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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)194Please respect copyright.ＰＥＮＡＮＡdXLjGPUx6Y
// better than use DFS as it just need to find out the shortest path.
class Solution {194Please respect copyright.ＰＥＮＡＮＡOYPD25dXH8
public int minMutation(String start, String end, String[] bank) {194Please respect copyright.ＰＥＮＡＮＡ1gng6Pjiez
// 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.194Please respect copyright.ＰＥＮＡＮＡveX4zBioja
Queue<String> queue = new LinkedList<>();194Please respect copyright.ＰＥＮＡＮＡn4nckjZ2nJ
Set<String> seen = new HashSet<>();194Please respect copyright.ＰＥＮＡＮＡEDFBKsoYkC
queue.add(start);194Please respect copyright.ＰＥＮＡＮＡ29R6eqaHRy
seen.add(start);194Please respect copyright.ＰＥＮＡＮＡazRZmUDQnX
194Please respect copyright.ＰＥＮＡＮＡnzyWyjkYry
int steps = 0;194Please respect copyright.ＰＥＮＡＮＡ3DnhgYZM54
194Please respect copyright.ＰＥＮＡＮＡAMfEIHMRTF
while (!queue.isEmpty()) {194Please respect copyright.ＰＥＮＡＮＡpSUTCaRbpO
int nodesInQueue = queue.size();194Please respect copyright.ＰＥＮＡＮＡBNl48eVTBK
for (int j = 0; j < nodesInQueue; j++) {194Please respect copyright.ＰＥＮＡＮＡA6jVkw3tuW
String node = queue.remove();194Please respect copyright.ＰＥＮＡＮＡXQw4HvL24o
// 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)) {194Please respect copyright.ＰＥＮＡＮＡMuzXpxqJr2
return steps;194Please respect copyright.ＰＥＮＡＮＡYGm7byAju8
}
for (char c: new char[] {'A', 'C', 'G', 'T'}) {194Please respect copyright.ＰＥＮＡＮＡz6GWmztJQ0
for (int i = 0; i < node.length(); i++) {194Please respect copyright.ＰＥＮＡＮＡAe7KcEsd5J
String neighbor = node.substring(0, i) + c + node.substring(i + 1);194Please respect copyright.ＰＥＮＡＮＡqm1T0qn5mU
if (!seen.contains(neighbor) && Arrays.asList(bank).contains(neighbor)) {194Please respect copyright.ＰＥＮＡＮＡ1loEIXrJ6q
queue.add(neighbor);194Please respect copyright.ＰＥＮＡＮＡMIXNYFuoih
seen.add(neighbor);194Please respect copyright.ＰＥＮＡＮＡs4mzlBxH68
}194Please respect copyright.ＰＥＮＡＮＡhRHXLEUP9e
}194Please respect copyright.ＰＥＮＡＮＡEcsxrTg1Gp
}194Please respect copyright.ＰＥＮＡＮＡ73G6SowrCp
}194Please respect copyright.ＰＥＮＡＮＡYvTVfYgaXd
194Please respect copyright.ＰＥＮＡＮＡnlvldcz5UN
steps++;194Please respect copyright.ＰＥＮＡＮＡExZRODQrYo
}194Please respect copyright.ＰＥＮＡＮＡRyV7l0Vn3Y
// If we finish the BFS and did not find end, return -1.194Please respect copyright.ＰＥＮＡＮＡ4YHOr5O4f3
return -1;194Please respect copyright.ＰＥＮＡＮＡZ3bjTrHHXC
}194Please respect copyright.ＰＥＮＡＮＡJIkUUZAlrW
}
