433. Minimum Genetic Mutation

Problem Statement

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

C++

Source file

class Solution {
public:
    int minMutation(string start, string end, vector<string>& bank) {
        //st holds all valid mutations
        unordered_set<string> st{bank.begin(),bank.end()};
        //if end mutaion is not in list return -1;
        if(!st.count(end)) return -1;
        //start BFS by pushing the starting mutation
        queue<string> Q;
        Q.push(start);
        int steps=0,s;
        string cur,t;
        while(!Q.empty()){
            s=Q.size();
            while(s--){
                cur=Q.front();
                Q.pop();
                //If we reach end mutation
                if(cur==end) return steps;
                //We erase the cur mutation in order to avoid redundant checking
                st.erase(cur);
                //as the length of mutation is 8 and it can take A,C,G,T
                //at each index we check the possibility of mutation by replcaing it with A,C,G,T
                for(int i=0;i<8;i++){
                    t=cur;
                    t[i]='A';
                    if(st.count(t)) Q.push(t);
                    t[i]='C';
                    if(st.count(t)) Q.push(t);
                    t[i]='G';
                    if(st.count(t)) Q.push(t);
                    t[i]='T';
                    if(st.count(t)) Q.push(t);
                }
            }
            steps++;
        }
        return -1;
    }
};