433. Minimum Genetic Mutation

Medium (Trung bình) C++ Python 🔗 Xem trên LeetCode

📋 Đề Bài

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, and bank[i] consist of only the characters ['A', 'C', 'G', 'T'].

🧠 Thuật Toán & Kỹ Thuật

Hash Table (Bảng băm)Union Find (Tập hợp rời rạc)String (Chuỗi)

⏱️ Thời gian O(n²)

💾 Không gian O(n)

💻 Lời Giải

C++ 0433-minimum-genetic-mutation.cpp

class Solution {
public:
    int minMutation(string start, string end, vector<string>& bank) {
        queue<string> mq;
        mq.push(start);
        unordered_map<string, bool> visited;
        int step = 0;
        while (!mq.empty()) {
            int n = mq.size();
            while (n--) {
                string word = mq.front();
                mq.pop();
                if (word == end) {
                    return step;
                }
                for (int i = 0; i < 8; ++i) {
                    char tmp = word[i];
                    for (int j = 0; j < 4; ++j) {
                        word[i] = "ACGT"[j];
                        if (!visited[word] and find(bank.begin(), bank.end(), word) != bank.end()) {
                            mq.push(word);
                            visited[word] = true;
                        }
                    }
                    word[i] = tmp;
                }
            }
            step++;
        }
        return -1;
    }
};

Python 0433-minimum-genetic-mutation.py

class Solution:
    def minMutation(self, start: str, end: str, bank: List[str]) -> int:
        dq = deque([start])
        visited = set([start])
        step = 0
        bank = set(bank)
        
        while dq:
            n = len(dq)
            for _ in range(n):
                gene = dq.popleft()
                if gene == end:
                    return step
                visited.add(gene)
                m = len(gene)
                gene = list(gene)
                for i in range(m):
                    tmp = gene[i]
                    for char in 'ACGT':
                        gene[i] = char
                        dummy = ''.join(gene)
                        if dummy in bank and dummy not in visited:
                            dq.append(dummy)
                    gene[i] = tmp
                    
            step += 1
            
        return -1