Java

• import java.util.*;

/**

433. Minimum Genetic Mutation

A gene string can be represented by an 8-character long string, with choices from "A", "C", "G", "T".

Suppose we need to investigate about a mutation (mutation from "start" to "end"),
where ONE mutation is defined as ONE single character changed in the gene string.

For example, "AACCGGTT" -> "AACCGGTA" is 1 mutation.

Also, there is a given gene "bank", which records all the valid gene mutations.
A gene must be in the bank to make it a valid gene string.

Now, given 3 things - start, end, bank,
your task is to determine what is the minimum number of mutations needed to mutate from "start" to "end".
If there is no such a mutation, return -1.

Note:

Starting point is assumed to be valid, so it might not be included in the bank.
If multiple mutations are needed, all mutations during in the sequence must be valid.
You may assume start and end string is not the same.

Example 1:

start: "AACCGGTT"
end:   "AACCGGTA"
bank: ["AACCGGTA"]

return: 1

Example 2:

start: "AACCGGTT"
end:   "AAACGGTA"
bank: ["AACCGGTA", "AACCGCTA", "AAACGGTA"]

return: 2

Example 3:

start: "AAAAACCC"
end:   "AACCCCCC"
bank: ["AAAACCCC", "AAACCCCC", "AACCCCCC"]

return: 3

@tag-backtracking

*/

public class Minimum_Genetic_Mutation {

// ref: https://leetcode.com/problems/minimum-genetic-mutation/discuss/91484/Java-Solution-using-BFS
public class Solution {
public int minMutation(String start, String end, String[] bank) {

if (start.equals(end)) {
return 0;
}

// construct set
Set<String> bankSet = new HashSet<>();

char[] charSet = new char[]{'A', 'C', 'G', 'T'};

int level = 0;
Set<String> visited = new HashSet<>();
queue.offer(start);

while (!queue.isEmpty()) {
int size = queue.size(); // get this level count
while (size-- > 0) { // @note: this will gurantee go over all this level
String curr = queue.poll();
if (curr.equals(end)) {
return level;
}

char[] currArray = curr.toCharArray();
for (int i = 0; i < currArray.length; i++) {

char old = currArray[i]; // save char in this index

for (char c : charSet) {
currArray[i] = c;
String next = new String(currArray);
if (!visited.contains(next) && bankSet.contains(next)) {
queue.offer(next);
}
}

currArray[i] = old; // restore after all above char trials
}
}
level++;
}

return -1;
}
}
}

• Todo

• while queue 不空：
cur = queue.pop()
if cur 有效且未被访问过：
进行处理
for 节点 in cur 的所有相邻节点：
if 该节点有效：
queue.push(该节点)