Algorithm
Problem Name: 433. Minimum Genetic Mutation
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 startGene to a gene string endGene 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 startGene and endGene and the gene bank bank, return the minimum number of mutations needed to mutate from startGene to endGene. 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: startGene = "AACCGGTT", endGene = "AACCGGTA", bank = ["AACCGGTA"] Output: 1
Example 2:
Input: startGene = "AACCGGTT", endGene = "AAACGGTA", bank = ["AACCGGTA","AACCGCTA","AAACGGTA"] Output: 2
Constraints:
0 <= bank.length <= 10startGene.length == endGene.length == bank[i].length == 8startGene,endGene, andbank[i]consist of only the characters['A', 'C', 'G', 'T'].
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
class Solution {
public:
int minMutation(string start, string end, vector<string>& bank) {
int step = 1;
deque < string>cur;
dequenext;
cur.push_back(start);
while(!cur.empty()){
string node = cur.front();
cur.pop_front();
for(auto& s: bank){
if(s == "" || !isNeighbor(node, s)) continue;
if(s == end) return step;
next.push_back(s);
s = "";
}
if(cur.empty()){
step++;
swap(cur, next);
}
}
return -1;
}
bool isNeighbor(const string& a, const string& b){
int diff = 0;
for(int i = 0; i < a.size(); i++> if(a[i] != b[i] && ++diff > 1) return false;
return diff == 1;
}
};
Input
Output
#2 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int minMutation(String start, String end, String[] bank) {
if (start.equals(end)) {
return 0;
}
Set < String> bankSet = new HashSet<>(Arrays.asList(bank));
char[] charSet = {'A', 'C', 'G', 'T'};
int currLevel = 0;
Set < String> visited = new HashSet<>();
Queue queue = new LinkedList<>();
queue.add(start);
visited.add(start);
while (!queue.isEmpty()) {
int size = queue.size();
while (size-- > 0) {
String curr = queue.remove();
if (curr.equals(end)) {
return currLevel;
}
char[] currCharArr = curr.toCharArray();
for (int i = 0; i < currCharArr.length; i++) {
char old = currCharArr[i];
for (char c : charSet) {
currCharArr[i] = c;
String newGene = String.valueOf(currCharArr);
if (!visited.contains(newGene) && bankSet.contains(newGene)) {
queue.add(newGene);
visited.add(newGene);
}
}
currCharArr[i] = old;
}
}
currLevel++;
}
return -1;
}
}
Input
Output
#3 Code Example with Javascript Programming
Code -
Javascript Programming
const minMutation = function(start, end, bank) {
const obj = { res: Number.MAX_VALUE }
dfs(start, end, bank, 0, obj, new Set())
return obj.res === Number.MAX_VALUE ? -1 : obj.res
}
function dfs(s, e, bank, num, obj, visited) {
if(s === e) {
obj.res = Math.min(obj.res, num)
return
}
for(let el of bank) {
let diff = 0
for(let i = 0, len = s.length; i < len; i++) {
if(s[i] !== el[i]> {
diff++
if(diff > 1) break
}
}
if(diff === 1 && !visited.has(el)) {
visited.add(el)
dfs(el, e, bank, num + 1, obj, visited)
visited.delete(el)
}
}
}
Input
Output
#4 Code Example with Python Programming
Code -
Python Programming
class Solution:
def minMutation(self, start: str, end: str, bank: List[str]) -> int:
bfs = [start]
genes = set(bank)
cnt = 0
while bfs:
arr = []
for g in bfs:
if g == end:
return cnt
for i, c in enumerate(g):
for new in 'AGTC':
if new != c:
s = g[:i] + new + g[i + 1:]
if s in genes:
arr.append(s)
genes.discard(s)
bfs = arr
cnt += 1
return -1
Input
Output