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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)217Please respect copyright.ＰＥＮＡＮＡfFhPT4qi3V
// better than use DFS as it just need to find out the shortest path.
class Solution {217Please respect copyright.ＰＥＮＡＮＡnpu1VZM4QI
public int minMutation(String start, String end, String[] bank) {217Please respect copyright.ＰＥＮＡＮＡOGLa4oOJxM
// 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.217Please respect copyright.ＰＥＮＡＮＡAeIVFawi5H
Queue<String> queue = new LinkedList<>();217Please respect copyright.ＰＥＮＡＮＡ0WcrvjKecx
Set<String> seen = new HashSet<>();217Please respect copyright.ＰＥＮＡＮＡE9v3ihQLnv
queue.add(start);217Please respect copyright.ＰＥＮＡＮＡZ7sVS55ePF
seen.add(start);217Please respect copyright.ＰＥＮＡＮＡQDRvO9gznj
217Please respect copyright.ＰＥＮＡＮＡH815SdZv0o
int steps = 0;217Please respect copyright.ＰＥＮＡＮＡ5r49guEt4K
217Please respect copyright.ＰＥＮＡＮＡUq8Wd1f0lb
while (!queue.isEmpty()) {217Please respect copyright.ＰＥＮＡＮＡMnfAYh6mmQ
int nodesInQueue = queue.size();217Please respect copyright.ＰＥＮＡＮＡfQloSJ0asU
for (int j = 0; j < nodesInQueue; j++) {217Please respect copyright.ＰＥＮＡＮＡw17ns9ryYb
String node = queue.remove();217Please respect copyright.ＰＥＮＡＮＡ4tLAOJmQ1M
// 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)) {217Please respect copyright.ＰＥＮＡＮＡU2vnEuBszF
return steps;217Please respect copyright.ＰＥＮＡＮＡsbzmErGLMV
}
for (char c: new char[] {'A', 'C', 'G', 'T'}) {217Please respect copyright.ＰＥＮＡＮＡSrpCBU71Y4
for (int i = 0; i < node.length(); i++) {217Please respect copyright.ＰＥＮＡＮＡT1yfJJyGyU
String neighbor = node.substring(0, i) + c + node.substring(i + 1);217Please respect copyright.ＰＥＮＡＮＡx1MF3lKdWS
if (!seen.contains(neighbor) && Arrays.asList(bank).contains(neighbor)) {217Please respect copyright.ＰＥＮＡＮＡrFlL0Tbzzq
queue.add(neighbor);217Please respect copyright.ＰＥＮＡＮＡ9HVeCkmAhx
seen.add(neighbor);217Please respect copyright.ＰＥＮＡＮＡBw2Z86P6bX
}217Please respect copyright.ＰＥＮＡＮＡPnY9SzcS6D
}217Please respect copyright.ＰＥＮＡＮＡs0iCQlGVJG
}217Please respect copyright.ＰＥＮＡＮＡIfr84ArlsX
}217Please respect copyright.ＰＥＮＡＮＡiwCFH1fdka
217Please respect copyright.ＰＥＮＡＮＡVHutmJbRDw
steps++;217Please respect copyright.ＰＥＮＡＮＡxMkRX5YZ3D
}217Please respect copyright.ＰＥＮＡＮＡXTtrsYvhdO
// If we finish the BFS and did not find end, return -1.217Please respect copyright.ＰＥＮＡＮＡtY4ekdaROb
return -1;217Please respect copyright.ＰＥＮＡＮＡ2s36GUf63n
}217Please respect copyright.ＰＥＮＡＮＡYyctKLlI0s
}
