Shuffle Two Strings

If A = “aab”, B = “abc”, C = “aaabbc” Here C is an interleaving string of A and B. Because C contains all the characters of A and B and the order of all these characters is also the same in all three strings.

The first line contains an integer 'T' which denotes the number of test cases or queries to be run. Then, the T test cases follow. The first and only line of each test case contains three strings A, B, and C each separated by a single space.

1 <= T <= 100 1 <= |A|, |B| <= 1000 1 <= |C| <= 2000 where |A|, |B|, |C| denotes the length of string A, B and C respectively. All the characters of the strings A, B, and C contain lowercase English letters only. Time limit: 1 second

For the first test case, all the characters of A and B are present in C, and All the characters of A are present in C in the same order as “abfddef”, and All the characters of B are present in C in the same order as “abfddef”. For the second case, length of C < (length of A + length of B).

Brute-force, Recursion

Approach:

To solve this problem, let us solve a smaller problem first. Let’s assume that the length of the string A and B is one and the length of the string C is two. Now to check whether C is formed by interleaving A and B or not, we consider the last character of string C, and if this character matches neither with the character of the string A nor with the character of string B, then we return false. But if it matches with any of the characters either from A or from B, we check the same for the other character in C. So, let's further expand this idea, for bigger problems. So, suppose the last character of C, matches with either the last character of A or B. In that case, we check recursively for the second last element and then the third last element and so on, until we finally reach the base case when all the strings eventually become empty.

Steps:

We create a recursive function let’s say isInterleaveUtil and pass A, B, C, n1, n2, n3 as arguments, where n1,n2, and n3 are the length of A, B, and C.

bool isInterleaveUtil( A, B, C, n1, n2, n3):

If all the strings become empty i.e. n1,n2.and n3 become 0, then simply return true.
If the length of C is not equal to (length of A + length of B),i.e n1+n2 != n3 then we simply return false.
If the last character of both A and B matches with the last character of C, then we check for both the cases i.e. retreating A by one character and B by one character i.e. call isInterleaveUtil(A, B, C, n1-1, n2, n3-1) and isInterleaveUtil(A, B, C, n1, n2-1, n3-1), and return true, if either of them returns true.
If the last character of C matches with the last character of A, but not with the last character of B, we move one character back in A and check recursively. I.e. call isInterleaveUtil(A, B, C, n1-1, n2, n3-1);
If the last character of C matches with the last character of B, but not with the last character of A, we move one character back in B and check recursively. I.e. call isInterleaveUtil(A, B, C, n1, n2-1, n3-1).
If the last character of C matches neither with the last character of A nor with the last character of B, then we simply return false.

Time Complexity

O(2^(N + M)), where N is the length of the string A and M is the length of string B.

In the worst case, for every character of the input strings A and B, there can be two choices. Hence, the time complexity will be O(2^(N + M)).

Space Complexity

O(N + M), where N is the length of the string A and M is the length of string B.

The number of recursive calls can go up to (N + M) in the worst case. Hence, the overall complexity will be O(N + M).

Problem statement

Sample Input 1:

Sample Output 1:

Explanation for sample 1:

Sample Input 2:

Sample output 2: