Problem Description

In this problem, we are given a start gene string and an end gene string, each 8 characters long composed of the characters 'A', 'C', 'G', and 'T'. We also have a bank of valid gene mutations. A mutation is a change of a single character in the gene string, and a gene must be in the bank to be considered a valid gene mutation.

The task is to determine the minimum number of mutations required to change the start gene into the end gene. If it's not possible to reach the end gene from the start gene by only using valid mutations from the bank, we must return -1. It is worth noting that the start gene is considered valid even if it's not present in the bank, but every step after must be a mutation existing in the bank.

To clarify: a gene string mutation is represented as a single character change from the set of possible characters. For example, changing 'AACCGGTT' to 'AACCGGTA' is considered one mutation.

Flowchart Walkthrough

To analyze Leetcode 433. Minimum Genetic Mutation using the Flowchart, let's proceed with a step-by-step approach to locate the suitable algorithm:

Is it a graph?

• Yes: The problem involves converting a genetic sequence to another using mutations that correspond to nodes in a graph, where edges represent valid mutations between sequences.

Is it a tree?

• No: Although there's a starting sequence and the mutations stem from it, mutations can potentially revert or branch out in ways that aren't strictly hierarchical like a tree.

Is the problem related to directed acyclic graphs (DAGs)?

• No: The problem relates to transforming one genetic sequence into another which doesn't specifically deal with acyclicity in a directed graph context.

Is the problem related to shortest paths?

• Yes: The problem asks for the minimum number of mutations needed to transform one genetic sequence into another, which is a shortest path problem in terms of steps from one node to another.

Is the graph weighted?

• No: Each mutation step from one sequence to another is essentially uniform; one mutation is considered one step.

Conclusion: The flowchart suggests using BFS for this unweighted shortest path problem, as BFS effectively explores all nodes at the present "depth" before moving on to nodes at the next depth level, thereby ensuring the minimum steps (mutations) are calculated first.

Intuition

To solve this problem, we can use either Breadth-First Search (BFS) or Depth-First Search (DFS), but BFS is more convenient for finding the shortest path in an unweighted graph, which is analogous to finding the minimum number of mutations needed.

The intuition behind the solution is to treat each gene string as a node in a graph and each valid mutation as an edge connecting two nodes. We use BFS to explore the graph level by level, starting from the start gene and working towards the end gene. BFS is ideal for this scenario because it finds the shortest path (minimum mutations) from the start to the end node.

Here are the key steps of the BFS approach:

1. Initialize a set from the bank to quickly check if a mutation is valid.
2. Begin with a queue initialized with the start gene and a mutation count of 0.
3. Use a dictionary to define mutation possibilities for each character.
4. Dequeue an element and iterate over its characters, changing one character at a time according to the mutation possibilities.
5. If a new mutation is valid (in the bank), enqueue it with a mutation count incremented by one.
6. If we reach the end gene, return the mutation count.
7. If the queue is exhausted without finding the end gene, return -1.

By using this BFS approach, we explore all possible gene mutations in the shortest number of steps, ensuring the minimum number of mutations needed to achieve the end gene, if possible.

Learn more about Breadth-First Search patterns.

Solution Approach

The solution uses Breadth-First Search (BFS), which is an algorithm for traversing or searching tree or graph data structures. It explores the neighbor nodes at the present depth prior to moving on to nodes at the next depth level. The implementation includes the following algorithms, data structures, and patterns:

1. Set: A set is initialized from the bank, which provides O(1) complexity for checking the existence of a gene string.

2. Queue (Deque): The deque data structure is used as a queue to implement BFS. Items are appended at one end and popped from the other end, mimicking a queue's First In, First Out (FIFO) behavior.

3. Hash Map: A dictionary mp maps each character to the other three characters, indicating the possible mutations from one character. This is used to easily determine the viable mutations at any position of a gene string.

4. Level-by-Level Traversal: The algorithm traverses the graph level by level. It considers all gene strings that can be reached from the current gene string in a single mutation as neighbors and explores them in the BFS manner.

5. Early Stopping: If at any point the end gene is found during the exploration, the function immediately returns the number of mutations (steps) taken to reach this state.

6. Graph Pruning: To avoid revisiting gene strings and thus entering cycles, every time a new valid gene string is found in the bank, it is removed from the set. This ensures that each gene string is visited only once.

