Minimum Genetic Mutation in C++

C++Server Side ProgrammingProgramming

Suppose we have a gene string. That can be represented by a string whose length is 8, This string is consists of these letters [A, C, G, T]. Now consider we want to investigate about a mutation, where ONE mutation is actually ONE single character changed in the gene string. As an example, "AACCGTTT" is changed like "AACCGTTA" is 1 mutation.

We also have a given gene "bank", where all the valid gene mutations are present. A gene must be in the bank to make it a valid gene string.

Now suppose we have given 3 things - start, end, bank, our task is to determine what is the minimum number of mutations needed to mutate from "start" to "end". If conversion from start to end is not possible, then return -1.

So, if the input is like start = "AACCGGTT", end = "AAACGGTA", bank = ["AACCGGTA", "AACCGCTA", "AAACGGTA"], then the output will be 2.

To solve this, we will follow these steps −

  • Define a function putStar(), this will take s,

  • Define an array ret

  • for initialize i := 0, when i ≪ size of s, update (increase i by 1), do −

    • temp := substring of s from 0 to i-1 concatenate " * " + substring of s from index i + 1 to end

    • insert temp at the end of ret

  • return ret

  • From the main method do the following −

  • Define one map called graph.

  • for initialize i := 0, when i < size of bank, update (increase i by 1), do −

    • s := bank[i]

    • out = putStar(bank[i])

    • for initialize j := 0, when j < size of out, update (increase j by 1), do

      • insert s at the end of graph[out[j]]

  • Define one queue q

  • insert start into q

  • Define one set visited

  • insert start into visited

  • for initialize lvl := 1, when not q is empty, update (increase lvl by 1), do −

    • sz := size of q

    • while sz is non-zero, decrease sz in each iteration, do −

      • node := first element of q

      • delete element from q

      • out = putStar(node)

      • for initialize i := 0, when i < size of out, update (increase i by 1), do−

        • u := out[i]

        • for initialize j := 0, when j < size of graph[u], update (increase j by 1), do −

          • v := graph[u, j]

          • if vi is in visited, then come out from the loop

          • if v is same as end, then return lvl

          • insert v into visited

          • insert v into q

  • return -1

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
   public:
   vector <string> putStar(string s){
      vector <string> ret;
      for(int i = 0; i < s.size(); i++){
         string temp = s.substr(0, i) + "*" + s.substr(i + 1);
         ret.push_back(temp);
      }
      return ret;
   }
   int minMutation(string start, string end, vector<string>& bank) {
      unordered_map < string, vector <string> > graph;
      for(int i = 0; i < bank.size(); i++){
         string s = bank[i];
         vector <string> out = putStar(bank[i]);
         for(int j = 0; j < out.size(); j++){
            graph[out[j]].push_back(s);
         }
      }
      queue <string> q;
      q.push(start);
      set <string> visited;
      visited.insert(start);
      for(int lvl = 1; !q.empty(); lvl++){
         int sz = q.size();
         while(sz--){
            string node = q.front();
            q.pop();
            vector <string> out = putStar(node);
            for(int i = 0; i < out.size(); i++){
               string u = out[i];
               for(int j = 0; j < graph[u].size(); j++){
                  string v = graph[u][j];
                  if(visited.count(v)) continue;
                  if(v == end) return lvl;
                  visited.insert(v);
                  q.push(v);
               }
            }
         }
      }
      return -1;
   }
};
main(){
   Solution ob;
   vector<string> v = {"AACCGGTA", "AACCGCTA", "AAACGGTA"};
   cout << (ob.minMutation("AACCGGTT", "AAACGGTA", v));
}

Input

"AACCGGTT", "AAACGGTA", {"AACCGGTA", "AACCGCTA", "AAACGGTA"}

Output

2
raja
Updated on 05-Jun-2020 07:01:54

Advertisements