Class Solution
Medium
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.
Example 1:
Input: start = “AACCGGTT”, end = “AACCGGTA”, bank = [“AACCGGTA”]
Output: 1
Example 2:
Input: start = “AACCGGTT”, end = “AAACGGTA”, bank = [“AACCGGTA”,“AACCGCTA”,“AAACGGTA”]
Output: 2
Example 3:
Input: start = “AAAAACCC”, end = “AACCCCCC”, bank = [“AAAACCCC”,“AAACCCCC”,“AACCCCCC”]
Output: 3
Constraints:
start.length == 8
end.length == 8
0 <= bank.length <= 10
bank[i].length == 8
start
,end
, andbank[i]
consist of only the characters['A', 'C', 'G', 'T']
.
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Constructor Summary
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Method Summary
Modifier and TypeMethodDescriptionint
minMutation
(java.lang.String start, java.lang.String end, java.lang.String[] bank) Methods inherited from class java.lang.Object
clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait
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Constructor Details
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Solution
public Solution()
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Method Details
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minMutation
public int minMutation(java.lang.String start, java.lang.String end, java.lang.String[] bank)
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