# Print all interleavings of given two strings

Given two strings str1 and str2, write a function that prints all interleavings of the given two strings. You may assume that all characters in both strings are different

Example:

``````Input: str1 = "AB",  str2 = "CD"
Output:
ABCD
ACBD
ACDB
CABD
CDAB

Input: str1 = "AB",  str2 = "C"
Output:
ABC
ACB
CAB
``````

The idea comes from Ray Toal:

The base case is when one of the two strings are empty:

interleave(s1, "") = {s1}

interleave("", s2) = {s2}

Notice the order of the arguments doesn't really matter, because

interleave("ab", "12") = {"ab12", "a1b2", "1ab2", "a12b", "1a2b", "12ab"} = interleave("12", "ab") So since the order doesn't matter we'll look at recursing on the length of the first string.

Okay so let's see how one case leads to the next. I'll just use a concrete example, and you can generalize this to real code.

interleave("", "abc") = {"abc"} interleave("1", "abc") = {"1abc", "a1bc", "ab1c", "abc1"} interleave("12", "abc") = {"12abc", "1a2bc", "1ab2c", "1abc2", "a12bc", "a1b2c", "a1bc2", "ab12c", "ab1c2" "abc12"} So everytime we added a character to the first string, we formed the new result set by adding the new character to all possible positions in the old result set. Let's look at exactly how we formed the third result above from the second. How did each element in the second result turn into elements in the third result when we added the "2"?

"1abc" => "12abc", "1a2bc", "1ab2c", "1abc2" "a1bc" => "a12bc", "a1b2c", "a1bc2" "ab1c" => "ab12c", "ab1c2" "abc1" => "abc2" Now look at things this way:

"1abc" => {1 w | w = interleave("2", "abc")} "a1bc" => {a1 w | w = interleave("2", "bc")} "ab1c" => {ab1 w | w = interleave("2", "c")} "abc1" => {abc1 w | w = interleave("2", "")}

The following is my code building on top of the above idea: Can anyone help me verify it?

``````void interleaving(const string& s2,
string result, int start, int depth)
{
if(depth == s2.size())
cout << result << endl;
else
{
for(int i = start; i <= result.size(); ++i)
{
result.insert(i, 1, s2[depth]);
interleaving(s2, result, i+1, depth+1);
result.erase(i, 1);
}
}
}

int main()
{
string s1("");
string s2("12");
string result(s1);
interleaving(s2, result, 0, 0);
system("pause");
return 0;
}
``````

There is another much more beautiful solution for this problem comes from an unknown friend akash01.

``````void printIlsRecur (char *str1, char *str2, char *iStr, int m, int n, int i)
{
// Base case: If all characters of str1 and str2 have been included in
// output string, then print the output string
if ( m==0 && n ==0 )
{
printf("%s\n", iStr) ;
}

// If some characters of str1 are left to be included, then include the
// first character from the remaining characters and recur for rest
if ( m != 0 )
{
iStr[i] = str1[0];
printIlsRecur (str1 + 1, str2, iStr, m-1, n, i+1);
}

// If some characters of str2 are left to be included, then include the
// first character from the remaining characters and recur for rest
if ( n != 0 )
{
iStr[i] = str2[0];
printIlsRecur (str1, str2+1, iStr, m, n-1, i+1);
}
}
``````

Anyway, really appreciate anyone can help me check my code!

-
There is a standard function to do this: `std::next_permutation()` –  Loki Astari Oct 25 '12 at 14:05
@LokiAstari, thanks very much. If it was an interview question and ask you to write the code by yourself, can you help me check it? –  FihopZz Oct 25 '12 at 14:48
Why not test it with the examples you gave in the question? Also, does it have to be recursive? Your last question used recursion too... (perhaps you should be learning Haskell). –  William Morris Oct 25 '12 at 23:54
@WilliamMorris, I've tested a lot of cases including the examples given in the question. It seems it's right. I'm doing practice in recursive algorithm. :-) –  FihopZz Oct 26 '12 at 0:17
@LokiAstari Given that the characters from either input string occur in the same order in the output as in the respective input, how does `next_permutation` help? –  Seg Fault Oct 27 '12 at 7:07