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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)220Please respect copyright.ＰＥＮＡＮＡ06HzfKh9QV
// better than use DFS as it just need to find out the shortest path.
class Solution {220Please respect copyright.ＰＥＮＡＮＡaqOr9aTr1Y
public int minMutation(String start, String end, String[] bank) {220Please respect copyright.ＰＥＮＡＮＡocvjx1JplV
// 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.220Please respect copyright.ＰＥＮＡＮＡqNGkivhpqu
Queue<String> queue = new LinkedList<>();220Please respect copyright.ＰＥＮＡＮＡdXMg7Yajxl
Set<String> seen = new HashSet<>();220Please respect copyright.ＰＥＮＡＮＡunWWU01Mel
queue.add(start);220Please respect copyright.ＰＥＮＡＮＡLlcS4taUTh
seen.add(start);220Please respect copyright.ＰＥＮＡＮＡZJ34crXcgk
220Please respect copyright.ＰＥＮＡＮＡqK7SattPB1
int steps = 0;220Please respect copyright.ＰＥＮＡＮＡuFJa0z18M4
220Please respect copyright.ＰＥＮＡＮＡ1nhOgOny4z
while (!queue.isEmpty()) {220Please respect copyright.ＰＥＮＡＮＡaEMVh9EINd
int nodesInQueue = queue.size();220Please respect copyright.ＰＥＮＡＮＡxSqTmE8ZKj
for (int j = 0; j < nodesInQueue; j++) {220Please respect copyright.ＰＥＮＡＮＡoa0swq8Ek4
String node = queue.remove();220Please respect copyright.ＰＥＮＡＮＡaIy6q1MIf6
// 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)) {220Please respect copyright.ＰＥＮＡＮＡ1MIW8eGClO
return steps;220Please respect copyright.ＰＥＮＡＮＡ0QKLBL9bWs
}
for (char c: new char[] {'A', 'C', 'G', 'T'}) {220Please respect copyright.ＰＥＮＡＮＡ4D7yqbdjPR
for (int i = 0; i < node.length(); i++) {220Please respect copyright.ＰＥＮＡＮＡDxjFnJDTyl
String neighbor = node.substring(0, i) + c + node.substring(i + 1);220Please respect copyright.ＰＥＮＡＮＡ4y8Lt0nDaM
if (!seen.contains(neighbor) && Arrays.asList(bank).contains(neighbor)) {220Please respect copyright.ＰＥＮＡＮＡL7cszX3j4o
queue.add(neighbor);220Please respect copyright.ＰＥＮＡＮＡ5klC3MgSoE
seen.add(neighbor);220Please respect copyright.ＰＥＮＡＮＡOE0mhEBmbW
}220Please respect copyright.ＰＥＮＡＮＡfEM6eltQnB
}220Please respect copyright.ＰＥＮＡＮＡf8L7lX5Eje
}220Please respect copyright.ＰＥＮＡＮＡgfiNORYv92
}220Please respect copyright.ＰＥＮＡＮＡwshKW4miqL
220Please respect copyright.ＰＥＮＡＮＡO3H4iM5Oya
steps++;220Please respect copyright.ＰＥＮＡＮＡY9tua4eRAo
}220Please respect copyright.ＰＥＮＡＮＡXAsmvUuCGq
// If we finish the BFS and did not find end, return -1.220Please respect copyright.ＰＥＮＡＮＡBIwPVz64rj
return -1;220Please respect copyright.ＰＥＮＡＮＡ6mwR8BXVSE
}220Please respect copyright.ＰＥＮＡＮＡ032oUK6zuG
}
