## Google Interview Question

Software Engineers**Country:**United States

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Interviewee's solution:

```
#include <iostream>
#include <string>
#include <vector>
using namespace std;
long factorial(long n){
long fact = 1;
for (long i = 1; i<= n; ++i)
fact *= i;
return fact;
}
void DFS(const string &s1, int m, int i,
const string&s2, int n, int j,
string path,vector<string> &ret){
if(i==m &&j==n) {
ret.push_back(path);
return;
}
if(i < m) DFS(s1, m, i+1, s2, n, j, path+s1, ret);
if(j < n) DFS(s1, m, i, s2, n, j+1, path+s2[j], ret);
}
vector<string> combineTowStrings(const string &s1,const string &s2) {
int m =s1.length(), n = s2.length();
long count =factorial(m+n)/factorial(n)/factorial(m);
cout <<"count:" << count << endl;
vector<string> ret;
DFS(s1, m, 0, s2,n, 0, "", ret);
return ret;
}
int main() {
vector<string> ret = combineTowStrings("AB","CDF");
cout <<ret.size() << endl;
for(string &s: ret) cout << s << endl;
return 0;
}
```

```
int find(String s1, String s2, int i, int j, String out){
if(i == s1.length() && j == s2.length())
return 0;
else if(i == s1.length()){
out += s2.substring(j, s2.length());
System.out.println(out);
return 1;
}
else if(j == s2.length()){
out += s1.substring(i, s1.length());
System.out.println(out);
return 1;
}
else{
int re = find(s1, s2, i+1, j, out + s1.charAt(i)) + find(s1, s2, i, j+1, out + s2.charAt(j));
return re;
}
}
```

- krbchd April 16, 2017