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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)140Please respect copyright.ＰＥＮＡＮＡBnFJ3OyJmh
// better than use DFS as it just need to find out the shortest path.
class Solution {140Please respect copyright.ＰＥＮＡＮＡAytyNBy4Jr
public int minMutation(String start, String end, String[] bank) {140Please respect copyright.ＰＥＮＡＮＡb8HNN31n40
// 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.140Please respect copyright.ＰＥＮＡＮＡwSMeTLXoBg
Queue<String> queue = new LinkedList<>();140Please respect copyright.ＰＥＮＡＮＡ2QOxMalUxf
Set<String> seen = new HashSet<>();140Please respect copyright.ＰＥＮＡＮＡEJXMGpzc9f
queue.add(start);140Please respect copyright.ＰＥＮＡＮＡe7URf2rDgG
seen.add(start);140Please respect copyright.ＰＥＮＡＮＡKSkdNGLEcX
140Please respect copyright.ＰＥＮＡＮＡaoyKksmmuL
int steps = 0;140Please respect copyright.ＰＥＮＡＮＡEkefnyZxeE
140Please respect copyright.ＰＥＮＡＮＡUoVvGjPixv
while (!queue.isEmpty()) {140Please respect copyright.ＰＥＮＡＮＡv62IRxtFdh
int nodesInQueue = queue.size();140Please respect copyright.ＰＥＮＡＮＡrs4tGSpvnY
for (int j = 0; j < nodesInQueue; j++) {140Please respect copyright.ＰＥＮＡＮＡvVxlPbk3KI
String node = queue.remove();140Please respect copyright.ＰＥＮＡＮＡlEP4PQczWh
// 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)) {140Please respect copyright.ＰＥＮＡＮＡFhT88UOXui
return steps;140Please respect copyright.ＰＥＮＡＮＡ7S2lmvVgQt
}
for (char c: new char[] {'A', 'C', 'G', 'T'}) {140Please respect copyright.ＰＥＮＡＮＡm9aDD0iDkp
for (int i = 0; i < node.length(); i++) {140Please respect copyright.ＰＥＮＡＮＡg9pVXntz7c
String neighbor = node.substring(0, i) + c + node.substring(i + 1);140Please respect copyright.ＰＥＮＡＮＡwmHQHJUcwq
if (!seen.contains(neighbor) && Arrays.asList(bank).contains(neighbor)) {140Please respect copyright.ＰＥＮＡＮＡhLdIgIXS58
queue.add(neighbor);140Please respect copyright.ＰＥＮＡＮＡ5rrSIYFnBN
seen.add(neighbor);140Please respect copyright.ＰＥＮＡＮＡCzjV6B97GD
}140Please respect copyright.ＰＥＮＡＮＡr8N7daXNW8
}140Please respect copyright.ＰＥＮＡＮＡ55OkpcjDGK
}140Please respect copyright.ＰＥＮＡＮＡJnUiq2JQnt
}140Please respect copyright.ＰＥＮＡＮＡkb1501nvHZ
140Please respect copyright.ＰＥＮＡＮＡPX03kmcSKp
steps++;140Please respect copyright.ＰＥＮＡＮＡ9dIgCgImkD
}140Please respect copyright.ＰＥＮＡＮＡqXQibbaKHg
// If we finish the BFS and did not find end, return -1.140Please respect copyright.ＰＥＮＡＮＡDAjofsDBqy
return -1;140Please respect copyright.ＰＥＮＡＮＡo1VItE7PTN
}140Please respect copyright.ＰＥＮＡＮＡ4D4XLfabb6
}
