Minimum Genetic Mutation

To solve this coding challenge, we need to develop a method that will navigate through the possible mutations of a gene string from the

startGene

to the

endGene

. The gene string must evolve one mutation at a time, with each step being a valid gene from the provided gene bank.

Explanation

1. Initialization and Input Checks : We'll verify whether the
   endGene
   exists within the
   bank
   . If not, it's impossible to reach the
   endGene
   from the
   startGene
   as per the problem's restrictions, and we'll return
   -1
   .
2. Breadth-First Search (BFS) : This search technique is ideal here because it explores all possible mutations at each level before moving on to mutations from the next level. This ensures that we find the shortest possible path from
   startGene
   to
   endGene
   . We'll use a queue to implement BFS.
3. Queue Operation :
• The queue will store tuples containing the current gene string and the number of mutations taken to reach it.
• We initiate the queue with the
  startGene
  and a mutation count of
  0
  .
4. Visited Set : To avoid revisiting gene strings and potential infinite loops, we'll maintain a set of visited gene strings.
5. Processing Each Gene :
• For each gene string and mutation count in the queue, we will check if the current gene matches the
  endGene
  .
• If it matches, return the current mutation count as it represents the minimum mutations needed.
6. Generating Mutations :
• For each character in the gene string, we will attempt to mutate it to one of the other valid characters ('A', 'C', 'G', 'T').
• For each mutation, we will check if the new gene string is in the bank and hasn't been visited.
• If valid, add the new gene string to the queue with an incremented mutation count and mark it as visited.
7. End Condition : If the queue is exhausted and no path to
   endGene
   is found, return
   -1
   .

Detailed Steps in Pseudocode

8. Initialize an empty set
   visited
   to keep track of visited gene strings.
9. Initialize the queue with the tuple
   (startGene, 0)
   and add
   startGene
   to visited.
10. Check if
    endGene
    is in the
    bank
    . If not, return
    -1
    .
11. While the queue is not empty, do the following:
• Dequeue the front element (currentGene, currentMutations).
• If
  currentGene == endGene
  , return
  currentMutations
  .
• For each character position in
  currentGene
  :
• Try mutating to each possible nucleotide ('A', 'C', 'G', 'T').
• If the resulting new gene string is valid (in
  bank
  and not visited):
• Mark it as visited.
• Enqueue the new gene string with the incremented mutation count.
12. If the queue is exhausted without finding
    endGene
    , return
    -1
    .

Pseudocode

                                            
# Check for endGene in bank
if endGene not in bank:
    # If endGene is not in bank, return -1
    return -1

# Initialize variables
# Queue to store current gene string and its mutation count
queue = [(startGene, 0)]
# Set to store visited gene strings
visited = set()
# Add startGene to visited
visited.add(startGene)

# BFS loop while queue is not empty
while queue is not empty:
    # Dequeue the front element (current gene string and mutation count)
    currentGene, currentMutations = queue.pop(0)
    
    # Check if current gene matches the end gene
    if currentGene == endGene:
        # If matched, return the current number of mutations
        return currentMutations
    
    # Try mutating every character of the current gene
    for i in range(len(currentGene)):
        # For every nucleotide, attempt mutation
        for nucleotide in ['A', 'C', 'G', 'T']:
            # Create a new gene string by changing one nucleotide
            newGene = currentGene[:i] + nucleotide + currentGene[i+1:]
            
            # Check if new gene string is valid and not visited
            if newGene in bank and newGene not in visited:
                # Mark new gene as visited
                visited.add(newGene)
                # Enqueue the new gene string with incremented mutation count
                queue.append((newGene, currentMutations + 1))

# If the end gene is not reached, return -1
return -1

                                        

By using the described BFS approach, we ensure that we investigate all possible mutations leading us to the minimum number of mutations needed. This solution is efficient given the constraints and guarantees that the shortest path to the target gene is found.