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: 3Code (Java):
class Solution { public int minMutation(String start, String end, String[] bank) { Queue<String> queue = new LinkedList<>(); int steps = 0; Set<String> bankSet = new HashSet<>(); for (String s : bank) { bankSet.add(s); } queue.offer(start); bankSet.remove(start); char[] gene = new char[]{'A', 'C', 'G', 'T'}; while (!queue.isEmpty()) { steps += 1; int size = queue.size(); for (int k = 0; k < size; k++) { char[] curr = queue.poll().toCharArray(); for (int i = 0; i < curr.length; i++) { char ori = curr[i]; for (char c : gene) { if (c == ori) { continue; } curr[i] = c; String mutated = new String(curr); if (bankSet.contains(mutated)) { if (mutated.equals(end)) { return steps; } queue.offer(mutated); bankSet.remove(mutated); } } curr[i] = ori; } } } return -1; } }
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