Problem Description

You have a gene string represented by an 8-character string containing only the characters 'A', 'C', 'G', and 'T'.

You need to find the minimum number of mutations required to change a starting gene string (startGene) into a target gene string (endGene). A mutation is defined as changing exactly one character in the gene string.

For example: "AACCGGTT" → "AACCGGTA" is one mutation (the last character changed from 'T' to 'A').

There's an additional constraint: you're given a gene bank (bank) which contains all valid gene mutations. Each intermediate gene string during the mutation process must exist in this bank to be considered valid. The starting gene is assumed to be valid and may or may not be in the bank.

The task is to return the minimum number of mutations needed to transform startGene to endGene. If it's impossible to reach the target gene through valid mutations, return -1.

Key points:

• Each gene string is exactly 8 characters long
• Only characters 'A', 'C', 'G', 'T' are allowed
• Each mutation changes exactly one character
• All intermediate genes must be in the bank (except possibly the starting gene)
• Find the shortest mutation path from start to end

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: This problem can be modeled as a graph where each gene string is a node, and edges exist between gene strings that differ by exactly one character. We need to find the shortest path from startGene to endGene through valid intermediate genes in the bank.

Is it a tree?

• No: The graph is not a tree. Multiple gene strings can potentially mutate into the same gene string, and there can be cycles (gene A → gene B → gene C → gene A is possible if all mutations are valid).

Is the problem related to directed acyclic graphs?

• No: The graph can have cycles, so it's not a DAG. A gene mutation can potentially be reversed through a series of other mutations.

Is the problem related to shortest paths?

• Yes: We need to find the minimum number of mutations (shortest path) from the starting gene to the ending gene.

Is the graph Weighted?

• No: Each mutation (edge) has the same cost of 1. Every single character change counts as exactly one mutation, making this an unweighted graph.

BFS

• Yes: For unweighted shortest path problems, BFS is the optimal choice.

Conclusion: The flowchart suggests using BFS (Breadth-First Search) for finding the minimum number of mutations from startGene to endGene. BFS is perfect here because:

1. We're dealing with an unweighted graph (each mutation has equal cost)
2. We need the shortest path between two nodes
3. BFS explores nodes level by level, guaranteeing the first time we reach the target is via the shortest path

Intuition

Think of this problem as finding the shortest path in a maze where each room (gene string) has doors to other rooms that differ by exactly one character. We start in one room (startGene) and want to reach another room (endGene) using the minimum number of steps.

The key insight is that each gene string can be connected to other gene strings that differ by exactly one character. This creates a graph structure where:

• Nodes = gene strings
• Edges = exist between genes that differ by exactly one character
• Edge weight = 1 (each mutation counts as one step)

Since we want the minimum number of mutations and each mutation has the same cost (1), we're essentially looking for the shortest path in an unweighted graph. BFS is perfect for this because it explores all genes at distance 1 first, then distance 2, and so on. This level-by-level exploration guarantees that the first time we encounter the target gene, we've found it via the shortest path.

The approach becomes clear:

1. Start from startGene and explore all valid genes that are one mutation away
2. From each of those, explore genes that are one more mutation away
3. Continue until we find endGene or exhaust all possibilities

We use a queue to maintain the BFS order and a visited set to avoid revisiting the same gene (which would create unnecessary cycles). For each gene we process, we check all genes in the bank to see which ones differ by exactly one character - these become our next candidates to explore.

The beauty of BFS here is that it naturally tracks the mutation count through its level-by-level traversal. When we finally reach endGene, the depth level tells us exactly how many mutations were needed.

Solution Approach

The implementation uses BFS with a queue and visited set to find the shortest mutation path:

Data Structures Used:

• deque (double-ended queue): Stores tuples of (gene_string, mutation_count) for BFS traversal
• set (vis): Tracks visited gene strings to avoid revisiting and creating cycles

Algorithm Steps:

1. Initialize BFS structures:

   • Create a queue q with the starting gene and mutation count 0: [(startGene, 0)]
   • Create a visited set vis containing the starting gene to mark it as explored

2. BFS traversal: While the queue is not empty:

   • Dequeue the front element to get current gene and its depth (number of mutations so far)
   • Check if we've reached the target: if gene == endGene, return depth

3. Explore neighbors: For each gene string nxt in the bank:

   • Calculate the difference between current gene and bank gene using: sum(a != b for a, b in zip(gene, nxt))
   • If the difference is exactly 1 (one mutation away) AND this gene hasn't been visited:
     • Add (nxt, depth + 1) to the queue for future exploration
     • Mark nxt as visited by adding to vis

4. Return result:

   • If we exit the while loop without finding endGene, return -1 (no valid mutation path exists)

Key Implementation Details:

• The difference calculation sum(a != b for a, b in zip(gene, nxt)) efficiently counts character mismatches between two genes by zipping them character by character and summing the boolean differences.

• Using depth + 1 when enqueueing ensures each level of BFS correctly tracks the mutation count from the start.

• The visited set prevents exploring the same gene multiple times, which is crucial for efficiency and avoiding infinite loops.

• We don't need to generate all possible one-character mutations explicitly; instead, we check against the bank directly, which is more efficient when the bank is small.

The time complexity is O(N * M * K) where N is the bank size, M is the gene length (8), and K is the maximum queue size. The space complexity is O(N) for the visited set and queue.

Example Walkthrough

Let's walk through a concrete example to illustrate the BFS solution approach:

Input:

• startGene = "AACCGGTT"
• endGene = "AACCGGTA"
• bank = ["AACCGGTA", "AACCGCTT", "AACCGGAT"]

Step-by-step BFS Process:

Initial Setup:

• Queue: [("AACCGGTT", 0)]
• Visited: {"AACCGGTT"}

Iteration 1:

• Dequeue: ("AACCGGTT", 0)
• Check if this is endGene? No, continue
• Check neighbors in bank:
  • "AACCGGTA": differs by 1 char (position 7: T→A) ✓ Valid neighbor
  • "AACCGCTT": differs by 2 chars (positions 5,6: G→C, G→T) ✗ Not a neighbor
  • "AACCGGAT": differs by 1 char (position 6: T→A) ✓ Valid neighbor
• Add valid unvisited neighbors to queue
• Queue: [("AACCGGTA", 1), ("AACCGGAT", 1)]
• Visited: {"AACCGGTT", "AACCGGTA", "AACCGGAT"}

Iteration 2:

• Dequeue: ("AACCGGTA", 1)
• Check if this is endGene? Yes! Return depth = 1

Result: The minimum number of mutations is 1.

The mutation path was: "AACCGGTT" → "AACCGGTA" (changed last character from T to A).

Why BFS guarantees the shortest path: Notice that BFS explored all genes at distance 1 before moving to distance 2. Since we found the target at distance 1, we know this is the shortest possible path. If there were a longer path (like going through "AACCGGAT" first), BFS wouldn't explore it because it found the target earlier.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n * m^2) where n is the length of each gene string and m is the size of the gene bank.

• The BFS explores at most m genes from the bank (worst case: visiting all genes in the bank)
• For each gene dequeued, we compare it with all genes in the bank, which takes O(m) iterations
• For each comparison, we calculate the difference between two genes using zip and sum, which takes O(n) time
• Therefore, the total time complexity is O(m * m * n) = O(n * m^2)

Note: The reference answer suggests O(C * n * m) where C = 4 represents generating all possible mutations by changing each position to one of 4 possible characters. However, this code takes a different approach by directly comparing with genes in the bank rather than generating mutations, resulting in O(n * m^2) complexity.

Space Complexity: O(m) where m is the size of the gene bank.

• The queue can contain at most m genes (all genes from the bank)
• The visited set can also contain at most m + 1 genes (all genes from the bank plus the start gene)
• Each gene string has length n, but we're storing references to existing strings rather than creating new ones
• Therefore, the space complexity is O(m) for the queue and visited set

Common Pitfalls

1. Not Checking if Target Gene Exists in Bank

A critical pitfall is assuming the target gene (endGene) must exist in the bank for a valid mutation path to exist. The algorithm will fail to find a valid path if endGene is not in the bank, even when it should return a valid mutation count.

Why this happens: The BFS only explores genes that are in the bank, so if the target isn't in the bank, it will never be reached through the neighbor exploration loop.

Example scenario:

• startGene = "AACCGGTT"
• endGene = "AACCGGTA"
• bank = ["AACCGGTT"]

The algorithm would return -1 even though the target is just one mutation away from the start.

Solution: Add an early check to ensure the target gene exists in the bank (unless it's the same as the start gene):

# Add this check at the beginning of the function
if endGene not in bank and endGene != startGene:
    return -1

2. Inefficient Neighbor Generation

Another pitfall is generating all possible mutations (8 positions × 4 characters = 32 possibilities) and then checking if each exists in the bank. This becomes inefficient when the bank is small.

Inefficient approach:

# Don't do this - generates unnecessary mutations
for i in range(8):
    for char in ['A', 'C', 'G', 'T']:
        if char != current_gene[i]:
            mutated = current_gene[:i] + char + current_gene[i+1:]
            if mutated in bank and mutated not in visited:
                # Process mutation

Better approach (as shown in the solution): Iterate through the bank directly and check if each gene is exactly one mutation away. This is more efficient when the bank size is smaller than the number of possible mutations.

3. Using List Instead of Set for Bank Lookup

If the bank remains as a list, the in operation for checking membership becomes O(N) for each check, significantly slowing down the algorithm.

Solution: Convert the bank to a set for O(1) lookup if you need to check membership frequently:

bank_set = set(bank)
# Now checking 'gene in bank_set' is O(1) instead of O(N)