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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)235Please respect copyright.ＰＥＮＡＮＡvpTumNtJKQ
// better than use DFS as it just need to find out the shortest path.
class Solution {235Please respect copyright.ＰＥＮＡＮＡlCL6ZuZIeX
public int minMutation(String start, String end, String[] bank) {235Please respect copyright.ＰＥＮＡＮＡ8Gf55CbvPJ
// 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.235Please respect copyright.ＰＥＮＡＮＡFqU1JESg32
Queue<String> queue = new LinkedList<>();235Please respect copyright.ＰＥＮＡＮＡtVRrU36pmL
Set<String> seen = new HashSet<>();235Please respect copyright.ＰＥＮＡＮＡBZqpM6Ds2p
queue.add(start);235Please respect copyright.ＰＥＮＡＮＡrwfEMh8BsZ
seen.add(start);235Please respect copyright.ＰＥＮＡＮＡkeDBcGKHVc
235Please respect copyright.ＰＥＮＡＮＡpXU2AZsvkI
int steps = 0;235Please respect copyright.ＰＥＮＡＮＡEOjBZtiRxt
235Please respect copyright.ＰＥＮＡＮＡlo04RLOYA3
while (!queue.isEmpty()) {235Please respect copyright.ＰＥＮＡＮＡm5oFSRi7KN
int nodesInQueue = queue.size();235Please respect copyright.ＰＥＮＡＮＡRO5FPkH3wm
for (int j = 0; j < nodesInQueue; j++) {235Please respect copyright.ＰＥＮＡＮＡzBSVe17mm7
String node = queue.remove();235Please respect copyright.ＰＥＮＡＮＡWyOWHugODH
// 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)) {235Please respect copyright.ＰＥＮＡＮＡJ3Rjo3NASB
return steps;235Please respect copyright.ＰＥＮＡＮＡ43Iyeu88Hv
}
for (char c: new char[] {'A', 'C', 'G', 'T'}) {235Please respect copyright.ＰＥＮＡＮＡuSKuYWzHlB
for (int i = 0; i < node.length(); i++) {235Please respect copyright.ＰＥＮＡＮＡYJvKvcPIxE
String neighbor = node.substring(0, i) + c + node.substring(i + 1);235Please respect copyright.ＰＥＮＡＮＡEo5NU7yK0u
if (!seen.contains(neighbor) && Arrays.asList(bank).contains(neighbor)) {235Please respect copyright.ＰＥＮＡＮＡr82ujyXEyB
queue.add(neighbor);235Please respect copyright.ＰＥＮＡＮＡaqcFwH9ODf
seen.add(neighbor);235Please respect copyright.ＰＥＮＡＮＡXaDqSms6MS
}235Please respect copyright.ＰＥＮＡＮＡV2IuoSoPBc
}235Please respect copyright.ＰＥＮＡＮＡ3q8O4bGRWg
}235Please respect copyright.ＰＥＮＡＮＡpWgAhkA9Q4
}235Please respect copyright.ＰＥＮＡＮＡj2qbnGid1M
235Please respect copyright.ＰＥＮＡＮＡEKlABObP7d
steps++;235Please respect copyright.ＰＥＮＡＮＡ52krOz8o4k
}235Please respect copyright.ＰＥＮＡＮＡwTawSAXo3O
// If we finish the BFS and did not find end, return -1.235Please respect copyright.ＰＥＮＡＮＡqRYEK7pq0K
return -1;235Please respect copyright.ＰＥＮＡＮＡ2RA0y1ybjo
}235Please respect copyright.ＰＥＮＡＮＡGM8D7BFsM7
}
