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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)266Please respect copyright.ＰＥＮＡＮＡfdzEkVzaxO
// better than use DFS as it just need to find out the shortest path.
class Solution {266Please respect copyright.ＰＥＮＡＮＡiuRmoGXtbw
public int minMutation(String start, String end, String[] bank) {266Please respect copyright.ＰＥＮＡＮＡyPOQH5lyZ6
// 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.266Please respect copyright.ＰＥＮＡＮＡsGtzoYgIql
Queue<String> queue = new LinkedList<>();266Please respect copyright.ＰＥＮＡＮＡiG0iXsbrcT
Set<String> seen = new HashSet<>();266Please respect copyright.ＰＥＮＡＮＡ0t52TaixYL
queue.add(start);266Please respect copyright.ＰＥＮＡＮＡ0993ouQlCz
seen.add(start);266Please respect copyright.ＰＥＮＡＮＡTpC4UKsFjt
266Please respect copyright.ＰＥＮＡＮＡL9lhwG1QcU
int steps = 0;266Please respect copyright.ＰＥＮＡＮＡPTk6riIAEm
266Please respect copyright.ＰＥＮＡＮＡAZYTu1B0Br
while (!queue.isEmpty()) {266Please respect copyright.ＰＥＮＡＮＡPs1tZIRLYs
int nodesInQueue = queue.size();266Please respect copyright.ＰＥＮＡＮＡg8DOBeK9UM
for (int j = 0; j < nodesInQueue; j++) {266Please respect copyright.ＰＥＮＡＮＡTX01xoBczv
String node = queue.remove();266Please respect copyright.ＰＥＮＡＮＡROsq9mk3kW
// 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)) {266Please respect copyright.ＰＥＮＡＮＡuI4q0cGUuJ
return steps;266Please respect copyright.ＰＥＮＡＮＡYkjJJmmFQO
}
for (char c: new char[] {'A', 'C', 'G', 'T'}) {266Please respect copyright.ＰＥＮＡＮＡVkIBEHWP27
for (int i = 0; i < node.length(); i++) {266Please respect copyright.ＰＥＮＡＮＡ7nb00jR6bW
String neighbor = node.substring(0, i) + c + node.substring(i + 1);266Please respect copyright.ＰＥＮＡＮＡfuIpXFCctP
if (!seen.contains(neighbor) && Arrays.asList(bank).contains(neighbor)) {266Please respect copyright.ＰＥＮＡＮＡZ4vzdHA91Y
queue.add(neighbor);266Please respect copyright.ＰＥＮＡＮＡxIZwspnhCS
seen.add(neighbor);266Please respect copyright.ＰＥＮＡＮＡWmkb8zvxxP
}266Please respect copyright.ＰＥＮＡＮＡ6WGk8vk12S
}266Please respect copyright.ＰＥＮＡＮＡPWc9dLXLbM
}266Please respect copyright.ＰＥＮＡＮＡBhQ0PwFdvc
}266Please respect copyright.ＰＥＮＡＮＡAjRYQRydme
266Please respect copyright.ＰＥＮＡＮＡqNcFYaFX26
steps++;266Please respect copyright.ＰＥＮＡＮＡSF4hPoETTy
}266Please respect copyright.ＰＥＮＡＮＡzWw2xdQS4A
// If we finish the BFS and did not find end, return -1.266Please respect copyright.ＰＥＮＡＮＡbajG5VxJLG
return -1;266Please respect copyright.ＰＥＮＡＮＡIHchPNhDfA
}266Please respect copyright.ＰＥＮＡＮＡkkn9XiD6Cv
}