The implementation steps are as follows:

• Create a set of the gene bank to facilitate quick look-up.
• Initialize a deque with a tuple containing the start gene and an initial step count of 0.
• Define a mp dictionary representing the mutation map from each character to the other three possible characters.
• Start the BFS loop by popping an element from the queue:
  • For each character v in the current gene string t, loop through the possible characters j in mp[v] to generate new strings.
  • Construct a new gene string next by replacing the character at position i with j.
  • If next is found in the set, append it to the queue with the step count incremented and remove it from the set to prevent future visits.
• Continue the loop until the queue is empty or the end gene is reached.

Here is the complete BFS implementation in code:

s = set(bank)  # Convert bank into a set for O(1) access
q = deque([(start, 0)])  # Initialize the queue with a tuple of start gene and step counter
mp = {'A': 'TCG', 'T': 'ACG', 'C': 'ATG', 'G': 'ATC'}  # Mutation map

# BFS loop
while q:
    t, step = q.popleft()  # Pop the next gene string and current step count
    if t == end:  # End reached
        return step
    for i, v in enumerate(t):  # Iterate through each character in the string
        for j in mp[v]:  # Go through all possible mutations for the character
            next = t[:i] + j + t[i + 1:]  # Create the new mutated string
            if next in s:  # Check if the mutation is valid
                q.append((next, step + 1))  # Enqueue the new gene with incremented steps
                s.remove(next)  # Remove the visited gene from the set
return -1  # Return -1 if end can't be reached

This code utilizes BFS to find the shortest path from startGene to endGene, which corresponds to the minimum mutations required to transform one into the other.

Example Walkthrough

Let's demonstrate the solution approach with an example:

Assume our start gene is AACCGGTT and our end gene is AACCGGTA. The bank of valid gene mutations includes AACCGGTA, AACCGGTC, and AACCGGCT.

1. The set of bank mutations is {AACCGGTA, AACCGGTC, AACCGGCT}. Since we can check the presence of a gene in O(1) time, this helps us quickly validate new mutations.

2. We initialize our queue (deque) with the start gene and a mutation count of 0: deque([('AACCGGTT', 0)]).

3. Our mutation map mp is {'A': 'TCG', 'T': 'ACG', 'C': 'ATG', 'G': 'ATC'}, showing all possible single character mutations for each character.

Now, we begin the BFS:

• We dequeue the first element: ('AACCGGTT', 0).
• Beginning with the first character, we check all other characters, but since no mutation is present in the bank for a change in the first character, we move to the second character and so on.
• We finally reach the last character 'T'. The possible mutations for 'T' are 'A', 'C', and 'G'.
• We mutate the last character to 'A' (resulting in AACCGGTA) and check if it's in the bank. It is, so we enqueue it with a mutation count incremented by one: deque([('AACCGGTA', 1)]).
• We remove AACCGGTA from the set to avoid revisiting this gene string.
• Continuing the BFS, we dequeue ('AACCGGTA', 1) and find it's the end gene. The search is over, and we return 1.

Hence, we have found that the minimum number of mutations to get from AACCGGTT to AACCGGTA using the given gene bank is 1.

Solution Implementation

Time and Space Complexity

The given code snippet is a solution to the genetic mutation problem, which finds the minimum number of mutations needed to transform the start gene sequence into the end gene sequence using a given set of gene mutations from the bank.

Time Complexity

To analyze time complexity, let's look at the components involved:

1. The BFS traversal: The algorithm uses a queue to perform a breadth-first search.

2. Checking each character and trying all possible mutations: For each gene sequence, the algorithm checks each of its N characters (where N is the length of the gene sequence, typically 8 in the problem constraint) and attempts to mutate it to all other possible characters (3 possibilities per character, since there are 4 possible nucleotides).

3. Checking and removing gene sequences from the set: Upon each successful mutation, it checks if the new sequence is in the set and removes it if present.

Considering that there are M gene sequences in the bank:

• In the worst case, the algorithm might have to visit all M sequences in the bank.
• For each sequence, we perform N character checks and 3 possible mutations.
• Each mutation check involves an O(1) set containment check and O(1) removal operation (since it's a set).

Thus, the total time complexity can be estimated as O(M * N * 3), which simplifies to O(M * N) as constants can be ignored in Big O notation.

Space Complexity

The primary space-consuming structures are:

1. The queue used for BFS which, in the worst case, might hold all M sequences before mutation at the same level.
2. The set s, also holding the M gene sequences from the bank.
3. The mp dictionary with a constant size of 4 (not dependent on N or M), therefore can be considered O(1).

So the space complexity is dominated by the queue and the set, both of which may hold up to M gene sequences. Hence, the space complexity is O(M).

Learn more about how to find time and space complexity quickly using problem constraints.