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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)275Please respect copyright.ＰＥＮＡＮＡqa8WPQTooT
// better than use DFS as it just need to find out the shortest path.
class Solution {275Please respect copyright.ＰＥＮＡＮＡNcaGjoILiM
public int minMutation(String start, String end, String[] bank) {275Please respect copyright.ＰＥＮＡＮＡXQSulOoLgu
// 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.275Please respect copyright.ＰＥＮＡＮＡFoiK61h7gX
Queue<String> queue = new LinkedList<>();275Please respect copyright.ＰＥＮＡＮＡGeIWTJ3OQp
Set<String> seen = new HashSet<>();275Please respect copyright.ＰＥＮＡＮＡozBG1v0dP9
queue.add(start);275Please respect copyright.ＰＥＮＡＮＡwxL8LmLGAg
seen.add(start);275Please respect copyright.ＰＥＮＡＮＡS7DbgPR8aF
275Please respect copyright.ＰＥＮＡＮＡMt71Ina4PZ
int steps = 0;275Please respect copyright.ＰＥＮＡＮＡ59vYM0qVu9
275Please respect copyright.ＰＥＮＡＮＡwfGaG6AIpv
while (!queue.isEmpty()) {275Please respect copyright.ＰＥＮＡＮＡzZz5lpmqJI
int nodesInQueue = queue.size();275Please respect copyright.ＰＥＮＡＮＡ6qQpSenYrr
for (int j = 0; j < nodesInQueue; j++) {275Please respect copyright.ＰＥＮＡＮＡOIT1AdeITO
String node = queue.remove();275Please respect copyright.ＰＥＮＡＮＡXB1kwyjW3d
// 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)) {275Please respect copyright.ＰＥＮＡＮＡddjzXux6DH
return steps;275Please respect copyright.ＰＥＮＡＮＡBgan2nq47U
}
for (char c: new char[] {'A', 'C', 'G', 'T'}) {275Please respect copyright.ＰＥＮＡＮＡbEgtfPrhTA
for (int i = 0; i < node.length(); i++) {275Please respect copyright.ＰＥＮＡＮＡYiYv95RBVn
String neighbor = node.substring(0, i) + c + node.substring(i + 1);275Please respect copyright.ＰＥＮＡＮＡut3PmiPv0I
if (!seen.contains(neighbor) && Arrays.asList(bank).contains(neighbor)) {275Please respect copyright.ＰＥＮＡＮＡEIiP1I3JIc
queue.add(neighbor);275Please respect copyright.ＰＥＮＡＮＡrUd4dvUShm
seen.add(neighbor);275Please respect copyright.ＰＥＮＡＮＡhQPJb3qfeC
}275Please respect copyright.ＰＥＮＡＮＡ0GpgBZya2z
}275Please respect copyright.ＰＥＮＡＮＡh9qWT5GeLF
}275Please respect copyright.ＰＥＮＡＮＡ5aSXVeWcoJ
}275Please respect copyright.ＰＥＮＡＮＡlbhmSV1FUT
275Please respect copyright.ＰＥＮＡＮＡa571fF7GCE
steps++;275Please respect copyright.ＰＥＮＡＮＡNhLBEWsSPA
}275Please respect copyright.ＰＥＮＡＮＡjefiwHNd1S
// If we finish the BFS and did not find end, return -1.275Please respect copyright.ＰＥＮＡＮＡWMVYjsUhv8
return -1;275Please respect copyright.ＰＥＮＡＮＡ9DfmHWmYx6
}275Please respect copyright.ＰＥＮＡＮＡ3RRLDOCI1S
}
