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)294Please respect copyright.ＰＥＮＡＮＡheMBt0iAtY
// better than use DFS as it just need to find out the shortest path.
class Solution {294Please respect copyright.ＰＥＮＡＮＡLJijp53z3i
public int minMutation(String start, String end, String[] bank) {294Please respect copyright.ＰＥＮＡＮＡ6jBFXF5Ic6
// 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.294Please respect copyright.ＰＥＮＡＮＡCvFFeu2nHH
Queue<String> queue = new LinkedList<>();294Please respect copyright.ＰＥＮＡＮＡcsNfQQcdsv
Set<String> seen = new HashSet<>();294Please respect copyright.ＰＥＮＡＮＡXnD3hDNDVB
queue.add(start);294Please respect copyright.ＰＥＮＡＮＡ5FQWYUYop9
seen.add(start);294Please respect copyright.ＰＥＮＡＮＡgJQr2kP8v7
294Please respect copyright.ＰＥＮＡＮＡYZUkRXhIek
int steps = 0;294Please respect copyright.ＰＥＮＡＮＡmqUIjTFTaQ
294Please respect copyright.ＰＥＮＡＮＡ2dSLktTSEp
while (!queue.isEmpty()) {294Please respect copyright.ＰＥＮＡＮＡNKgHFODFdA
int nodesInQueue = queue.size();294Please respect copyright.ＰＥＮＡＮＡkUp8fYiPjL
for (int j = 0; j < nodesInQueue; j++) {294Please respect copyright.ＰＥＮＡＮＡR8ebEXHjiD
String node = queue.remove();294Please respect copyright.ＰＥＮＡＮＡoHNuYPRFlc
// 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)) {294Please respect copyright.ＰＥＮＡＮＡhqHb2AaiMX
return steps;294Please respect copyright.ＰＥＮＡＮＡctzf5H8Ybl
}
for (char c: new char[] {'A', 'C', 'G', 'T'}) {294Please respect copyright.ＰＥＮＡＮＡH8bJlHKpR5
for (int i = 0; i < node.length(); i++) {294Please respect copyright.ＰＥＮＡＮＡoSNXzZ4rHO
String neighbor = node.substring(0, i) + c + node.substring(i + 1);294Please respect copyright.ＰＥＮＡＮＡkTLmXZi7RJ
if (!seen.contains(neighbor) && Arrays.asList(bank).contains(neighbor)) {294Please respect copyright.ＰＥＮＡＮＡnGwxtjU1zg
queue.add(neighbor);294Please respect copyright.ＰＥＮＡＮＡ5iecCP4W0H
seen.add(neighbor);294Please respect copyright.ＰＥＮＡＮＡv7prinkWhK
}294Please respect copyright.ＰＥＮＡＮＡms2eIldQxb
}294Please respect copyright.ＰＥＮＡＮＡxU11kZnep6
}294Please respect copyright.ＰＥＮＡＮＡOft3jH3XhI
}294Please respect copyright.ＰＥＮＡＮＡyvk7M1CDg9
294Please respect copyright.ＰＥＮＡＮＡy31ZUrm5Il
steps++;294Please respect copyright.ＰＥＮＡＮＡL8DItmMVBg
}294Please respect copyright.ＰＥＮＡＮＡPFCFZiVg84
// If we finish the BFS and did not find end, return -1.294Please respect copyright.ＰＥＮＡＮＡDvEFYvlwPk
return -1;294Please respect copyright.ＰＥＮＡＮＡd2FVgmw6hd
}294Please respect copyright.ＰＥＮＡＮＡPqO5f8GuMA
}
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