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Minimum Genetic Mutation in C++
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
#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
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