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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)260Please respect copyright.ＰＥＮＡＮＡ5mRJUGtcLo
// better than use DFS as it just need to find out the shortest path.
class Solution {260Please respect copyright.ＰＥＮＡＮＡW5XHOnQCca
public int minMutation(String start, String end, String[] bank) {260Please respect copyright.ＰＥＮＡＮＡiLE1WKmBgL
// 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.260Please respect copyright.ＰＥＮＡＮＡeu3Uf9Zbwg
Queue<String> queue = new LinkedList<>();260Please respect copyright.ＰＥＮＡＮＡrbHBsYglwt
Set<String> seen = new HashSet<>();260Please respect copyright.ＰＥＮＡＮＡd6wcWBxytb
queue.add(start);260Please respect copyright.ＰＥＮＡＮＡi1dsNzG7QR
seen.add(start);260Please respect copyright.ＰＥＮＡＮＡHpK3QbQem4
260Please respect copyright.ＰＥＮＡＮＡMf4URhAeeF
int steps = 0;260Please respect copyright.ＰＥＮＡＮＡhFMx5pABT1
260Please respect copyright.ＰＥＮＡＮＡhIBMVaygTs
while (!queue.isEmpty()) {260Please respect copyright.ＰＥＮＡＮＡesbYGmDlfW
int nodesInQueue = queue.size();260Please respect copyright.ＰＥＮＡＮＡEK4IusjhNh
for (int j = 0; j < nodesInQueue; j++) {260Please respect copyright.ＰＥＮＡＮＡwkb2kvugkX
String node = queue.remove();260Please respect copyright.ＰＥＮＡＮＡDBBUyKeXMX
// 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)) {260Please respect copyright.ＰＥＮＡＮＡcQZ1d0iM4C
return steps;260Please respect copyright.ＰＥＮＡＮＡXxBmETrngb
}
for (char c: new char[] {'A', 'C', 'G', 'T'}) {260Please respect copyright.ＰＥＮＡＮＡnYnVExGPft
for (int i = 0; i < node.length(); i++) {260Please respect copyright.ＰＥＮＡＮＡ9H8XyJQPcy
String neighbor = node.substring(0, i) + c + node.substring(i + 1);260Please respect copyright.ＰＥＮＡＮＡZ7awXISu4x
if (!seen.contains(neighbor) && Arrays.asList(bank).contains(neighbor)) {260Please respect copyright.ＰＥＮＡＮＡLcMywsl24k
queue.add(neighbor);260Please respect copyright.ＰＥＮＡＮＡbttYU0qSUf
seen.add(neighbor);260Please respect copyright.ＰＥＮＡＮＡeJcmWMCkUM
}260Please respect copyright.ＰＥＮＡＮＡdYFb8OHKYn
}260Please respect copyright.ＰＥＮＡＮＡmHGzwPeAcm
}260Please respect copyright.ＰＥＮＡＮＡQRj055EAfu
}260Please respect copyright.ＰＥＮＡＮＡr0dh3TxJit
260Please respect copyright.ＰＥＮＡＮＡcMbT4v8Rtt
steps++;260Please respect copyright.ＰＥＮＡＮＡ5gB137hsgw
}260Please respect copyright.ＰＥＮＡＮＡZ3ZHH0nAgu
// If we finish the BFS and did not find end, return -1.260Please respect copyright.ＰＥＮＡＮＡLc3vhJnEbO
return -1;260Please respect copyright.ＰＥＮＡＮＡ8JpW3pcMhC
}260Please respect copyright.ＰＥＮＡＮＡVKecbpG7iB
}
