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Q: 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 start to a gene string end 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 start and end and the gene bank bank, return the minimum number of mutations needed to mutate from start to end. If there is no such a mutation, return -1.
Note that the starting point is assumed to be valid, so it might not be included in the bank.
A: BFS (Breadth-First Search)239Please respect copyright.ＰＥＮＡＮＡHQSNu7vWH7
// better than use DFS as it just need to find out the shortest path.
class Solution {239Please respect copyright.ＰＥＮＡＮＡOsyPGVnpxI
public int minMutation(String start, String end, String[] bank) {239Please respect copyright.ＰＥＮＡＮＡ4sLX7xSyJ7
// Initialize a queue queue and a set seen. The queue will be used for BFS and the set will be used to prevent visiting a node more than once. Initially, the queue and set should hold start.239Please respect copyright.ＰＥＮＡＮＡuTIssSDAjP
Queue<String> queue = new LinkedList<>();239Please respect copyright.ＰＥＮＡＮＡTFDQ45JgKx
Set<String> seen = new HashSet<>();239Please respect copyright.ＰＥＮＡＮＡ53WqiPoxMz
queue.add(start);239Please respect copyright.ＰＥＮＡＮＡOXkFjRPhhi
seen.add(start);239Please respect copyright.ＰＥＮＡＮＡ3jEKnGrkTT
239Please respect copyright.ＰＥＮＡＮＡa6B6tWmnJy
int steps = 0;239Please respect copyright.ＰＥＮＡＮＡ7HFuhp0OmW
239Please respect copyright.ＰＥＮＡＮＡcLa16SWitT
while (!queue.isEmpty()) {239Please respect copyright.ＰＥＮＡＮＡxBK8E3SZh5
int nodesInQueue = queue.size();239Please respect copyright.ＰＥＮＡＮＡYXQKedd79i
for (int j = 0; j < nodesInQueue; j++) {239Please respect copyright.ＰＥＮＡＮＡCMQt9GkK3P
String node = queue.remove();239Please respect copyright.ＰＥＮＡＮＡGft6ymoxec
// Perform a BFS. At each node, if node == end, return the number of steps so far. Otherwise, iterate over all the neighbors. For each neighbor, if neighbor is not in seen and neighbor is in bank, add it to queue and seen.
if (node.equals(end)) {239Please respect copyright.ＰＥＮＡＮＡJzqZPfcyYc
return steps;239Please respect copyright.ＰＥＮＡＮＡ8xFaXbzdPz
}
for (char c: new char[] {'A', 'C', 'G', 'T'}) {239Please respect copyright.ＰＥＮＡＮＡzy7sLxXpu6
for (int i = 0; i < node.length(); i++) {239Please respect copyright.ＰＥＮＡＮＡmKkza02KsD
String neighbor = node.substring(0, i) + c + node.substring(i + 1);239Please respect copyright.ＰＥＮＡＮＡHbcZByXYna
if (!seen.contains(neighbor) && Arrays.asList(bank).contains(neighbor)) {239Please respect copyright.ＰＥＮＡＮＡLGHyCCBqS6
queue.add(neighbor);239Please respect copyright.ＰＥＮＡＮＡybAieP96fp
seen.add(neighbor);239Please respect copyright.ＰＥＮＡＮＡjdLPXIjfrI
}239Please respect copyright.ＰＥＮＡＮＡc5Kk1Rz0pU
}239Please respect copyright.ＰＥＮＡＮＡMQ0WJLBqem
}239Please respect copyright.ＰＥＮＡＮＡT35AGcPV75
}239Please respect copyright.ＰＥＮＡＮＡ2VPht6gKSY
239Please respect copyright.ＰＥＮＡＮＡuTyI1dLha3
steps++;239Please respect copyright.ＰＥＮＡＮＡyBMZ8VPhj0
}239Please respect copyright.ＰＥＮＡＮＡldoeTy7S3d
// If we finish the BFS and did not find end, return -1.239Please respect copyright.ＰＥＮＡＮＡKvN1UIQpNg
return -1;239Please respect copyright.ＰＥＮＡＮＡlyXoObM25z
}239Please respect copyright.ＰＥＮＡＮＡWZmDMsgdA0
}
