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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)267Please respect copyright.ＰＥＮＡＮＡrK1I90yUxM
// better than use DFS as it just need to find out the shortest path.
class Solution {267Please respect copyright.ＰＥＮＡＮＡDRO9zslkIv
public int minMutation(String start, String end, String[] bank) {267Please respect copyright.ＰＥＮＡＮＡ0bsd05PccD
// 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.267Please respect copyright.ＰＥＮＡＮＡKq5anqfLSf
Queue<String> queue = new LinkedList<>();267Please respect copyright.ＰＥＮＡＮＡnY1So2Hiwc
Set<String> seen = new HashSet<>();267Please respect copyright.ＰＥＮＡＮＡimTnUIGir8
queue.add(start);267Please respect copyright.ＰＥＮＡＮＡArHM765AxM
seen.add(start);267Please respect copyright.ＰＥＮＡＮＡbcQlgqiVz6
267Please respect copyright.ＰＥＮＡＮＡRePq4Z98Mq
int steps = 0;267Please respect copyright.ＰＥＮＡＮＡsyK9AYNP7H
267Please respect copyright.ＰＥＮＡＮＡu1ujVwv69h
while (!queue.isEmpty()) {267Please respect copyright.ＰＥＮＡＮＡkrIn1Xlz58
int nodesInQueue = queue.size();267Please respect copyright.ＰＥＮＡＮＡQd6iHX7OYT
for (int j = 0; j < nodesInQueue; j++) {267Please respect copyright.ＰＥＮＡＮＡUrdWqChIw8
String node = queue.remove();267Please respect copyright.ＰＥＮＡＮＡvi8CJld9c2
// 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)) {267Please respect copyright.ＰＥＮＡＮＡB8qc0FrA7m
return steps;267Please respect copyright.ＰＥＮＡＮＡVSIFvM8KAL
}
for (char c: new char[] {'A', 'C', 'G', 'T'}) {267Please respect copyright.ＰＥＮＡＮＡfIkSRX3Nta
for (int i = 0; i < node.length(); i++) {267Please respect copyright.ＰＥＮＡＮＡotshuRKvP2
String neighbor = node.substring(0, i) + c + node.substring(i + 1);267Please respect copyright.ＰＥＮＡＮＡA5DGhRRMwt
if (!seen.contains(neighbor) && Arrays.asList(bank).contains(neighbor)) {267Please respect copyright.ＰＥＮＡＮＡJn5fPFXAyc
queue.add(neighbor);267Please respect copyright.ＰＥＮＡＮＡxDMwDxV90d
seen.add(neighbor);267Please respect copyright.ＰＥＮＡＮＡUCoWmFYGJO
}267Please respect copyright.ＰＥＮＡＮＡ3RX6R9o4H2
}267Please respect copyright.ＰＥＮＡＮＡPauePPMI5l
}267Please respect copyright.ＰＥＮＡＮＡOdaCjNcFDQ
}267Please respect copyright.ＰＥＮＡＮＡJsbf6W4rMq
267Please respect copyright.ＰＥＮＡＮＡRGWovzxthf
steps++;267Please respect copyright.ＰＥＮＡＮＡ2tUQ5nq8vy
}267Please respect copyright.ＰＥＮＡＮＡ5kGKVgcMVV
// If we finish the BFS and did not find end, return -1.267Please respect copyright.ＰＥＮＡＮＡgkB5gijWWl
return -1;267Please respect copyright.ＰＥＮＡＮＡMgZmFilsZo
}267Please respect copyright.ＰＥＮＡＮＡ7R5eGoAQRn
}
