433. Minimum Genetic Mutation

Description

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 <= 10
• startGene.length == endGene.length == bank[i].length == 8
• startGene, endGene, and bank[i] consist of only the characters ['A', 'C', 'G', 'T'].

Solutions

Solution 1: BFS

We define a queue q to store the current gene sequence and the number of changes, and a set vis to store the visited gene sequences. Initially, we add the starting gene sequence start to the queue q and the set vis.

Then, we continuously take out a gene sequence from the queue q. If this gene sequence equals the target gene sequence, we return the current number of changes. Otherwise, we iterate through the gene bank bank, calculate the difference value between the current gene sequence and the gene sequence in the gene bank. If the difference value is $1$ and the gene sequence in the gene bank has not been visited, we add it to the queue q and the set vis.

If the queue q is empty, it means that the gene change cannot be completed, so we return $-1$.

The time complexity is $O(C \times n \times m)$, and the space complexity is $O(n \times m)$. Where $n$ and $m$ are the lengths of the gene sequence and the gene bank respectively, and $C$ is the size of the character set of the gene sequence. In this problem, $C = 4$.

Python3JavaC++GoTypeScriptRust

class Solution:
    def minMutation(self, startGene: str, endGene: str, bank: List[str]) -> int:
        q = deque([(startGene, 0)])
        vis = {startGene}
        while q:
            gene, depth = q.popleft()
            if gene == endGene:
                return depth
            for nxt in bank:
                diff = sum(a != b for a, b in zip(gene, nxt))
                if diff == 1 and nxt not in vis:
                    q.append((nxt, depth + 1))
                    vis.add(nxt)
        return -1

class Solution {
    public int minMutation(String startGene, String endGene, String[] bank) {
        Deque<String> q = new ArrayDeque<>();
        q.offer(startGene);
        Set<String> vis = new HashSet<>();
        vis.add(startGene);
        int depth = 0;
        while (!q.isEmpty()) {
            for (int m = q.size(); m > 0; --m) {
                String gene = q.poll();
                if (gene.equals(endGene)) {
                    return depth;
                }
                for (String next : bank) {
                    int c = 2;
                    for (int k = 0; k < 8 && c > 0; ++k) {
                        if (gene.charAt(k) != next.charAt(k)) {
                            --c;
                        }
                    }
                    if (c > 0 && !vis.contains(next)) {
                        q.offer(next);
                        vis.add(next);
                    }
                }
            }
            ++depth;
        }
        return -1;
    }
}

class Solution {
public:
    int minMutation(string startGene, string endGene, vector<string>& bank) {
        queue<pair<string, int>> q{{{startGene, 0}}};
        unordered_set<string> vis = {startGene};
        while (!q.empty()) {
            auto [gene, depth] = q.front();
            q.pop();
            if (gene == endGene) {
                return depth;
            }
            for (const string& next : bank) {
                int c = 2;
                for (int k = 0; k < 8 && c; ++k) {
                    c -= gene[k] != next[k];
                }
                if (c && !vis.contains(next)) {
                    vis.insert(next);
                    q.push({next, depth + 1});
                }
            }
        }
        return -1;
    }
};

func minMutation(startGene string, endGene string, bank []string) int {
    type pair struct {
        s     string
        depth int
    }
    q := []pair{pair{sta