in an unweighted graph. We can represent each gene as a node, and mutations between genes as edges.
Insight 2: The gene bank acts as a constraint; we can only explore nodes (genes) that are present in the bank.
Insight 3: To prevent infinite loops, we need to keep track of visited genes. A hash set is perfect for efficient lookup.

Complexity Analysis

Time Complexity: O(B*L)

Space Complexity: O(B)

Topics

This problem involves: Hash Table, String, Breadth-First Search.

Companies

Asked at: X.

Minimum Genetic Mutation | LeetCode 433

Medium

Hash Table String Breadth-First Search

Time: O(B*L)Space: O(B)

Solution Code

Available in:

import java.util.*;

class Solution {
    public int minMutation(String startGene, String endGene, String[] bank) {
        if (startGene.equals(endGene)) {
            return 0;
        }

        Set<String> bankSet = new HashSet<>(Arrays.asList(bank));
        if (!bankSet.contains(endGene)) {
            return -1;
        }

        Queue<Pair<String, Integer>> queue = new LinkedList<>();
        queue.offer(new Pair<>(startGene, 0));
        Set<String> visited = new HashSet<>();
        visited.add(startGene);

        while (!queue.isEmpty()) {
            Pair<String, Integer> current = queue.poll();
            String currentGene = current.getKey();
            int mutations = current.getValue();

            if (currentGene.equals(endGene)) {
                return mutations;
            }

            for (int i = 0; i < currentGene.length(); i++) {
                char originalChar = currentGene.charAt(i);
                for (char c : "ACGT".toCharArray()) {
                    if (c == originalChar) {
                        continue;
                    }

                    String nextGene = currentGene.substring(0, i) + c + currentGene.substring(i + 1);

                    if (bankSet.contains(nextGene) && !visited.contains(nextGene)) {
                        visited.add(nextGene);
                        queue.offer(new Pair<>(nextGene, mutations + 1));
                    }
                }
            }
        }

        return -1;
    }

    private static class Pair<K, V> {
        private final K key;
        private final V value;

        public Pair(K key, V value) {
            this.key = key;
            this.value = value;
        }

        public K getKey() {
            return key;
        }

        public V getValue() {
            return value;
        }
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 433: Minimum Genetic Mutation - Detailed explanation and solution approaches.

Topics

Hash Table String Breadth-First Search

Companies

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:433

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

13 steps

1
Step 1: Check if `startGene` is the same as `endGene`. If so, return 0 (no mutations needed).
2
Step 2: Convert the `bank` list to a set (`bank_set`) for faster lookups.
3
Step 3: Check if `endGene` is present in the `bank_set`. If not, return -1 because the target is unreachable.
4
Step 4: Initialize a queue with `startGene` and a mutation count of 0. Also, initialize a `visited` set to keep track of visited genes.
5
Step 5: While the queue is not empty, dequeue a gene and its mutation count.
6
Step 6: If the dequeued gene is the `endGene`, return the mutation count.
7
Step 7: For each character in the current gene (8 characters):
8
Step 8: Iterate through all possible characters ('A', 'C', 'G', 'T').
9
Step 9: If the current character is the same as the original character at that position, skip it.
10
Step 10: Create a new gene string by replacing the character at the current position with the new character.
11
Step 11: Check if the new gene is in `bank_set` and not in `visited`. If both are true:
12
Step 12: Add the new gene to `visited`. Enqueue the new gene and increment the mutation count.
13
Step 13: If the queue becomes empty without finding the `endGene`, return -1 (no path exists).

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(B*L)

💾

Space Complexity

O(B)

📝

Why?

The time complexity arises from potentially exploring each of the B genes in the `bank` and creating possible mutations from each gene of length L. The space complexity comes from storing the `bank` elements in the `bank_set` (size B), the genes visited in the `visited` set (up to size B), and the genes in the queue during the BFS process (up to size B).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Bidirectional BFS

Perform BFS simultaneously from both the `startGene` and the `endGene`. This can potentially reduce the search space, especially when the optimal path is relatively short.

Time: O(B^(d/2)), where B is the branching factor and d is the distance between start and end.Space: O(B^(d/2))

Trade-offs: Bidirectional BFS can be faster for certain graph structures, but it can be more complex to implement correctly. It's best used when the shortest path is likely to be significantly shorter than the total number of nodes. The space complexity can still be substantial.

✨

Key Insights

1Insight 1: Breadth-First Search (BFS) is ideal for finding the shortest path in an unweighted graph. We can represent each gene as a node, and mutations between genes as edges.
2Insight 2: The gene bank acts as a constraint; we can only explore nodes (genes) that are present in the bank.
3Insight 3: To prevent infinite loops, we need to keep track of visited genes. A hash set is perfect for efficient lookup.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: `startGene` is equal to `endGene`. The solution correctly returns 0 because no mutations are needed.
Edge case 2: `endGene` is not in the `bank`. The solution returns -1, indicating that no valid path exists.
Edge case 3: No path exists between `startGene` and `endGene` using genes in the `bank`. The BFS will terminate, and the solution will return -1.
Edge case 4: The `bank` is empty. The solution handles this gracefully by returning -1, assuming endGene is not startGene and since endGene must be in bank.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check if the `endGene` is present in the `bank`. This leads to an incorrect result if the target is unreachable.

✗Common mistake 2: Not keeping track of visited genes, which results in an infinite loop and incorrect mutation counts.

✗Common mistake 3: Failing to handle the edge case where the startGene and endGene are equal. Missing this results in the wrong answer.

✗Common mistake 4: Not using a set for the bank leads to O(n) lookups in the 'bank' instead of O(1) lookups in a set, impacting the time complexity significantly for larger banks.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Bioinformatics - Finding the minimum number of mutations required for a virus to evolve resistance to a drug. The gene bank represents the space of viable mutations.

🏦Real-world scenario 2: Compiler optimization - Transforming one code representation into another with minimal changes. Each transformation step must be a valid compiler optimization rule.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 13, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 29

❓

Frequently Asked Questions

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Minimum Genetic Mutation"?

For this medium-level Hash Table problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?