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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)254Please respect copyright.ＰＥＮＡＮＡbXEFqnHLC1
// better than use DFS as it just need to find out the shortest path.
class Solution {254Please respect copyright.ＰＥＮＡＮＡWl2oRkPym3
public int minMutation(String start, String end, String[] bank) {254Please respect copyright.ＰＥＮＡＮＡg3s2HFbqHn
// 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.254Please respect copyright.ＰＥＮＡＮＡcyf2uCMqeS
Queue<String> queue = new LinkedList<>();254Please respect copyright.ＰＥＮＡＮＡ84r0Wdgd2p
Set<String> seen = new HashSet<>();254Please respect copyright.ＰＥＮＡＮＡ2gxWLHMfno
queue.add(start);254Please respect copyright.ＰＥＮＡＮＡZxnFE008EP
seen.add(start);254Please respect copyright.ＰＥＮＡＮＡdi8hnJNSho
254Please respect copyright.ＰＥＮＡＮＡHNNILdqbLq
int steps = 0;254Please respect copyright.ＰＥＮＡＮＡC6KXNmqc2W
254Please respect copyright.ＰＥＮＡＮＡjpScn4Mlpy
while (!queue.isEmpty()) {254Please respect copyright.ＰＥＮＡＮＡ0556XeRpal
int nodesInQueue = queue.size();254Please respect copyright.ＰＥＮＡＮＡRzgCQrsAkv
for (int j = 0; j < nodesInQueue; j++) {254Please respect copyright.ＰＥＮＡＮＡ043KXJfYhi
String node = queue.remove();254Please respect copyright.ＰＥＮＡＮＡgAT9U2X8rP
// 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)) {254Please respect copyright.ＰＥＮＡＮＡOsFoJjoXtl
return steps;254Please respect copyright.ＰＥＮＡＮＡ8KO3K7eJp7
}
for (char c: new char[] {'A', 'C', 'G', 'T'}) {254Please respect copyright.ＰＥＮＡＮＡdQ1jFAMaHa
for (int i = 0; i < node.length(); i++) {254Please respect copyright.ＰＥＮＡＮＡiD94rIggEp
String neighbor = node.substring(0, i) + c + node.substring(i + 1);254Please respect copyright.ＰＥＮＡＮＡ9SV7WRfils
if (!seen.contains(neighbor) && Arrays.asList(bank).contains(neighbor)) {254Please respect copyright.ＰＥＮＡＮＡho7AtQ36Rt
queue.add(neighbor);254Please respect copyright.ＰＥＮＡＮＡxm6CIFclYn
seen.add(neighbor);254Please respect copyright.ＰＥＮＡＮＡ3VhLbhc30e
}254Please respect copyright.ＰＥＮＡＮＡvEiQAmiXTB
}254Please respect copyright.ＰＥＮＡＮＡrKd7xukHr5
}254Please respect copyright.ＰＥＮＡＮＡkyo734lXDW
}254Please respect copyright.ＰＥＮＡＮＡLXP4lSxVa3
254Please respect copyright.ＰＥＮＡＮＡzVYJpMcO0P
steps++;254Please respect copyright.ＰＥＮＡＮＡ9pxeisJGFo
}254Please respect copyright.ＰＥＮＡＮＡSdSBFAsaPQ
// If we finish the BFS and did not find end, return -1.254Please respect copyright.ＰＥＮＡＮＡ83bqmiz8s0
return -1;254Please respect copyright.ＰＥＮＡＮＡbIoSEneqWI
}254Please respect copyright.ＰＥＮＡＮＡcOlAFycwgY
}
