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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)268Please respect copyright.ＰＥＮＡＮＡyd5QWW5Rq4
// better than use DFS as it just need to find out the shortest path.
class Solution {268Please respect copyright.ＰＥＮＡＮＡ49BMuQN21Q
public int minMutation(String start, String end, String[] bank) {268Please respect copyright.ＰＥＮＡＮＡIF9YyNX3XB
// 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.268Please respect copyright.ＰＥＮＡＮＡ8MY2xN0Yvt
Queue<String> queue = new LinkedList<>();268Please respect copyright.ＰＥＮＡＮＡ7N348inrLl
Set<String> seen = new HashSet<>();268Please respect copyright.ＰＥＮＡＮＡS5MSyBYIdd
queue.add(start);268Please respect copyright.ＰＥＮＡＮＡKqEZL9a7t2
seen.add(start);268Please respect copyright.ＰＥＮＡＮＡtESJm2k7AN
268Please respect copyright.ＰＥＮＡＮＡ5eD6XyHtf4
int steps = 0;268Please respect copyright.ＰＥＮＡＮＡQxuyDCMxBW
268Please respect copyright.ＰＥＮＡＮＡ6csPdkwTxB
while (!queue.isEmpty()) {268Please respect copyright.ＰＥＮＡＮＡ2YFl5nqlzi
int nodesInQueue = queue.size();268Please respect copyright.ＰＥＮＡＮＡvOea4doeFh
for (int j = 0; j < nodesInQueue; j++) {268Please respect copyright.ＰＥＮＡＮＡXslZYJKYOc
String node = queue.remove();268Please respect copyright.ＰＥＮＡＮＡsisw47hiuC
// 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)) {268Please respect copyright.ＰＥＮＡＮＡb3edi66CCf
return steps;268Please respect copyright.ＰＥＮＡＮＡ5rjdA1FLv3
}
for (char c: new char[] {'A', 'C', 'G', 'T'}) {268Please respect copyright.ＰＥＮＡＮＡbuEw7gcZSb
for (int i = 0; i < node.length(); i++) {268Please respect copyright.ＰＥＮＡＮＡMD1xjkcn8h
String neighbor = node.substring(0, i) + c + node.substring(i + 1);268Please respect copyright.ＰＥＮＡＮＡ9hcFX1540t
if (!seen.contains(neighbor) && Arrays.asList(bank).contains(neighbor)) {268Please respect copyright.ＰＥＮＡＮＡYbomT2s3JQ
queue.add(neighbor);268Please respect copyright.ＰＥＮＡＮＡ9DwiMslvbQ
seen.add(neighbor);268Please respect copyright.ＰＥＮＡＮＡIu3U38FK5J
}268Please respect copyright.ＰＥＮＡＮＡUBFjOcQsoP
}268Please respect copyright.ＰＥＮＡＮＡnDMzoJIi6r
}268Please respect copyright.ＰＥＮＡＮＡhJBRaYP6tT
}268Please respect copyright.ＰＥＮＡＮＡiwp9UXFDVQ
268Please respect copyright.ＰＥＮＡＮＡnzFKCtwqEF
steps++;268Please respect copyright.ＰＥＮＡＮＡ38nNx0pjMn
}268Please respect copyright.ＰＥＮＡＮＡEj83ATF1DB
// If we finish the BFS and did not find end, return -1.268Please respect copyright.ＰＥＮＡＮＡ6ecQbL2IwV
return -1;268Please respect copyright.ＰＥＮＡＮＡn7Kzdwqab1
}268Please respect copyright.ＰＥＮＡＮＡazX6ZNvOQy
}
