/*

Interleaving String

Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.

For example,

Given:

s1 = “aabcc”,

s2 = “dbbca”,

When s3 = “aadbbcbcac”, return true.

When s3 = “aadbbbaccc”, return false.

*/

#include<string>

usingnamespace std;

classInterleavingString {

public:

bool isInterleave(std::strings1,strings2,strings3) {

if ((s1.length() + s2.length()) != s3.length())returnfalse;

if (s1.length() == 0)

{

if (s3 == s2)returntrue;

elsereturnfalse;

}

bool* results = newbool[(s1.length() + 1)*(s3.length() + 1)];

results[0] =true;

for (unsignedint indx = 1; indx < (s1.length() + 1)*(s3.length() + 1); indx++)

results[indx] =false;

for (unsignedint len3 = 1; len3 <= s3.length(); len3++)

{

unsignedint startIndx = 0;

unsignedint endIndx = len3;

if (len3 > s2.length()) startIndx = len3 – s2.length();

if (endIndx>s1.length()) endIndx = s1.length();

for (unsignedint len1 = startIndx; len1 <= endIndx; len1++)

{

if (len1>0 && len1 <= s1.length() && s3[len3 – 1] == s1[len1 – 1] && results[(len3 – 1)*(s1.length() + 1) + len1 – 1])

results[len3*(s1.length() + 1) + len1] = true;

if (len1<len3 &&len3 – len1 <= s2.length() && s3[len3 – 1] == s2[len3 – len1 – 1] && results[(len3 – 1)*(s1.length() + 1) + len1])

results[len3*(s1.length() + 1) + len1] = true;

}

}

bool result = results[s3.length()*(s1.length() + 1) + s1.length()];

delete[] results;

return result;

}

};