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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)256Please respect copyright.ＰＥＮＡＮＡXZ6O3d1Y15
// better than use DFS as it just need to find out the shortest path.
class Solution {256Please respect copyright.ＰＥＮＡＮＡXiu0NHxhGt
public int minMutation(String start, String end, String[] bank) {256Please respect copyright.ＰＥＮＡＮＡlnDUFB0SUO
// 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.256Please respect copyright.ＰＥＮＡＮＡbNarROohaw
Queue<String> queue = new LinkedList<>();256Please respect copyright.ＰＥＮＡＮＡVH8cgPTcbA
Set<String> seen = new HashSet<>();256Please respect copyright.ＰＥＮＡＮＡ6olEMjQ8ft
queue.add(start);256Please respect copyright.ＰＥＮＡＮＡ5jqhu2VTf4
seen.add(start);256Please respect copyright.ＰＥＮＡＮＡFI8MfmGBGH
256Please respect copyright.ＰＥＮＡＮＡ37hL6vaN3L
int steps = 0;256Please respect copyright.ＰＥＮＡＮＡLl4yvo6THt
256Please respect copyright.ＰＥＮＡＮＡgpCv3zMhgp
while (!queue.isEmpty()) {256Please respect copyright.ＰＥＮＡＮＡfE0fA0d6xe
int nodesInQueue = queue.size();256Please respect copyright.ＰＥＮＡＮＡGZrwvuLi1n
for (int j = 0; j < nodesInQueue; j++) {256Please respect copyright.ＰＥＮＡＮＡdi8Qv8DMmI
String node = queue.remove();256Please respect copyright.ＰＥＮＡＮＡzyGFRrugLl
// 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)) {256Please respect copyright.ＰＥＮＡＮＡ3sCStWJf8V
return steps;256Please respect copyright.ＰＥＮＡＮＡ80BAQ1mxbp
}
for (char c: new char[] {'A', 'C', 'G', 'T'}) {256Please respect copyright.ＰＥＮＡＮＡAfSzGuxRKW
for (int i = 0; i < node.length(); i++) {256Please respect copyright.ＰＥＮＡＮＡObXQUs5BlN
String neighbor = node.substring(0, i) + c + node.substring(i + 1);256Please respect copyright.ＰＥＮＡＮＡlilCS4aWfa
if (!seen.contains(neighbor) && Arrays.asList(bank).contains(neighbor)) {256Please respect copyright.ＰＥＮＡＮＡYd1UJ6qdHx
queue.add(neighbor);256Please respect copyright.ＰＥＮＡＮＡVhjUqatKlm
seen.add(neighbor);256Please respect copyright.ＰＥＮＡＮＡdEROOvNOht
}256Please respect copyright.ＰＥＮＡＮＡGyF1SBM6rG
}256Please respect copyright.ＰＥＮＡＮＡkytjZKVxFP
}256Please respect copyright.ＰＥＮＡＮＡ8uL4PdHvF4
}256Please respect copyright.ＰＥＮＡＮＡaaFB0mP38v
256Please respect copyright.ＰＥＮＡＮＡ1V12lDaQOE
steps++;256Please respect copyright.ＰＥＮＡＮＡzWTpy9h6Fx
}256Please respect copyright.ＰＥＮＡＮＡGNSI2jruDI
// If we finish the BFS and did not find end, return -1.256Please respect copyright.ＰＥＮＡＮＡZzKqSFQGPH
return -1;256Please respect copyright.ＰＥＮＡＮＡ36AuxxoLrj
}256Please respect copyright.ＰＥＮＡＮＡRy1I54sEOp
}
